text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Tags: math
Correct Solution:
```
from math import ceil
q = int(input())
for i in range(q):
n = int(input())
x = [int(i) for i in input().split()]
print(ceil(sum(x)/n))
```
| 13,400 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Tags: math
Correct Solution:
```
import math
t=int(input())
for i in range(t):
n=int(input())
sum=0
ar=list(map(int,input().split()))
for j in range(n):
sum=sum+ar[j]
if(sum%n==0):
print(math.floor(sum/n))
else:
print(math.ceil(sum/n))
```
| 13,401 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Tags: math
Correct Solution:
```
for t in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
sum=0
for k in a:
sum+=k
i=0
while(True):
if((sum+i)%n==0):
print((sum+i)//n)
break;
else:
i+=1
```
| 13,402 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
import math
t=int(input())
for _ in range(t):
n=int(input())
l=list(map(int,input().split()))
l.sort()
print(math.ceil((sum(l))/n))
```
Yes
| 13,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
q = int(input())
for _ in range(q):
n = int(input())
dat = list(map(int, input().split()))
import math
print(math.ceil(sum(dat)/n))
```
Yes
| 13,404 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
import math
for i in range(int(input())):
n = int(input())
print(math.ceil(sum(list(map(int, input().split(' ')))) / n))
```
Yes
| 13,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
q = int (input ())
q1 = q;
while q1 > 0:
n = int (input ())
l = map(int, input().split(" "))
sumatotal = sum(l)
rpta = (int) (sumatotal/n);
if (sumatotal % n != 0):
rpta = rpta +1
print (rpta)
q1 = q1 - 1
```
Yes
| 13,406 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
import sys
T = int(sys.stdin.readline())
for i in range(T):
N = int(sys.stdin.readline())
nums = list(map(int, sys.stdin.readline().split()))
print("%.f" %(sum(nums)/N))
```
No
| 13,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
q = int(input())
for i in range(q):
n = int(input())
a = list(map(int, input().split()))
s = sum(a)
print(s / n + (s % n != 0))
```
No
| 13,408 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
t=int(input())
#l=list()
min1=99999
for q in range(t):
n=int(input())
a=list(map(int,input().split()))
s=sum(a)
for i in range(1,max(a)+1):
if(n*i>=s):
if(min1>i):
min1=i
print(min1)
#print(l)
#x=len(l)
#print(l[x])
```
No
| 13,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
import math
quer = input()
quer = int(quer)
for _ in range(1, quer+1):
num = input()
num = int(num)
inp_lst = [int(i) for i in input().split()]
inp_lst.sort()
avg = math.ceil(sum(inp_lst) / num)
#if avg. not present in list, find the immediate larger one than this
ans_lst = sorted(i for i in inp_lst if i >= avg)
print(ans_lst[0])
```
No
| 13,410 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
def main():
n = int(input())
ss = list(zip(input(), range(0,n)))
res = [1]*n
max_colour = 1
nextx = [0] * 26
colours_after = [[True]+[False]*26 for _ in range(26)]
for i in range(n):
ch = ord(ss[i][0]) - ord('a')
opos = ss[i][1]
nx = nextx[ch]
# ss[nx:i + 1] = [ss[i]] + ss[nx:i]
col = colours_after[ch].index(False)
if col > max_colour:
max_colour = col
res[opos] = col
for c in range(ch):
colours_after[c][col] = True
for c in range(ch, 26):
nextx[c] = min(nextx[c], nx+1)
print(max_colour)
print(' '.join(map(str, res)))
if __name__ == "__main__":
main()
```
| 13,411 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
n = int(input())
s = input()
ans = [-1] * n
f = True
for i in range(n):
if ans[i] == -1:
ans[i] = 0
for j in range(i + 1, n):
if (s[j] < s[i]):
if ans[j] == -1:
ans[j] = (ans[i] - 1) ** 2
else:
now = ans[j]
new = (ans[i] - 1) ** 2
if new != now:
f = False
else:
ans[j] = new
if not f:
print("NO")
else:
print("YES")
for i in range(n):
print(ans[i], end='')
```
| 13,412 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
n=int(input())
s=input()
last,last1='a','a'
ans=''
flag=1
for i in s:
if i<last and i<last1:
flag=0
break
elif i>=last:
last=i
ans+='0'
else:
last1=i
ans+='1'
if flag:
print('YES')
print(ans)
else:
print('NO')
```
| 13,413 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
import sys
def main():
n = int(sys.stdin.readline().split()[0])
s = sys.stdin.readline().split()[0]
color = [None]*n
color[0] = 0
for i in range(n):
if color[i] == None:
color[i] = 0
for j in range(i+1, n):
if ord(s[j]) < ord(s[i]):
if color[j] == color[i]:
print("NO")
return
if color[j] == None:
color[j] = color[i]^1
print("YES")
print(*color, sep = "")
main()
```
| 13,414 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**30, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] > k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
n=int(input())
d=dict()
for i in range(97,124):
d[chr(i)]=0
l=input()
d[l[0]]+=1
ans=[0]*n
ans[0]=d[l[0]]
for i in range(1,n):
for j in sorted(d.keys(),reverse=True):
if j>l[i]:
ans[i]=max(ans[i],d[j]+1)
else:
break
d[l[i]]=ans[i]
print(max(ans))
print(*ans,sep=" ")
```
| 13,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
#t = int(input())
t=1
while t!=0:
t-=1
n = int(input())
s = input()
l1, l2 = 'a' , 'a'
ans = ''
for i in range(n):
if s[i]>=l1:
l1 = s[i]
ans+='0'
elif s[i]>=l2:
l2 = s[i]
ans+='1'
else:
ans='NO'
break
if ans=='NO':
print(ans)
else:
print('YES')
print(ans)
```
| 13,416 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
def main():
n = int(input())
s = list(input())
colors = [-1]*n
for i in range(len(s)):
if colors[i] == -1:
colors[i] = 0
for j in range(i+1,n):
if s[j] < s[i]:
if colors[j] == colors[i]:
print('NO')
return
if colors[j] == -1:
if colors[i] == 0:
colors[j] = 1
else:
colors[j] = 0
print('YES')
for i in colors:
print(i,end = '')
main()
```
| 13,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
n=int(input())
s=input()
rama=[]
for i in range(n):
rama.append(ord(s[i])-96)
visit=[1 for i in range(27)]
a=[]
maxi=0
for i in range(n):
p=visit[rama[i]]
a.append(p)
maxi=max(maxi,p)
for j in range(rama[i]):
visit[j]=max(visit[j],p+1)
print(maxi)
for i in a:
print(i,end=' ')
```
| 13,418 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
# sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7
def bisearch_min(mn, mx, func):
ok = mx
ng = mn
while ng+1 < ok:
mid = (ok+ng) // 2
if func(mid):
ok = mid
else:
ng = mid
return ok
def check(m):
if m == len(B):
return True
if B[m][-1][0] <= a:
return True
else:
return False
N = INT()
A = [ord(c)-97 for c in input()]
B = [[] for i in range(1)]
B[0].append((A[0], 0))
for i, a in enumerate(A[1:], 1):
idx = bisearch_min(-1, len(B), check)
if idx == len(B):
B.append([(a, i)])
else:
B[idx].append((a, i))
ans = [0] * N
for a, li in enumerate(B):
for _, idx in li:
ans[idx] = str(a)
if len(B) <= 2:
YES()
print(''.join(ans))
else:
NO()
```
Yes
| 13,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Submitted Solution:
```
n = int(input())
s = input()
g = {}
for i in range(len(s)):
g[i] = []
for i in range(len(s)):
for j in range(i):
if s[j] > s[i]:
g[i].append(j)
g[j].append(i)
colored = [False] * len(s)
color = [None] * len(s)
def brush(v, c):
colored[v] = True
color[v] = c
for n in g[v]:
if colored[n]:
if color[n] == c:
return False
else:
if not brush(n, c ^ 1):
return False
return True
failed = False
for i in range(len(s)):
if not colored[i]:
if not brush(i, 0):
failed = True
break
if failed:
print ('NO')
else:
print ('YES')
print (''.join( map(lambda i: '1' if color[i] == 1 else '0', range(len(s))) ))
```
Yes
| 13,420 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Submitted Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
def solve(N, S):
# Same color must be already sorted since they can't be swapped with each other
# Greedily build increasing subsequences
indices = [[0]]
# last value -> which list
for i, x in enumerate(S[1:], 1):
for l in indices:
if S[l[-1]] <= x:
l.append(i)
break
else:
indices.append([i])
ans = [None for i in range(N)]
for color, l in enumerate(indices):
for i in l:
ans[i] = color
# Format for E1
if len(indices) <= 2:
return "YES\n" + "".join(map(str, ans))
else:
return 'NO'
# Format for E2
possible = str(len(indices))
return possible + "\n" + " ".join(str(x + 1) for x in ans)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
(N,) = [int(x) for x in input().split()]
S = input().decode().rstrip()
ans = solve(N, S)
print(ans)
```
Yes
| 13,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Submitted Solution:
```
def find(x):
r = x
while r != f[r]:
r = f[r]
while x != r:
pre = f[x]
f[x] = r
x = pre
return r
a = input()
n = int(a)
a = input()
str = list(a)
id = [0] * n
f = [0] * 2 * n
name = [0] * 2 * n
for i in range(n):
id[i] = i
for i in range(2 * n):
f[i] = i
name[i] = -1
Find = 0
for i in range(n):
if Find:
break
for j in range(n - i - 1):
if str[j] > str[j + 1]:
str[j], str[j + 1] = str[j + 1], str[j]
id[j], id[j + 1] = id[j + 1], id[j]
fx = find(id[j])
fnx = find(id[j] + n)
fy = find(id[j + 1])
fny = find(id[j + 1] + n)
if fx == fy or fnx == fny:
Find = 1
break
else:
f[fx] = fny
f[fy] = fnx
if Find:
print('NO')
else:
print('YES')
for i in range(n):
fx = find(i)
if name[fx] < 0:
name[fx] = 0
if(fx >= n):
name[find(fx - n)] = 1
else:
name[find(fx + n)] = 1
print('{:d}'.format(name[fx]), end = '')
```
Yes
| 13,422 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Submitted Solution:
```
import math
import sys
from collections import defaultdict
def stdinWrapper():
data = '''8
aaabbcbb
'''
for line in data.split('\n'):
yield line
if '--debug' not in sys.argv:
def stdinWrapper():
while True:
yield input()
inputs = stdinWrapper()
def inputWrapper():
return next(inputs)
def getType(_type):
return _type(inputWrapper())
def getArray(_type):
return [_type(x) for x in inputWrapper().split()]
''' Solution '''
l = getType(int)
_str = getType(str)
dp = [1] * l
maxdp = [0] * 26
dp = [1] * len(_str)
p = [-1] * len(_str)
for i in range(1, len(_str)):
for c in range(ord(_str[i])-ord('a')+1, 26):
dp[i] = max(dp[i], maxdp[c]+1)
maxdp[ord(_str[i]) - ord('a')] = max(dp[i], maxdp[ord(_str[i]) - ord('a')])
pos = -1
_max = 0
for i in range(len(_str)):
if dp[i] > _max:
_max = dp[i]
pos = i
result = []
while True:
if pos == -1:
break
result.append(_str[pos])
pos = p[pos]
print(max(dp))
print(' '.join([str(x) for x in dp]))
```
No
| 13,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Submitted Solution:
```
import sys
import math
from collections import defaultdict,Counter
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
# sys.stdout=open("CP1/output.txt",'w')
# sys.stdin=open("CP1/input.txt",'r')
# m=pow(10,9)+7
n=int(input())
s=list(input())
s2=s[:]
s1=sorted(s)
c=[0]*n
# c1=[0]*n
for j in range(1,n):
if s[j]<s[j-1]:
c[j]=1-c[j-1]
# c1[j]=1-c1[j-1]
elif s[j]==s[j-1]:
c[j]=c[j-1]
c2=''.join(list(map(str,c)))
flag=0
# print(c)
while s!=s1:
for j in range(1,n):
if s[j]<s[j-1]:
if c[j]!=c[j-1]:
c[j],c[j-1]=c[j-1],c[j]
s[j],s[j-1]=s[j-1],s[j]
else:
flag=1
break
if flag==1:
break
# for j in range(1,n):
# if s[j]<s[j-1]:
# c[j]=1-c[j-1]
# else:
# c[j]=c[j-1]
# print(s)
# print(c)
if flag==0:
print("YES")
print(c2)
else:
print("NO")
# else:
# flag=0
# c2=''.join(list(map(str,c1)))
# # c2=c1[:]
# s=s2[:]
# while s!=s1:
# for j in range(1,n):
# if s[j]<s[j-1]:
# if c1[j]!=c1[j-1]:
# c1[j],c1[j-1]=c1[j-1],c1[j]
# s[j],s[j-1]=s[j-1],s[j]
# else:
# flag=1
# break
# if flag==1:
# break
# if flag==0:
# print("YES")
# print(c2)
# else:
# print("NO")
```
No
| 13,424 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Submitted Solution:
```
from sys import stdin, stdout
def main():
n = int(stdin.readline())
s = stdin.readline().rstrip()
h1 = 'A'
h2 = 'B'
res = []
for i,x in enumerate(s):
if i == 0:
res.append("0")
h1 = s[i]
continue
if s[i] < h1 and s[i] < h2:
stdout.write("NO")
return
if s[i] > h1:
h2 = 'B'
h1 = s[i]
res.append("0")
elif s[i] > h2:
h2 = s[i]
res.append("1")
else:
res.append("1")
stdout.write("YES\n" + "".join(res))
main()
```
No
| 13,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 β€ n β€ 200) β the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
Submitted Solution:
```
def binSearch(a, n):
i, j = 0, len(a) - 1
while i != j:
if (j - i) % 2 == 1:
if a[i + ((j - i) // 2)] > n:
i = i + (j - i) // 2 + 1
else:
j = i + (j - i) // 2
else:
if a[i + (j - i) // 2] > n:
i = i + (j - i) // 2
else:
j = i + (j - i) // 2
return i
input()
k = ['a', 'a']
ans = ''
maxColor = 1
for symb in input():
i = binSearch(k, symb)
ans += str(i + 1)
if i + 1 > maxColor:
maxColor = i + 1
k.append('a')
k[i] = symb
print(maxColor)
print(*ans)
```
No
| 13,426 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Tags: sortings
Correct Solution:
```
def main():
n, m = map(int, input().split())
l = list(tuple(map(int, input().split())) for _ in range(m))
h = [False for _ in range(n + 1)]
v = [False for _ in range(n + 1)]
d1 = [False for _ in range(n * 2 + 1)]
d2 = [False for _ in range(n * 2 + 1)]
for x, y in l:
line = h[y]
if not line:
h[y] = (x, x)
else:
mi, ma = line
if mi > x:
h[y] = (x, ma)
elif ma < x:
h[y] = (mi, x)
line = v[x]
if not line:
v[x] = (y, y)
else:
mi, ma = line
if mi > y:
v[x] = (y, ma)
elif ma < y:
v[x] = (mi, y)
xy = x + y
line = d1[xy]
if not line:
d1[xy] = (x, x)
else:
mi, ma = line
if mi > x:
d1[xy] = (x, ma)
elif ma < x:
d1[xy] = (mi, x)
xy = x - y + n
line = d2[xy]
if not line:
d2[xy] = (x, x)
else:
mi, ma = line
if mi > x:
d2[xy] = (x, ma)
elif ma < x:
d2[xy] = (mi, x)
res = [0] * 9
for x, y in l:
tot = 0
mi, ma = v[x]
if mi < y:
tot += 1
if y < ma:
tot += 1
for mi, ma in h[y], d1[x + y], d2[x - y + n]:
if mi < x:
tot += 1
if x < ma:
tot += 1
res[tot] += 1
print(*res)
if __name__ == '__main__':
main()
```
| 13,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Tags: sortings
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------------------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=10**10, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
n,m=list(map(int,input().split()))
f=[]
maxx=[-999999999]*n
minx=[999999999]*n
maxy=[-999999999]*n
miny=[999999999]*n
maxm=[-999999999]*(2*n-1)
minm=[999999999]*(2*n-1)
maxs=[-999999999]*(2*n-1)
mins=[999999999]*(2*n-1)
r=[0 for i in range(9)]
for i in range(m):
y,x=list(map(int,input().split()))
a,b,c,d=y-1,x-1,y+x-2,y-x+n-1
f.append((a,b,c,d))
if a>maxx[b]:maxx[b]=a
if a<minx[b]:minx[b]=a
if x-1>maxy[a]:maxy[a]=x-1
if x-1<miny[a]:miny[a]=x-1
if c>maxs[d]:maxs[d]=c
if c<mins[d]:mins[d]=c
if d>maxm[c]:maxm[c]=d
if d<minm[c]:minm[c]=d
for i in f:
k=0
if i[0]<maxx[i[1]]:k+=1
if i[0]>minx[i[1]]:k+=1
if i[1]<maxy[i[0]]:k+=1
if i[1]>miny[i[0]]:k+=1
if i[2]<maxs[i[3]]:k+=1
if i[2]>mins[i[3]]:k+=1
if i[3]<maxm[i[2]]:k+=1
if i[3]>minm[i[2]]:k+=1
r[k]+=1
print(*r)
```
| 13,428 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Tags: sortings
Correct Solution:
```
def main():
n, m = map(int, input().split())
l = list(tuple(map(int, input().split())) for _ in range(m))
h = [False] * (n + 1)
v = [False] * (n + 1)
d1 = [False] * (n * 2 + 1)
d2 = [False] * (n * 2 + 1)
for x, y in l:
line = h[y]
if line:
mi, ma = line
if mi > x:
h[y] = (x, ma)
elif ma < x:
h[y] = (mi, x)
else:
h[y] = (x, x)
line = v[x]
if line:
mi, ma = line
if mi > y:
v[x] = (y, ma)
elif ma < y:
v[x] = (mi, y)
else:
v[x] = (y, y)
xy = x + y
line = d1[xy]
if line:
mi, ma = line
if mi > x:
d1[xy] = (x, ma)
elif ma < x:
d1[xy] = (mi, x)
else:
d1[xy] = (x, x)
xy = x - y + n
line = d2[xy]
if line:
mi, ma = line
if mi > x:
d2[xy] = (x, ma)
elif ma < x:
d2[xy] = (mi, x)
else:
d2[xy] = (x, x)
res = [0] * 9
for x, y in l:
tot = 0
mi, ma = v[x]
if mi < y:
tot += 1
if y < ma:
tot += 1
for mi, ma in h[y], d1[x + y], d2[x - y + n]:
if mi < x:
tot += 1
if x < ma:
tot += 1
res[tot] += 1
print(*res)
if __name__ == '__main__':
main()
```
| 13,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Tags: sortings
Correct Solution:
```
def main():
n, m = map(int, input().split())
l = list(tuple(map(int, input().split())) for _ in range(m))
h = [False] * (n + 1)
v = [False] * (n + 1)
d1 = [False] * (n * 2 + 1)
d2 = [False] * (n * 2 + 1)
for x, y in l:
line = h[y]
if line:
mi, ma = line
if mi > x:
line[0] = x
elif ma < x:
line[1] = x
else:
h[y] = [x, x]
line = v[x]
if line:
mi, ma = line
if mi > y:
line[0] = y
elif ma < y:
line[1] = y
else:
v[x] = [y, y]
xy = x + y
line = d1[xy]
if line:
mi, ma = line
if mi > x:
line[0] = x
elif ma < x:
line[1] = x
else:
d1[xy] = [x, x]
xy = x - y + n
line = d2[xy]
if line:
mi, ma = line
if mi > x:
line[0] = x
elif ma < x:
line[1] = x
else:
d2[xy] = [x, x]
res = [0] * 9
for x, y in l:
tot = 0
mi, ma = v[x]
if mi < y:
tot += 1
if y < ma:
tot += 1
for mi, ma in h[y], d1[x + y], d2[x - y + n]:
if mi < x:
tot += 1
if x < ma:
tot += 1
res[tot] += 1
print(*res)
if __name__ == '__main__':
main()
```
| 13,430 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Tags: sortings
Correct Solution:
```
n,m=list(map(int,input().split()))
f=[]
maxx=[-999999999]*n
minx=[999999999]*n
maxy=[-999999999]*n
miny=[999999999]*n
maxm=[-999999999]*(2*n-1)
minm=[999999999]*(2*n-1)
maxs=[-999999999]*(2*n-1)
mins=[999999999]*(2*n-1)
r=[0 for i in range(9)]
for i in range(m):
y,x=list(map(int,input().split()))
a,b,c,d=y-1,x-1,y+x-2,y-x+n-1
f.append((a,b,c,d))
if a>maxx[b]:maxx[b]=a
if a<minx[b]:minx[b]=a
if x-1>maxy[a]:maxy[a]=x-1
if x-1<miny[a]:miny[a]=x-1
if c>maxs[d]:maxs[d]=c
if c<mins[d]:mins[d]=c
if d>maxm[c]:maxm[c]=d
if d<minm[c]:minm[c]=d
for i in f:
k=0
if i[0]<maxx[i[1]]:k+=1
if i[0]>minx[i[1]]:k+=1
if i[1]<maxy[i[0]]:k+=1
if i[1]>miny[i[0]]:k+=1
if i[2]<maxs[i[3]]:k+=1
if i[2]>mins[i[3]]:k+=1
if i[3]<maxm[i[2]]:k+=1
if i[3]>minm[i[2]]:k+=1
r[k]+=1
print(*r)
```
| 13,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Tags: sortings
Correct Solution:
```
n, m = map(int, input().split())
k = 2 * n + 1
t = [[0] * k for i in range(8)]
p, r = [0] * m, [0] * 9
for k in range(m):
x, y = map(int, input().split())
a, b = n + y - x, y + x
p[k] = (x, y)
if t[0][x]: t[0][x], t[4][x] = min(t[0][x], y), max(t[4][x], y)
else: t[0][x] = t[4][x] = y
if t[1][y]: t[1][y], t[5][y] = min(t[1][y], x), max(t[5][y], x)
else: t[1][y] = t[5][y] = x
if t[2][a]: t[2][a], t[6][a] = min(t[2][a], x), max(t[6][a], x)
else: t[2][a] = t[6][a] = x
if t[3][b]: t[3][b], t[7][b] = min(t[3][b], x), max(t[7][b], x)
else: t[3][b] = t[7][b] = x
for x, y in p:
a, b = n + y - x, y + x
s = (t[0][x] == y) + (t[4][x] == y) \
+ (t[1][y] == x) + (t[5][y] == x) \
+ (t[2][a] == x) + (t[6][a] == x) \
+ (t[3][b] == x) + (t[7][b] == x)
r[8 - s] += 1
print(' '.join(map(str, r)))
```
| 13,432 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Tags: sortings
Correct Solution:
```
n,m=list(map(int,input().split()))
f=[]
maxy=[-999999999]*n#y-1
miny=[999999999]*n#y-1
maxx=[-999999999]*n#x-1
minx=[999999999]*n#x-1
maxm=[-999999999]*(2*n-1)#x-y+n-1
minm=[999999999]*(2*n-1)#x-y+n-1
maxs=[-999999999]*(2*n-1)#x+y-2
mins=[999999999]*(2*n-1)#x+y-2
for i in range(m):
y,x=list(map(int,input().split()))
a,b,c,d=y-1,x-1,x-y+n-1,x+y-2
f.append((a,x-1,x-y+n-1,x+y-2))
if maxy[a]<b:maxy[a]=b
if miny[a]>b:miny[a]=b
if maxx[b]<a:maxx[b]=a
if minx[b]>a:minx[b]=a
if maxm[c]<d:maxm[c]=d
if minm[c]>d:minm[c]=d
if maxs[d]<c:maxs[d]=c
if mins[d]>c:mins[d]=c
r=[0]*9
for e in f:
k=0
if maxy[e[0]]>e[1]:k+=1
if miny[e[0]]<e[1]:k+=1
if maxx[e[1]]>e[0]:k+=1
if minx[e[1]]<e[0]:k+=1
if maxm[e[2]]>e[3]:k+=1
if minm[e[2]]<e[3]:k+=1
if maxs[e[3]]>e[2]:k+=1
if mins[e[3]]<e[2]:k+=1
r[k]+=1
print(*r)
```
| 13,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Submitted Solution:
```
n,m=list(map(int,input().split()))
kol,fig,dos=[0,0,0,0,0,0,0,0,0],[],[[0]*n for i in range(n)]
for i in range(m):
x,y=list(map(int,input().split()))
fig.append([x,y])
dos[y-1][x-1]=2
def pr(x,y,v):
a,b=x-1,y-1
for i in range(x,n):
if dos[b][i]==2:v+=1;break
for i in range(0,x-1):
if dos[b][i]==2:v+=1;break
for i in range(y,n):
if dos[i][a]==2:v+=1;break
for i in range(0,b):
if dos[i][a]==2:v+=1;break
for i in range(1,n):
if b+i<n and a+i<n and dos[b+i][a+i]==2:v+=1;break
for i in range(1,n):
if a+i<n and b-i>=0 and dos[b-i][a+i]==2:v+=1;break
for i in range(1,n):
if a-1>=0 and b-1>=0 and dos[b-i][a-i]==2:v+=1;break
for i in range(1,n):
if b+i<n and a-1>=0 and dos[b+i][a-i]==2:v+=1;break
kol[v]+=1
for i in fig:pr(i[0],i[1],0)
print(*kol)
```
No
| 13,434 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Submitted Solution:
```
import sys
def place_compare(a, b):
if a[0] < b[0]:
return True
if a[0] > b[0]:
return False
return a[1] < b[1]
values = [int(x) for x in sys.stdin.readline().split()]
dimension, n_queens = values[0], values[1]
x_positions, y_positions, additions, subtractions = [], [], [], []
for i in range(n_queens):
values = [int(x) for x in sys.stdin.readline().split()]
x, y = values[0], values[1]
subtractions.append((x - y, i))
additions.append((x + y, i))
x_positions.append((x, i))
y_positions.append((y, i))
additions.sort(key=lambda x:x[0])
subtractions.sort(key=lambda x:x[0])
x_positions.sort(key=lambda x:x[0])
y_positions.sort(key=lambda x:x[0])
print(additions, subtractions, x_positions, y_positions)
tallies = [0 for x in range(n_queens)]
for cur_list in [additions, subtractions, x_positions, y_positions]:
for i in range(n_queens-1):
if cur_list[i][0] == cur_list[i+1][0]:
tallies[cur_list[i][1]] += 1
tallies[cur_list[i+1][1]] += 1
results = [0 for x in range(9)]
for count in tallies:
results[count] += 1
for index, result in enumerate(results):
sys.stdout.write(str(result) + " ")
print("")
```
No
| 13,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Submitted Solution:
```
from collections import defaultdict
def main():
n, m = map(int, input().split())
l = list(tuple(map(int, input().split())) for _ in range(m))
h, v, d1, d2 = (defaultdict(list) for _ in range(4))
for x, y in l:
h[y].append(x)
v[x].append(y)
d1[x + y].append(x)
d2[x - y].append(x)
map(list.sort, (h, v, d1, d2))
res = [0] * 9
for x, y in l:
tot = 0
line = v[x]
i = line.index(y)
if i:
tot += 1
if i < len(line) - 1:
tot += 1
for line in h[y], d1[x + y], d2[x - y]:
i = line.index(x)
if i:
tot += 1
if i < len(line) - 1:
tot += 1
res[tot] += 1
print(*res)
if __name__ == '__main__':
main()
```
No
| 13,436 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A queen is the strongest chess piece. In modern chess the queen can move any number of squares in any horizontal, vertical or diagonal direction (considering that there're no other pieces on its way). The queen combines the options given to the rook and the bishop.
There are m queens on a square n Γ n chessboard. You know each queen's positions, the i-th queen is positioned in the square (ri, ci), where ri is the board row number (numbered from the top to the bottom from 1 to n), and ci is the board's column number (numbered from the left to the right from 1 to n). No two queens share the same position.
For each queen one can count w β the number of other queens that the given queen threatens (attacks). For a fixed attack direction only the first queen in this direction is under attack if there are many queens are on the ray of the attack. Obviously, for any queen w is between 0 and 8, inclusive.
Print the sequence t0, t1, ..., t8, where ti is the number of queens that threaten exactly i other queens, i.e. the number of queens that their w equals i.
Input
The first line of the input contains a pair of integers n, m (1 β€ n, m β€ 105), where n is the size of the board and m is the number of queens on the board. Then m following lines contain positions of the queens, one per line. Each line contains a pair of integers ri, ci (1 β€ ri, ci β€ n) β the queen's position. No two queens stand on the same square.
Output
Print the required sequence t0, t1, ..., t8, separating the numbers with spaces.
Examples
Input
8 4
4 3
4 8
6 5
1 6
Output
0 3 0 1 0 0 0 0 0
Input
10 3
1 1
1 2
1 3
Output
0 2 1 0 0 0 0 0 0
Submitted Solution:
```
from collections import defaultdict
from bisect import bisect_left
def main():
n, m = map(int, input().split())
l = list(tuple(map(int, input().split())) for _ in range(m))
h, v, d1, d2 = (defaultdict(list) for _ in range(4))
for x, y in l:
h[y].append(x)
v[x].append(y)
d1[x + y].append(x)
d2[x - y].append(x)
map(list.sort, (h, v, d1, d2))
res = [0] * 9
for x, y in l:
tot = 0
line = v[x]
i = bisect_left(line, y)
if i:
tot += 1
if i < len(line) - 1:
tot += 1
for line in h[y], d1[x + y], d2[x - y]:
i = bisect_left(line, x)
if i:
tot += 1
if i < len(line) - 1:
tot += 1
res[tot] += 1
print(*res)
if __name__ == '__main__':
main()
```
No
| 13,437 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Tags: constructive algorithms, sortings
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
a.sort()
m = (n + 1) // 2 - 1
for x, y in zip(a[m::-1], a[m + 1::]):
print(x, y, end=' ')
if n % 2:
print(a[0], end=' ')
print()
```
| 13,438 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Tags: constructive algorithms, sortings
Correct Solution:
```
import math
t=int(input())
for g in range(0,t):
n=int(input())
a=list(map(int,input().split(" ")))
a.sort()
if(n%2==0):
k=int(n/2)-1
else:
k=int(n/2)
b=[]
b.append(a[k])
i=k-1
j=k+1
while(i>=0 and j<=n-1):
b.append(a[j])
j=j+1
b.append(a[i])
i=i-1
if(n%2==0):
b.append(a[n-1])
for i in range(0,n):
print(b[i],end=" ")
print("")
```
| 13,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Tags: constructive algorithms, sortings
Correct Solution:
```
import sys
import math as mt
input=sys.stdin.buffer.readline
import bisect
t=int(input())
#tot=0
for __ in range(t):
n=int(input())
a=list(map(int,input().split()))
a.sort()
ans=[]
i=(n//2)-1
j=(n//2)+1
ans.append(a[n//2])
n1=n
n-=1
ch=0
if (n1-n//2-1)>n//2:
ch=1
while n:
if j>=n1:
ans.append(a[i])
i-=1
else:
if ch%2==0:
#print(ch,j,n,a[j])
ans.append(a[j])
j+=1
else:
ans.append(a[i])
i-=1
ch+=1
n-=1
print(*ans)
```
| 13,440 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Tags: constructive algorithms, sortings
Correct Solution:
```
t=int(input())
p=[]
for ikn in range(0,t):
n=int(input())
s=input().split( )
for i in range(0,n):
s[i]=int(s[i])
s.sort()
d=[]
g=0
while len(d)<len(s):
d.append(s[g])
d.append(s[n-g-1])
g+=1
if len(d)==n+1:
del(d[n])
d.reverse()
p.append(d)
for i in p:
for j in i:
print(j,end=' ')
print()
```
| 13,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Tags: constructive algorithms, sortings
Correct Solution:
```
for t in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
a.sort()
l=[]
for i in range(n):
if i%2:
l+=[a.pop(0)]
else:
l+=[a.pop()]
print(*l[::-1])
```
| 13,442 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Tags: constructive algorithms, sortings
Correct Solution:
```
t = int(input())
for q in range(t):
n=int(input())
arr = list(map(int, input().split()))
arr.sort()
ans=[]
if n%2==0:
for i in range(n//2):
ans.append(arr[n-i-1])
ans.append(arr[i])
else:
for i in range(n//2):
ans.append(arr[n - i - 1])
ans.append(arr[i])
ans.append(arr[n//2])
for k in range(len(ans)-1,-1,-1):
print(ans[k],end=' ')
print()
```
| 13,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Tags: constructive algorithms, sortings
Correct Solution:
```
for t in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
a.sort()
b = [0] * n
j = 0
for i in range(n // 2, n):
b[j] = a[i]
if i != n - 1 or n % 2 == 0:
b[j + 1] = a[n - i - 1 - n % 2]
j += 2
print(*b)
```
| 13,444 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Tags: constructive algorithms, sortings
Correct Solution:
```
def solve():
n = int(input())
a = sorted([int(x) for x in input().split()])
l, r = 0, len(a) - 1
cur = 0
res = []
while(l <= r):
if(cur == 0):
res.append(str(a[l]))
l += 1
else:
res.append(str(a[r]))
r -= 1
cur = 1 - cur
print(" ".join(res[::-1]))
t = int(input())
for _ in range(t):
solve()
```
| 13,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Submitted Solution:
```
testcases = int(input())
while(testcases):
testcases -= 1
n = int(input())
arr = list(map(int,input().split()))
arr.sort()
while arr:
print(arr.pop(len(arr)//2),end = " ")
print("")
```
Yes
| 13,446 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Submitted Solution:
```
import sys
for _ in range(int(sys.stdin.readline())):
n = int(sys.stdin.readline())
a = list(map(int, sys.stdin.readline().split()))
a.sort()
b = [None] * n
j = (n - 1) >> 1
for i in range(n):
b[i] = a[j]
if i & 1 == 0:
j += i + 1
else:
j -= i + 1
print(*b)
```
Yes
| 13,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Submitted Solution:
```
for _ in range(int(input())):
total_numbers = int(input())
numbers = list(map(int, input().split()))
numbers.sort()
final = []
for i in range(total_numbers//2):
final.append(numbers[i])
position = (-1)*(i+1)
final.append(numbers[position])
if total_numbers % 2 == 1:
middle = (total_numbers//2)
final.append(numbers[middle])
final.reverse()
print(" ".join(map(str, final)))
```
Yes
| 13,448 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
i = n // 2 + n % 2
j = i + 1
p = []
for k in range(n // 2 + n % 2):
p.append(i)
p.append(j)
i -= 1
j += 1
if n % 2:
p.pop()
print(' '.join(str(a[i - 1]) for i in p))
```
Yes
| 13,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Submitted Solution:
```
def solution():
t = int(input())
for _ in range(t):
n = int(input())
nums = list(map(int,input().split()))
nums.sort()
if n % 2 == 0:
j = n // 2
i = j - 1
print(nums[i],end=" ")
print(nums[j], end="")
i -= 1
j += 1
else:
i = n // 2
print(nums[i],end="")
j = i + 1
i -= 1
while i >= 0:
print(" " + str(nums[i]), end="")
print(" " + str(nums[j]), end="")
i -= 1
j += 1
solution()
```
No
| 13,450 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Submitted Solution:
```
import re
def neg(x):
return abs(int(x[0])-int(x[1]))
def solve():
n=int(input())
stri=input()
temp=stri.split(" ")
res=[]
for i in range(0,n,2):
#print(i)
res.append([temp[i],temp[i+1]])
temp.clear()
#print(res)
#print(res)
res = sorted(res,key=lambda x: neg(x))
#print(res)
ans=""
for indx,i in enumerate(res):
print(i[0], end = " ")
print(i[1], end=" ")
t=int(input())
for i in range(t):
solve()
```
No
| 13,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Submitted Solution:
```
from functools import cmp_to_key
num = int(input())
def cmp(a,b):
global prev
if abs(int(a)-int(b)) < prev:
prev = abs(int(a)-int(b))
return -1
else:
return 1
for _ in range(num):
l = int(input())
lis = input().split(' ')
prev = 9e999
lis = sorted(lis,key=cmp_to_key(cmp))
print(' '.join(lis))
```
No
| 13,452 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have array of n numbers a_{1}, a_{2}, β¦, a_{n}.
Rearrange these numbers to satisfy |a_{1} - a_{2}| β€ |a_{2} - a_{3}| β€ β¦ β€ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.
Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.
You have to answer independent t test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
The first line of each test case contains single integer n (3 β€ n β€ 10^{5}) β the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.
The second line of each test case contains n integers a_{1}, a_{2}, β¦, a_{n} (-10^{9} β€ a_{i} β€ 10^{9}).
Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.
Example
Input
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output
5 5 4 6 8 -2
1 2 4 8
Note
In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 β€ |a_{2} - a_{3}| = 1 β€ |a_{3} - a_{4}| = 2 β€ |a_{4} - a_{5}| = 2 β€ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".
In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 β€ |a_{2} - a_{3}| = 2 β€ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".
Submitted Solution:
```
import sys
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
N = int(input())
A = [int(x) for x in input().split()]
A.sort()
ans = []
for i in range(N // 2):
ans.append(A[-i - 1])
ans.append(A[1])
if N % 2 == 1:
ans.append(A[N // 2 + 1])
print(*reversed(ans))
if __name__ == '__main__':
main()
```
No
| 13,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Tags: bitmasks, brute force, constructive algorithms, dp, hashing, strings
Correct Solution:
```
def solve(arr, n , m ) :
fix = arr[0]
for t in range(m) :
ch = fix[t]
for k in range(26) :
fix[t] = chr(ord('a') + k )
flag = True
for i in range(n) :
count = 0
for j in range(m) :
if arr[i][j] != fix[j] :
count +=1
if count >1 :
flag= False
break
if flag == True :
return ''.join(fix)
fix[t] = ch
return -1
for T in range(int(input())) :
n , m = list(map(int,input().split()))
arr = []
for i in range(n) :
arr.append(list(input()))
v = solve(arr,n , m)
print(v)
```
| 13,454 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Tags: bitmasks, brute force, constructive algorithms, dp, hashing, strings
Correct Solution:
```
t=int(input())
for _ in range(t):
n,m=map(int,input().split())
s=[]
for _ in range(n):
s.append(input())
ans=list(s[0])
l=[]
for i in range(m):
for j in range(26):
ans[i]=chr(j+97)
l.append(''.join(ans))
ans[i]=s[0][i]
nf=0
for i in l:
flag=0
for j in s:
ct=0
for k in range(m):
if j[k]!=i[k]:
ct+=1
if ct>1:
flag=1
break
if flag==0:
print(''.join(i))
nf=1
break
if nf==0:
print('-1')
```
| 13,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Tags: bitmasks, brute force, constructive algorithms, dp, hashing, strings
Correct Solution:
```
import math
import string
def change(s, i, symbol):
ans = s[0:i] + symbol + s[(i + 1):]
return ans
def check(a, n, m, s):
ans = 0
for i in range(n):
cnt = 0
for j in range(m):
if(a[i][j] != s[j]):
cnt += 1
if(cnt > 1):
return 0
return 1
t = int(input())
for ttt in range(t):
n, m = map(int, input().split())
a = []
for i in range(n):
a.append(str(input()))
s = a[0]
ans = 0
for j in range(m):
if(ans == 1):
break
for c in string.ascii_lowercase:
now = change(s, j, c)
if(check(a, n, m, now)):
ans = 1
print(now)
break
if(ans == 0):
print(-1)
```
| 13,456 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Tags: bitmasks, brute force, constructive algorithms, dp, hashing, strings
Correct Solution:
```
import string
def spystring(words):
def neigh(word):
candidates = {word}
for i in range(len(word)):
for c in string.ascii_lowercase:
candidates.add(word[:i] + c + word[i + 1:])
return candidates
intersection = neigh(words[0])
for word in words[1:]:
intersection &= neigh(word)
return intersection and intersection.pop() or -1
def solve():
n, m = map(int, input().split())
s = [input() for _ in range(n)]
print(spystring(s))
for _ in range(int(input())):
solve()
```
| 13,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Tags: bitmasks, brute force, constructive algorithms, dp, hashing, strings
Correct Solution:
```
import sys
import math
from collections import defaultdict,deque
import heapq
t = int(sys.stdin.readline())
for _ in range(t):
n,m = map(int,sys.stdin.readline().split())
l =[]
for i in range(n):
s = sys.stdin.readline()[:-1]
l.append(s)
#print(l,'l')
ind = -1
z = True
till = -1
for i in range(m):
#z = True
let = l[0][i]
for j in range(n):
if l[j][i] != let:
z = False
ind = j
till = i
break
if not z:
break
if z:
print(l[0])
continue
z = True
for k in range(26):
ans = l[0][:till] + chr(97+k) + l[0][till+1:]
z=True
for i in range(n):
cnt = 0
for j in range(m):
if l[i][j] != ans[j]:
cnt += 1
#print(cnt,'cnt',i,'i')
if cnt > 1:
z = False
break
if z:
print(ans)
break
if z:
continue
z = True
for k in range(26):
ans = l[ind][:till] + chr(97+k) + l[ind][till+1:]
z=True
for i in range(n):
cnt = 0
for j in range(m):
if l[i][j] != ans[j]:
cnt += 1
#print(cnt,'cnt',i,'i')
if cnt > 1:
z = False
break
if z:
print(ans)
break
if not z:
print(-1)
```
| 13,458 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Tags: bitmasks, brute force, constructive algorithms, dp, hashing, strings
Correct Solution:
```
#import math
#from functools import lru_cache
#import heapq
#from collections import defaultdict
#from collections import Counter
#from sys import stdout
#from sys import setrecursionlimit
from sys import stdin
input = stdin.readline
for Ti in range(int(input().strip())):
n, m = [int(x) for x in input().strip().split()]
sa = []
for ni in range(n):
sa.append(input().strip())
s1 = sa[0]
val = True
ans = ''
for m1 in range(m):
for ci in list('abcdefghijklmnopqrstuvwwxyz'):
s1 = sa[0]
s1 = s1[:m1] + ci + s1[m1+1:]
#print(s1)
val = True
for ni in range(n):
dc = 0
for mi in range(m):
if(sa[ni][mi]!=s1[mi]): dc+=1
if(dc>1):
val=False
break
if val:
ans = s1
break
else:
continue
break
if not ans:
print('-1')
else:
print(ans)
```
| 13,459 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Tags: bitmasks, brute force, constructive algorithms, dp, hashing, strings
Correct Solution:
```
# cook your dish here
import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
import math
import collections
from sys import stdin,stdout,setrecursionlimit
import bisect as bs
setrecursionlimit(2**20)
M = 10**9+7
T = int(stdin.readline())
# T = 1
def check(s1,s2):
co = 0
for i in range(m):
if(s1[i] != s2[i]):
co += 1
if(co > 1):
return False
return True
for _ in range(T):
# n = int(stdin.readline())
n,m = list(map(int,stdin.readline().split()))
# h = list(map(int,stdin.readline().split()))
# q = list(map(int,stdin.readline().split()))
# b = list(map(int,stdin.readline().split()))
# s = stdin.readline().strip('\n')
ss = []
for i in range(n):
ss.append(list(stdin.readline().strip('\n')))
res = False
for i in range(m):
for a in range(26):
c = chr(ord('a')+a)
t = ss[0].copy()
t[i] = c
cc = 0
for j in range(1,n):
if(not check(t,ss[j])):
break
cc += 1
if(cc == n-1):
res = True
break
if(res):
break
if(res):
print(''.join(t))
continue
print(-1)
```
| 13,460 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Tags: bitmasks, brute force, constructive algorithms, dp, hashing, strings
Correct Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
def gift():
for _ in range(t):
n,m = list(map(int,input().split()))
arrys=[]
for i in range(n):
kap=list(input())
arrys.append(kap)
ans=[]
checkdiff=True
for i in range(n):
for j in range(i,n):
if checkdiff:
counter=0
for k in range(m):
if arrys[i][k]!=arrys[j][k]:
counter+=1
if counter>2:
checkdiff=False
break
else:
break
if checkdiff:
for j in range(m):
counterdic={}
for i in range(n):
ele=arrys[i][j]
freq=counterdic.get(ele,0)
counterdic[ele]=freq+1
#print(counterdic,len(counterdic),arrys[0][j])
if len(counterdic)==1:
ans.append(arrys[0][j])
else:
ans.append(list(counterdic.keys()))
pans=ans[0]
for j in range(1,m):
#print(pans,ans)
tis=ans[j]
#print(tis,pans,ans)
tans=[]
if len(tis)==1:
for di in pans:
di+=tis
tans.append(di)
else:
for di in pans:
for ti in tis:
diu=di+ti
tans.append(diu)
#print(tans)
pans=tans
gotans=False
for ina in pans:
ans=True
for i in range(n):
counter=0
for j in range(m):
#print(arrys)
if ina[j]!=arrys[i][j]:
counter+=1
if counter>=2:
ans=False
break
if ans:
yield ina
gotans=True
break
if not gotans:
yield -1
else:
yield -1
#aaaab
#abaaa
#aaaab
if __name__ == '__main__':
t= int(input())
ans = gift()
print(*ans,sep='\n')
```
| 13,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Submitted Solution:
```
import string
def isvalid(word):
for i in range(n):
differences = 0
for j in range(m):
if word[j] != strings[i][j]:
differences += 1
if differences > 1:
return False
return True
def solve():
alphabet = string.ascii_lowercase
if isvalid(strings[0]):
return strings[0]
for i in range(m):
s = strings[0].copy()
for j in range(len(alphabet)):
s[i] = alphabet[j]
if isvalid(s):
return s
return '-1'
if __name__ == '__main__':
testcases = int(input())
for i in range(testcases):
n, m = map(int, input().split())
strings = []
for j in range(n):
strings.append(list(input()))
print(''.join(solve()))
```
Yes
| 13,462 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Submitted Solution:
```
#include <CodeforcesSolutions.h>
#include <ONLINE_JUDGE <solution.cf(contestID = "1360",problemID = "F",method = "GET")>.h>
"""
Author : thekushalghosh
Team : CodeDiggers
I prefer Python language over the C++ language :p :D
Visit my website : thekushalghosh.github.io
"""
import sys,math,cmath,time
start_time = time.time()
global tt
#---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
return(input().strip())
def invr():
return(map(int,input().split()))
##########################################################################
################# ---- THE ACTUAL CODE STARTS BELOW ---- #################
def solve():
n,m = invr()
a = []
for i in range(n):
s = insr()
a.append(s)
b = [["0"] * len(a) for i in range(m)]
for i in range(m):
for j in range(n):
b[i][j] = a[j][i]
q = a[0]
qw = 0
for i in range(m):
qq = string_counter(b[i])
for j in range(26):
w = q[:i] + chr(97 + j) + q[i + 1:]
for k in range(n):
ww = 0
for l in range(m):
if w[l] != a[k][l]:
ww = ww + 1
if ww < 2:
True
else:
break
else:
print(w)
qw = 1
break
if qw == 1:
break
if qw == 0:
print(-1)
################## ---- THE ACTUAL CODE ENDS ABOVE ---- ##################
##########################################################################
def main():
global tt
if not ONLINE_JUDGE:
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
t = 1
t = inp()
for tt in range(t):
solve()
if not ONLINE_JUDGE:
print("Time Elapsed :",time.time() - start_time,"seconds")
sys.stdout.close()
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#
def counter(a):
q = [0] * max(a)
for i in range(len(a)):
q[a[i] - 1] = q[a[i] - 1] + 1
return(q)
def string_counter(a):
q = [0] * 26
for i in range(len(a)):
q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
return(q)
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
input = sys.stdin.readline
main()
```
Yes
| 13,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Submitted Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
from math import *
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a,b):
if a == 0:
return b
return gcd(b%a, a)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return res
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if a[mid] < x:
lo = mid+1
else:
hi = mid
return lo
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()))
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
def isValid(this):
for i in range(n):
count = 0;
for j in range(m):
if this[j] != a[i][j]:
count += 1;
if count > 1:
return False
return True;
for _ in range(int(input())):
n, m = int_array(); a = [];
for __ in range(n):
a.append(list(input().strip()));
x = a[0]; f = 1;
for i in range(m):
if not f:
break;
for j in range(26):
this = x.copy(); this[i] = chr(97 + j);
if isValid(this):
print(''.join(this));
f = 0; break;
if f:
print(-1);
```
Yes
| 13,464 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Submitted Solution:
```
from copy import deepcopy
t=int(input())
for _ in range(t):
n,m=list(map(int,input().split()))
s=[]
for i in range(n):
s.append(list(input()))
def diff(a,b):
count=0
for i in range(0,len(a)):
if a[i]!=b[i]:
count+=1
return count
flag1=0
str=-1
temp = deepcopy(s[0])
for i in range(0,m):
temp=deepcopy(s[0])
for c in 'qwertyuiopasdfghjklzxcvbnm':
temp[i]=c
# print(temp)
flag=1
for l in s[1:]:
if diff(l,temp)>1:
flag=0
break
if flag==1:
flag1=1
break
if flag1==1:
break
if flag1==1:
print(''.join(temp))
else:
print(-1)
```
Yes
| 13,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Submitted Solution:
```
for T in range(int(input())):
n, m = [int(x) for x in input().split()]
arr = []
for i in range(n):
arr.append(list(input()))
if n == 1:
print(*arr[0], sep = '')
continue
first = arr[0]
done = 0
for i in range(m):
prev = first[i]
for j in range(97, 97 + 26):
ch = chr(j)
first[i] = ch
for k in range(1, n):
bad = 0
for pos in range(m):
if first[pos] != arr[k][pos]:
bad += 1
if bad > 1:
break
if bad > 1:
break
else:
print(*first, sep = '')
done = 1
break
if done:
break
if done:
break
else:
first[i] = prev
if not done:
print(-1)
```
No
| 13,466 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Submitted Solution:
```
import sys
try:
sys.stdin = open('Input.txt', 'r')
sys.stdout = open('Output.txt', 'w')
except:
pass
for testCases in range(int(input())):
s = []
n,m = map(int,input().split())
for i in range(n):
s.append(list(input()))
# print(s)
# break
ans = s[0]
index = -1
char = ''
limit = 0
flag = True
checkAns = True
for i in range(n):
limit = 0
diff1 = 0
diff2 = 0
for j in range(m):
# print(j,ans)
if flag:
if s[i][j] != ans[j]:
if limit == 2:
checkAns = False
break
if limit == 0:
index = j
char = s[i][j]
limit +=1
else:
temp1 = ans[:]
temp1[index] = char
temp2 = ans[:]
temp2[j] = s[i][j]
ans = [temp1,temp2]
flag = False
# print(ans)
else:
# print("YAY")
if ans[0]!=[] and s[i][j]!=ans[0][j]:
diff1 += 1
if ans[1]!=[] and s[i][j]!=ans[1][j]:
diff2 += 1
if diff1>=2:
ans[0] = []
if diff2>=2:
ans[1] = []
if checkAns == False:
break
if checkAns:
try:
print("".join(ans))
except:
if ans[0]!=[]:
print("".join(ans[0]))
elif ans[1]!=[]:
print("".join(ans[1]))
else:
print(-1)
else:
print(-1)
```
No
| 13,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Submitted Solution:
```
def stringgen(arr,n,m):
first=arr[0]
for i in range(m):
save=first[i]
for ch in "abcdefghijklmnopqrstuvxyz":
first[i]=ch
flag=True
for rest in arr[1:]:
cnt=0
for x in range(m):
if rest[x]!=first[x]:
cnt+=1
if cnt>1:
flag=False
break
if flag==True:
return "".join(first)
first[i]=save
return -1
for i in range(int(input())):
n,m=map(int,input().strip().split())
blanck=[]
for j in range(n):
blanck.append([i for i in input()])
print(stringgen(blanck,n,m))
```
No
| 13,468 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Submitted Solution:
```
def solve(n, m, arr):
if n == 1:
print(arr[0])
return
# s1 = arr[0]
# s2 = ""
s1, s2 = "", ""
for i in range(1, n):
diff = check(arr[0], arr[i])
if len(diff) <= 1:
continue
elif len(diff) == 2:
idx1, idx2 = diff
s1, s2 = arr[0], arr[i]
s1 = s1[:idx1] + arr[i][idx1] + s1[idx1 + 1:]
s2 = s2[:idx2] + arr[0][idx2] + s2[idx2 + 1:]
break
else:
print(-1)
return
if s1 == "":
print(arr[0])
return
f1, f2 = True, True
for s in arr:
diff = check(s1, s)
if len(diff) > 1:
f1 = False
break
if f1:
print(s1)
return
for s in arr:
diff = check(s2, s)
if len(diff) > 1:
f2 = False
break
if f2:
print(s2)
return
print(-1)
return
def check(s1, s2):
diff = []
for i in range(len(s1)):
if s1[i] != s2[i]:
diff.append(i)
return diff
t = int(input())
for i in range(0, t):
n, m = map(int, input().split())
arr = []
for j in range(n):
arr.append(input())
solve(n, m, arr)
```
No
| 13,469 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings a_1, a_2, β¦, a_n: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] β s[j].
Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.
Input
The first line contains an integer t (1 β€ t β€ 100) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1 β€ n β€ 10) and m (1 β€ m β€ 10) β the number of strings and their length.
Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.
Output
Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).
Example
Input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
Output
abab
-1
aaa
ab
z
Note
The first test case was explained in the statement.
In the second test case, the answer does not exist.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
# main code
for t in range(ni()):
n,m=li()
l=[]
for i in range(n):
l.append(list(raw_input().strip()))
f1=0
for i in range(n):
ans=list(l[i])
f2=0
f3=0
for j in range(n):
if i==j:
continue
f=0
for k in range(m):
if ans[k]!=l[j][k]:
#print i,j,k,'asdf'
if not f2:
ans[k]=l[j][k]
f2=1
else:
f+=1
if f>1:
break
if f>1:
f3=1
break
#print f,f3,f2,j,k,ans
if not f3:
f1=1
break
if f1:
pr(''.join(ans)+'\n')
else:
pn(-1)
```
No
| 13,470 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are creating a level for a video game. The level consists of n rooms placed in a circle. The rooms are numbered 1 through n. Each room contains exactly one exit: completing the j-th room allows you to go the (j+1)-th room (and completing the n-th room allows you to go the 1-st room).
You are given the description of the multiset of n chests: the i-th chest has treasure value c_i.
Each chest can be of one of two types:
* regular chest β when a player enters a room with this chest, he grabs the treasure and proceeds to the next room;
* mimic chest β when a player enters a room with this chest, the chest eats him alive, and he loses.
The player starts in a random room with each room having an equal probability of being chosen. The players earnings is equal to the total value of treasure chests he'd collected before he lost.
You are allowed to choose the order the chests go into the rooms. For each k from 1 to n place the chests into the rooms in such a way that:
* each room contains exactly one chest;
* exactly k chests are mimics;
* the expected value of players earnings is minimum possible.
Please note that for each k the placement is chosen independently.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Input
The first contains a single integer n (2 β€ n β€ 3 β
10^5) β the number of rooms and the number of chests.
The second line contains n integers c_1, c_2, ..., c_n (1 β€ c_i β€ 10^6) β the treasure values of each chest.
Output
Print n integers β the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Examples
Input
2
1 2
Output
499122177 0
Input
8
10 4 3 6 5 10 7 5
Output
499122193 249561095 249561092 873463811 499122178 124780545 623902721 0
Note
In the first example the exact values of minimum expected values are: \frac 1 2, \frac 0 2.
In the second example the exact values of minimum expected values are: \frac{132} 8, \frac{54} 8, \frac{30} 8, \frac{17} 8, \frac{12} 8, \frac 7 8, \frac 3 8, \frac 0 8.
Tags: greedy, math, probabilities
Correct Solution:
```
def add(a, b):
return (a + b) % mod
def mult(a, b):
tmp = (a % mod) * (b % mod)
return tmp % mod
n = int(input())
c = [int(i) for i in input().split()]
c.sort(reverse = True)
pref = [0] * (n + 1)
pref[0] = 0
for i in range(1, n + 1):
pref[i] = pref[i - 1] + c[i - 1]
mod = 998244353
invn = pow(n, mod - 2, mod)
for k in range(1, n + 1):
ans = 0
j = 0
for i in range(0, n + 1, k):
ans = add(ans, mult(j, (add(pref[min(n, i + k)], -pref[i]))))
j += 1
ans = mult(ans, invn)
print(ans, end = ' ')
```
| 13,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are creating a level for a video game. The level consists of n rooms placed in a circle. The rooms are numbered 1 through n. Each room contains exactly one exit: completing the j-th room allows you to go the (j+1)-th room (and completing the n-th room allows you to go the 1-st room).
You are given the description of the multiset of n chests: the i-th chest has treasure value c_i.
Each chest can be of one of two types:
* regular chest β when a player enters a room with this chest, he grabs the treasure and proceeds to the next room;
* mimic chest β when a player enters a room with this chest, the chest eats him alive, and he loses.
The player starts in a random room with each room having an equal probability of being chosen. The players earnings is equal to the total value of treasure chests he'd collected before he lost.
You are allowed to choose the order the chests go into the rooms. For each k from 1 to n place the chests into the rooms in such a way that:
* each room contains exactly one chest;
* exactly k chests are mimics;
* the expected value of players earnings is minimum possible.
Please note that for each k the placement is chosen independently.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Input
The first contains a single integer n (2 β€ n β€ 3 β
10^5) β the number of rooms and the number of chests.
The second line contains n integers c_1, c_2, ..., c_n (1 β€ c_i β€ 10^6) β the treasure values of each chest.
Output
Print n integers β the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Examples
Input
2
1 2
Output
499122177 0
Input
8
10 4 3 6 5 10 7 5
Output
499122193 249561095 249561092 873463811 499122178 124780545 623902721 0
Note
In the first example the exact values of minimum expected values are: \frac 1 2, \frac 0 2.
In the second example the exact values of minimum expected values are: \frac{132} 8, \frac{54} 8, \frac{30} 8, \frac{17} 8, \frac{12} 8, \frac 7 8, \frac 3 8, \frac 0 8.
Tags: greedy, math, probabilities
Correct Solution:
```
from sys import stdin
input = stdin.buffer.readline
def inv(x):
return pow(x, mod - 2, mod)
mod = 998244353
n = int(input())
a = sorted(map(int, input().split()), reverse=True)
p = [0] * (n + 1)
p[0] = a[0]
for i in range(1, n):
p[i] = p[i - 1] + a[i]
#print(p)
for k in range(1, n + 1):
ans = 0
i = 0
while i + k <= n:
ans = (ans + (p[i + k - 1] - p[i - 1]) * (i // k)) % mod
#print(ans, p[i + k - 1], p[i - 1])
i += k
#print(ans)
ans = (ans + (p[n - 1] - p[i - 1]) * (i // k)) % mod
#print(ans, i)
ans = ans * inv(n)
print(ans % mod)
```
| 13,472 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are creating a level for a video game. The level consists of n rooms placed in a circle. The rooms are numbered 1 through n. Each room contains exactly one exit: completing the j-th room allows you to go the (j+1)-th room (and completing the n-th room allows you to go the 1-st room).
You are given the description of the multiset of n chests: the i-th chest has treasure value c_i.
Each chest can be of one of two types:
* regular chest β when a player enters a room with this chest, he grabs the treasure and proceeds to the next room;
* mimic chest β when a player enters a room with this chest, the chest eats him alive, and he loses.
The player starts in a random room with each room having an equal probability of being chosen. The players earnings is equal to the total value of treasure chests he'd collected before he lost.
You are allowed to choose the order the chests go into the rooms. For each k from 1 to n place the chests into the rooms in such a way that:
* each room contains exactly one chest;
* exactly k chests are mimics;
* the expected value of players earnings is minimum possible.
Please note that for each k the placement is chosen independently.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Input
The first contains a single integer n (2 β€ n β€ 3 β
10^5) β the number of rooms and the number of chests.
The second line contains n integers c_1, c_2, ..., c_n (1 β€ c_i β€ 10^6) β the treasure values of each chest.
Output
Print n integers β the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Examples
Input
2
1 2
Output
499122177 0
Input
8
10 4 3 6 5 10 7 5
Output
499122193 249561095 249561092 873463811 499122178 124780545 623902721 0
Note
In the first example the exact values of minimum expected values are: \frac 1 2, \frac 0 2.
In the second example the exact values of minimum expected values are: \frac{132} 8, \frac{54} 8, \frac{30} 8, \frac{17} 8, \frac{12} 8, \frac 7 8, \frac 3 8, \frac 0 8.
Tags: greedy, math, probabilities
Correct Solution:
```
from sys import stdin
def inverse(a,mod):
return pow(a,mod-2,mod)
n = int(stdin.readline())
c = list(map(int,stdin.readline().split()))
mod = 998244353
c.sort()
d = [0]
for i in range(n):
d.append(d[-1] + c[i])
inv = inverse( n , mod )
ans = []
for k in range(1,n+1):
now = 0
ni = n-k
cnt = 1
while ni > k:
now += cnt * (d[ni]-d[ni-k])
cnt += 1
ni -= k
now += cnt * d[ni]
ans.append(now * inv % mod)
print (*ans)
```
| 13,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are creating a level for a video game. The level consists of n rooms placed in a circle. The rooms are numbered 1 through n. Each room contains exactly one exit: completing the j-th room allows you to go the (j+1)-th room (and completing the n-th room allows you to go the 1-st room).
You are given the description of the multiset of n chests: the i-th chest has treasure value c_i.
Each chest can be of one of two types:
* regular chest β when a player enters a room with this chest, he grabs the treasure and proceeds to the next room;
* mimic chest β when a player enters a room with this chest, the chest eats him alive, and he loses.
The player starts in a random room with each room having an equal probability of being chosen. The players earnings is equal to the total value of treasure chests he'd collected before he lost.
You are allowed to choose the order the chests go into the rooms. For each k from 1 to n place the chests into the rooms in such a way that:
* each room contains exactly one chest;
* exactly k chests are mimics;
* the expected value of players earnings is minimum possible.
Please note that for each k the placement is chosen independently.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Input
The first contains a single integer n (2 β€ n β€ 3 β
10^5) β the number of rooms and the number of chests.
The second line contains n integers c_1, c_2, ..., c_n (1 β€ c_i β€ 10^6) β the treasure values of each chest.
Output
Print n integers β the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Examples
Input
2
1 2
Output
499122177 0
Input
8
10 4 3 6 5 10 7 5
Output
499122193 249561095 249561092 873463811 499122178 124780545 623902721 0
Note
In the first example the exact values of minimum expected values are: \frac 1 2, \frac 0 2.
In the second example the exact values of minimum expected values are: \frac{132} 8, \frac{54} 8, \frac{30} 8, \frac{17} 8, \frac{12} 8, \frac 7 8, \frac 3 8, \frac 0 8.
Tags: greedy, math, probabilities
Correct Solution:
```
n=int(input());c=sorted(list(map(int,input().split())));mod=998244353;inv=pow(n,mod-2,mod);imos=[c[i] for i in range(n)];res=[0]*n
for i in range(1,n):imos[i]+=imos[i-1]
for i in range(1,n+1):
temp=0;L=n-i;R=n-1;count=0
while True:
temp+=count*(imos[R]-imos[L-1]*(L>=1));count+=1
if L==0:break
else:R-=i;L=max(0,L-i)
res[i-1]=(temp*inv)%mod
print(*res)
```
| 13,474 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are creating a level for a video game. The level consists of n rooms placed in a circle. The rooms are numbered 1 through n. Each room contains exactly one exit: completing the j-th room allows you to go the (j+1)-th room (and completing the n-th room allows you to go the 1-st room).
You are given the description of the multiset of n chests: the i-th chest has treasure value c_i.
Each chest can be of one of two types:
* regular chest β when a player enters a room with this chest, he grabs the treasure and proceeds to the next room;
* mimic chest β when a player enters a room with this chest, the chest eats him alive, and he loses.
The player starts in a random room with each room having an equal probability of being chosen. The players earnings is equal to the total value of treasure chests he'd collected before he lost.
You are allowed to choose the order the chests go into the rooms. For each k from 1 to n place the chests into the rooms in such a way that:
* each room contains exactly one chest;
* exactly k chests are mimics;
* the expected value of players earnings is minimum possible.
Please note that for each k the placement is chosen independently.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Input
The first contains a single integer n (2 β€ n β€ 3 β
10^5) β the number of rooms and the number of chests.
The second line contains n integers c_1, c_2, ..., c_n (1 β€ c_i β€ 10^6) β the treasure values of each chest.
Output
Print n integers β the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Examples
Input
2
1 2
Output
499122177 0
Input
8
10 4 3 6 5 10 7 5
Output
499122193 249561095 249561092 873463811 499122178 124780545 623902721 0
Note
In the first example the exact values of minimum expected values are: \frac 1 2, \frac 0 2.
In the second example the exact values of minimum expected values are: \frac{132} 8, \frac{54} 8, \frac{30} 8, \frac{17} 8, \frac{12} 8, \frac 7 8, \frac 3 8, \frac 0 8.
Tags: greedy, math, probabilities
Correct Solution:
```
from itertools import accumulate
def readline():
return map(int, input().split())
MOD = 998244353
def sum_mod(a, b):
return (a + b) % MOD
def main():
n = int(input())
c = sorted(readline())
assert len(c) == n
psums = list(accumulate(c, sum_mod))
psums.insert(0, 0)
assert len(psums) == n + 1
inv_n = pow(n, MOD-2, MOD)
for k in range(1, n + 1):
p = 0
for t in range(n % k, n, k):
p = (p + psums[t]) % MOD
p *= inv_n
print(p % MOD, end=' ')
print()
if __name__ == '__main__':
main()
```
| 13,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are creating a level for a video game. The level consists of n rooms placed in a circle. The rooms are numbered 1 through n. Each room contains exactly one exit: completing the j-th room allows you to go the (j+1)-th room (and completing the n-th room allows you to go the 1-st room).
You are given the description of the multiset of n chests: the i-th chest has treasure value c_i.
Each chest can be of one of two types:
* regular chest β when a player enters a room with this chest, he grabs the treasure and proceeds to the next room;
* mimic chest β when a player enters a room with this chest, the chest eats him alive, and he loses.
The player starts in a random room with each room having an equal probability of being chosen. The players earnings is equal to the total value of treasure chests he'd collected before he lost.
You are allowed to choose the order the chests go into the rooms. For each k from 1 to n place the chests into the rooms in such a way that:
* each room contains exactly one chest;
* exactly k chests are mimics;
* the expected value of players earnings is minimum possible.
Please note that for each k the placement is chosen independently.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Input
The first contains a single integer n (2 β€ n β€ 3 β
10^5) β the number of rooms and the number of chests.
The second line contains n integers c_1, c_2, ..., c_n (1 β€ c_i β€ 10^6) β the treasure values of each chest.
Output
Print n integers β the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Examples
Input
2
1 2
Output
499122177 0
Input
8
10 4 3 6 5 10 7 5
Output
499122193 249561095 249561092 873463811 499122178 124780545 623902721 0
Note
In the first example the exact values of minimum expected values are: \frac 1 2, \frac 0 2.
In the second example the exact values of minimum expected values are: \frac{132} 8, \frac{54} 8, \frac{30} 8, \frac{17} 8, \frac{12} 8, \frac 7 8, \frac 3 8, \frac 0 8.
Tags: greedy, math, probabilities
Correct Solution:
```
n=int(input())
c=list(map(int,input().split()))
mod=998244353
inv=pow(n,mod-2,mod)
c.sort()
imos=[c[i] for i in range(n)]
for i in range(1,n):
imos[i]+=imos[i-1]
res=[0]*n
for i in range(1,n+1):
temp=0
L=n-i
R=n-1
count=0
while True:
temp+=count*(imos[R]-imos[L-1]*(L>=1))
count+=1
if L==0:
break
else:
R-=i
L=max(0,L-i)
res[i-1]=(temp*inv)%mod
print(*res)
```
| 13,476 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are creating a level for a video game. The level consists of n rooms placed in a circle. The rooms are numbered 1 through n. Each room contains exactly one exit: completing the j-th room allows you to go the (j+1)-th room (and completing the n-th room allows you to go the 1-st room).
You are given the description of the multiset of n chests: the i-th chest has treasure value c_i.
Each chest can be of one of two types:
* regular chest β when a player enters a room with this chest, he grabs the treasure and proceeds to the next room;
* mimic chest β when a player enters a room with this chest, the chest eats him alive, and he loses.
The player starts in a random room with each room having an equal probability of being chosen. The players earnings is equal to the total value of treasure chests he'd collected before he lost.
You are allowed to choose the order the chests go into the rooms. For each k from 1 to n place the chests into the rooms in such a way that:
* each room contains exactly one chest;
* exactly k chests are mimics;
* the expected value of players earnings is minimum possible.
Please note that for each k the placement is chosen independently.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Input
The first contains a single integer n (2 β€ n β€ 3 β
10^5) β the number of rooms and the number of chests.
The second line contains n integers c_1, c_2, ..., c_n (1 β€ c_i β€ 10^6) β the treasure values of each chest.
Output
Print n integers β the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Examples
Input
2
1 2
Output
499122177 0
Input
8
10 4 3 6 5 10 7 5
Output
499122193 249561095 249561092 873463811 499122178 124780545 623902721 0
Note
In the first example the exact values of minimum expected values are: \frac 1 2, \frac 0 2.
In the second example the exact values of minimum expected values are: \frac{132} 8, \frac{54} 8, \frac{30} 8, \frac{17} 8, \frac{12} 8, \frac 7 8, \frac 3 8, \frac 0 8.
Tags: greedy, math, probabilities
Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 5)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def solve():
cs = [aa[0]]
for a in aa[1:]: cs.append((cs[-1] + a) % md)
# print(cs)
inv = pow(n, md - 2, md)
ans = []
for k in range(1, n):
cur = 0
for i in range(1, n):
if n - 1 - k * i < 0: break
cur = (cur + cs[n - 1 - k * i]) % md
cur = cur * inv % md
ans.append(cur)
ans.append(0)
print(*ans)
md=998244353
n=II()
aa=LI()
aa.sort()
solve()
```
| 13,477 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are creating a level for a video game. The level consists of n rooms placed in a circle. The rooms are numbered 1 through n. Each room contains exactly one exit: completing the j-th room allows you to go the (j+1)-th room (and completing the n-th room allows you to go the 1-st room).
You are given the description of the multiset of n chests: the i-th chest has treasure value c_i.
Each chest can be of one of two types:
* regular chest β when a player enters a room with this chest, he grabs the treasure and proceeds to the next room;
* mimic chest β when a player enters a room with this chest, the chest eats him alive, and he loses.
The player starts in a random room with each room having an equal probability of being chosen. The players earnings is equal to the total value of treasure chests he'd collected before he lost.
You are allowed to choose the order the chests go into the rooms. For each k from 1 to n place the chests into the rooms in such a way that:
* each room contains exactly one chest;
* exactly k chests are mimics;
* the expected value of players earnings is minimum possible.
Please note that for each k the placement is chosen independently.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Input
The first contains a single integer n (2 β€ n β€ 3 β
10^5) β the number of rooms and the number of chests.
The second line contains n integers c_1, c_2, ..., c_n (1 β€ c_i β€ 10^6) β the treasure values of each chest.
Output
Print n integers β the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Examples
Input
2
1 2
Output
499122177 0
Input
8
10 4 3 6 5 10 7 5
Output
499122193 249561095 249561092 873463811 499122178 124780545 623902721 0
Note
In the first example the exact values of minimum expected values are: \frac 1 2, \frac 0 2.
In the second example the exact values of minimum expected values are: \frac{132} 8, \frac{54} 8, \frac{30} 8, \frac{17} 8, \frac{12} 8, \frac 7 8, \frac 3 8, \frac 0 8.
Submitted Solution:
```
print(0)
```
No
| 13,478 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are creating a level for a video game. The level consists of n rooms placed in a circle. The rooms are numbered 1 through n. Each room contains exactly one exit: completing the j-th room allows you to go the (j+1)-th room (and completing the n-th room allows you to go the 1-st room).
You are given the description of the multiset of n chests: the i-th chest has treasure value c_i.
Each chest can be of one of two types:
* regular chest β when a player enters a room with this chest, he grabs the treasure and proceeds to the next room;
* mimic chest β when a player enters a room with this chest, the chest eats him alive, and he loses.
The player starts in a random room with each room having an equal probability of being chosen. The players earnings is equal to the total value of treasure chests he'd collected before he lost.
You are allowed to choose the order the chests go into the rooms. For each k from 1 to n place the chests into the rooms in such a way that:
* each room contains exactly one chest;
* exactly k chests are mimics;
* the expected value of players earnings is minimum possible.
Please note that for each k the placement is chosen independently.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Input
The first contains a single integer n (2 β€ n β€ 3 β
10^5) β the number of rooms and the number of chests.
The second line contains n integers c_1, c_2, ..., c_n (1 β€ c_i β€ 10^6) β the treasure values of each chest.
Output
Print n integers β the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.
It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q β 0. Report the values of P β
Q^{-1} \pmod {998244353}.
Examples
Input
2
1 2
Output
499122177 0
Input
8
10 4 3 6 5 10 7 5
Output
499122193 249561095 249561092 873463811 499122178 124780545 623902721 0
Note
In the first example the exact values of minimum expected values are: \frac 1 2, \frac 0 2.
In the second example the exact values of minimum expected values are: \frac{132} 8, \frac{54} 8, \frac{30} 8, \frac{17} 8, \frac{12} 8, \frac 7 8, \frac 3 8, \frac 0 8.
Submitted Solution:
```
for i in range(int(1e5)):
print('ΠΊΡ Π½Π΅ Π»Π°Π³Π°ΠΉ ΠΏΠΆ')
```
No
| 13,479 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plays a (yet another!) strategic computer game. In this game, he leads an army of mercenaries.
Polycarp wants to gather his army for a quest. There are n mercenaries for hire, and the army should consist of some subset of them.
The i-th mercenary can be chosen if the resulting number of chosen mercenaries is not less than l_i (otherwise he deems the quest to be doomed) and not greater than r_i (he doesn't want to share the trophies with too many other mercenaries). Furthermore, m pairs of mercenaries hate each other and cannot be chosen for the same quest.
How many non-empty subsets does Polycarp need to consider? In other words, calculate the number of non-empty subsets of mercenaries such that the size of this subset belongs to [l_i, r_i] for each chosen mercenary, and there are no two mercenaries in the subset that hate each other.
The answer may be large, so calculate it modulo 998244353.
Input
The first line contains two integers n and m (1 β€ n β€ 3 β
10^5, 0 β€ m β€ min(20, (n(n-1))/(2))) β the number of mercenaries and the number of pairs of mercenaries that hate each other.
Then n lines follow, the i-th of them contains two integers l_i and r_i (1 β€ l_i β€ r_i β€ n).
Then m lines follow, the i-th of them contains two integers a_i and b_i (1 β€ a_i < b_i β€ n) denoting that the mercenaries a_i and b_i hate each other. There are no two equal pairs in this list.
Output
Print one integer β the number of non-empty subsets meeting the constraints, taken modulo 998244353.
Examples
Input
3 0
1 1
2 3
1 3
Output
3
Input
3 1
1 1
2 3
1 3
2 3
Output
2
Submitted Solution:
```
from math import factorial as f
n,m=map(int,input().split())
l,r=map(int,input().split())
a,b=map(int,input().split())
print()
print(n -m)
```
No
| 13,480 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plays a (yet another!) strategic computer game. In this game, he leads an army of mercenaries.
Polycarp wants to gather his army for a quest. There are n mercenaries for hire, and the army should consist of some subset of them.
The i-th mercenary can be chosen if the resulting number of chosen mercenaries is not less than l_i (otherwise he deems the quest to be doomed) and not greater than r_i (he doesn't want to share the trophies with too many other mercenaries). Furthermore, m pairs of mercenaries hate each other and cannot be chosen for the same quest.
How many non-empty subsets does Polycarp need to consider? In other words, calculate the number of non-empty subsets of mercenaries such that the size of this subset belongs to [l_i, r_i] for each chosen mercenary, and there are no two mercenaries in the subset that hate each other.
The answer may be large, so calculate it modulo 998244353.
Input
The first line contains two integers n and m (1 β€ n β€ 3 β
10^5, 0 β€ m β€ min(20, (n(n-1))/(2))) β the number of mercenaries and the number of pairs of mercenaries that hate each other.
Then n lines follow, the i-th of them contains two integers l_i and r_i (1 β€ l_i β€ r_i β€ n).
Then m lines follow, the i-th of them contains two integers a_i and b_i (1 β€ a_i < b_i β€ n) denoting that the mercenaries a_i and b_i hate each other. There are no two equal pairs in this list.
Output
Print one integer β the number of non-empty subsets meeting the constraints, taken modulo 998244353.
Examples
Input
3 0
1 1
2 3
1 3
Output
3
Input
3 1
1 1
2 3
1 3
2 3
Output
2
Submitted Solution:
```
print("Haan man kiya, kar diye submit!")
```
No
| 13,481 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plays a (yet another!) strategic computer game. In this game, he leads an army of mercenaries.
Polycarp wants to gather his army for a quest. There are n mercenaries for hire, and the army should consist of some subset of them.
The i-th mercenary can be chosen if the resulting number of chosen mercenaries is not less than l_i (otherwise he deems the quest to be doomed) and not greater than r_i (he doesn't want to share the trophies with too many other mercenaries). Furthermore, m pairs of mercenaries hate each other and cannot be chosen for the same quest.
How many non-empty subsets does Polycarp need to consider? In other words, calculate the number of non-empty subsets of mercenaries such that the size of this subset belongs to [l_i, r_i] for each chosen mercenary, and there are no two mercenaries in the subset that hate each other.
The answer may be large, so calculate it modulo 998244353.
Input
The first line contains two integers n and m (1 β€ n β€ 3 β
10^5, 0 β€ m β€ min(20, (n(n-1))/(2))) β the number of mercenaries and the number of pairs of mercenaries that hate each other.
Then n lines follow, the i-th of them contains two integers l_i and r_i (1 β€ l_i β€ r_i β€ n).
Then m lines follow, the i-th of them contains two integers a_i and b_i (1 β€ a_i < b_i β€ n) denoting that the mercenaries a_i and b_i hate each other. There are no two equal pairs in this list.
Output
Print one integer β the number of non-empty subsets meeting the constraints, taken modulo 998244353.
Examples
Input
3 0
1 1
2 3
1 3
Output
3
Input
3 1
1 1
2 3
1 3
2 3
Output
2
Submitted Solution:
```
a, b = map(int, input().split())
q = []
for i in range(1, a+1):
l, r = map(int, input().split())
qwe = []
qwe.append(l)
qwe.append(r)
q.append(qwe)
w = []
if not b == 0:
for i in range(1, b+1):
m, n = map(int, input().split())
we = []
we.append(m)
we.append(n)
w.append(we)
k = 0
if b == 0:
k = a
else:
for i in range(a ):
for j in range(b):
if not q[i] == w[j]:
k += 1
print(k)
```
No
| 13,482 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plays a (yet another!) strategic computer game. In this game, he leads an army of mercenaries.
Polycarp wants to gather his army for a quest. There are n mercenaries for hire, and the army should consist of some subset of them.
The i-th mercenary can be chosen if the resulting number of chosen mercenaries is not less than l_i (otherwise he deems the quest to be doomed) and not greater than r_i (he doesn't want to share the trophies with too many other mercenaries). Furthermore, m pairs of mercenaries hate each other and cannot be chosen for the same quest.
How many non-empty subsets does Polycarp need to consider? In other words, calculate the number of non-empty subsets of mercenaries such that the size of this subset belongs to [l_i, r_i] for each chosen mercenary, and there are no two mercenaries in the subset that hate each other.
The answer may be large, so calculate it modulo 998244353.
Input
The first line contains two integers n and m (1 β€ n β€ 3 β
10^5, 0 β€ m β€ min(20, (n(n-1))/(2))) β the number of mercenaries and the number of pairs of mercenaries that hate each other.
Then n lines follow, the i-th of them contains two integers l_i and r_i (1 β€ l_i β€ r_i β€ n).
Then m lines follow, the i-th of them contains two integers a_i and b_i (1 β€ a_i < b_i β€ n) denoting that the mercenaries a_i and b_i hate each other. There are no two equal pairs in this list.
Output
Print one integer β the number of non-empty subsets meeting the constraints, taken modulo 998244353.
Examples
Input
3 0
1 1
2 3
1 3
Output
3
Input
3 1
1 1
2 3
1 3
2 3
Output
2
Submitted Solution:
```
def choose(mercenaries,pairs):
count = 0
for k in mercenaries:
chosen = 0
sub_count = 0
if chosen+1 >= k[0] and chosen <= k[1]:
count += 1
mx = k[1]
if chosen+1 < mx:
for kk in range(len(mercenaries)*3):
for i in mercenaries:
if i!=k and chosen+1 < i[1]:
if [mercenaries.index(k)+1,mercenaries.index(i)+1] not in pairs and [mercenaries.index(i)+1,mercenaries.index(k)+1] not in pairs:
chosen += 1
if chosen != 0:
count += 1
return (count)
n,m = list(map(int,input().split()))
mnaries = []
hate_pairs = []
for k in range(n):
mnaries.append(list(map(int,input().split())))
for k in range(m):
hate_pairs.append(list(map(int,input().split())))
print(choose(mnaries,hate_pairs))
```
No
| 13,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Tags: data structures, sortings
Correct Solution:
```
years = {}
for k in range(int(input())):
b, d = map(int, input().split())
if b in years:
years[b] += 1
else:
years[b] = 1
if d in years:
years[d] -= 1
else:
years[d] = -1
keys = sorted(years.keys())
maxLivingYear = 0
maxLivingCount = 0
currentCount = 0
for year in keys:
currentCount += years[year]
if currentCount > maxLivingCount:
maxLivingYear = year
maxLivingCount = currentCount
print(maxLivingYear, maxLivingCount)
```
| 13,484 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Tags: data structures, sortings
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz/'
M=1000000007
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
n=Int()
a=defaultdict(int)
for i in range(n):
x,y=value()
a[x]+=1
a[y]-=1
a=[(i,a[i]) for i in a]
a.sort()
MA=0
y=-1
cur=0
for c,f in a:
cur+=f
if(cur>MA):
MA=cur
y=c
print(y,MA)
```
| 13,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Tags: data structures, sortings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def print_list(l):
print(' '.join(map(str,l)))
# import heapq as hq
# import bisect as bs
# from collections import deque as dq
from collections import defaultdict as dc
# from math import ceil,floor,sqrt
# from collections import Counter
born = dc(int)
die = dc(int)
for _ in range(N()):
a,b = RL()
born[a]+=1
die[b]+=1
s = list(sorted(set(born.keys())|set(die.keys())))
m = 0
now = 0
for i in s:
now+=born[i]-die[i]
if now>m:
res = i
m = now
print(res,m)
```
| 13,486 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Tags: data structures, sortings
Correct Solution:
```
from sys import stdout, stdin, setrecursionlimit
from io import BytesIO, IOBase
from collections import *
from itertools import *
from random import *
from bisect import *
from string import *
from queue import *
from heapq import *
from math import *
from re import *
from os import *
####################################---fast-input-output----#########################################
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = read(self._fd, max(fstat(self._fd).st_size, 8192))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = read(self._fd, max(fstat(self._fd).st_size, 8192))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
stdin, stdout = IOWrapper(stdin), IOWrapper(stdout)
graph, mod, szzz = {}, 10**9 + 7, lambda: sorted(zzz())
def getStr(): return input()
def getInt(): return int(input())
def listStr(): return list(input())
def getStrs(): return input().split()
def isInt(s): return '0' <= s[0] <= '9'
def input(): return stdin.readline().strip()
def zzz(): return [int(i) for i in input().split()]
def output(answer, end='\n'): stdout.write(str(answer) + end)
def lcd(xnum1, xnum2): return (xnum1 * xnum2 // gcd(xnum1, xnum2))
dx = [-1, 1, 0, 0, 1, -1, 1, -1]
dy = [0, 0, 1, -1, 1, -1, -1, 1]
daysInMounth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
#################################################---Some Rule For Me To Follow---#################################
"""
--instants of Reading problem continuously try to understand them.
--If you Know some-one , Then you probably don't know him !
--Try & again try, maybe you're just one statement away!
"""
##################################################---START-CODING---###############################################
n = getInt()
lst = []
for i in range(n):
x,y=zzz()
lst.append((x,1))
lst.append((y,-1))
lst =sorted(lst)
ans,c=0,0
for i in lst:
c+=i[1]
if ans<c:
ans,y=c,i[0]
print(y,ans)
```
| 13,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Tags: data structures, sortings
Correct Solution:
```
"""
Author - Satwik Tiwari .
4th Oct , 2020 - Sunday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil
from copy import deepcopy
from collections import deque
# from collections import Counter as counter # Counter(list) return a dict with {key: count}
# from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
# from itertools import permutations as permutate
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
mod = 10**9+7
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#===============================================================================================
# code here ;))
def bucketsort(order, seq):
buckets = [0] * (max(seq) + 1)
for x in seq:
buckets[x] += 1
for i in range(len(buckets) - 1):
buckets[i + 1] += buckets[i]
new_order = [-1] * len(seq)
for i in reversed(order):
x = seq[i]
idx = buckets[x] = buckets[x] - 1
new_order[idx] = i
return new_order
def ordersort(order, seq, reverse=False):
bit = max(seq).bit_length() >> 1
mask = (1 << bit) - 1
order = bucketsort(order, [x & mask for x in seq])
order = bucketsort(order, [x >> bit for x in seq])
if reverse:
order.reverse()
return order
def long_ordersort(order, seq):
order = ordersort(order, [int(i & 0x7fffffff) for i in seq])
return ordersort(order, [int(i >> 31) for i in seq])
def multikey_ordersort(order, *seqs, sort=ordersort):
for i in reversed(range(len(seqs))):
order = sort(order, seqs[i])
return order
def solve(case):
n = int(inp())
b = []
type = []
for i in range(n):
x,y = sep()
b.append(x)
b.append(y)
type.append(1)
type.append(0)
order = multikey_ordersort(range(2*n),b,type)
have = []
for i in order:
have.append((b[i],type[i]))
# print(have)
year = 0
cnt =0
currin = 0
for i in range(len(have)):
curr = have[i]
if(curr[1] == 1):
currin+=1
else:
currin-=1
if(currin>cnt):
cnt = currin
year = curr[0]
# print(curr,currin,cnt,year)
print(year,cnt)
testcase(1)
# testcase(int(inp()))
```
| 13,488 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Tags: data structures, sortings
Correct Solution:
```
import sys
from collections import defaultdict
def load_sys():
return sys.stdin.readlines()
def load_local():
with open('input.txt','r') as f:
input=f.readlines()
return input
def years(arr):
events=[[x[0],'b'] for x in arr]+[[x[1],'d'] for x in arr]
events.sort(key=lambda x:x[0])
ctr=defaultdict(int)
for y,event in events:
if event=='b':
ctr[y]+=1
else:
ctr[y]-=1
prev=0
mx=float('-inf')
ans=0
for y in ctr:
prev+=ctr[y]
ctr[y]=prev
if ctr[y]>mx:
ans=y
mx=ctr[y]
return str(ans)+' '+str(mx)
#input=load_local()
input=load_sys()
N=int(input[0])
arr=[]
#test_cases=[]
for i in range(1,len(input)):
person=input[i].split()
person=[int(x) for x in person]
arr.append(person)
print(years(arr))
```
| 13,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Tags: data structures, sortings
Correct Solution:
```
n = int(input())
years = []
for i in range(n):
vals = input().split()
years.append((int(vals[0]), 1))
years.append((int(vals[1]), -1))
years.sort()
max_population = 0
population = 0
for year in years:
if year[1] == -1:
population -= 1
else:
population += 1
if population > max_population:
max_population = population
max_year = year[0]
print(max_year, max_population)
```
| 13,490 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Tags: data structures, sortings
Correct Solution:
```
dct = {}
for i in range(int(input())):
a,b = map(int,input().split())
dct[a] = dct.get(a,0)+1
dct[b] = dct.get(b,0)-1
cnt = curr = y = 0
for i in sorted(dct.keys()):
curr += dct[i]
if curr > cnt :
cnt = curr
y = i
print(y,cnt)
```
| 13,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Submitted Solution:
```
# 9 :24
def readInt():
return int(input())
def readLine():
return [int(s) for s in input().split(" ")]
def readString():
return input()
def ask(od):
allData = []
for a,b in od:
allData.append((a,True))
allData.append((b,False))
allData = sorted(allData)
maxL = 0
maxYear = 0
current = 0
for num,b in allData:
if(b):
current += 1
else:
current -= 1
if(current > maxL):
maxL = current
maxYear = num
return maxYear, maxL
# import random
n = readInt()
ret = []
for _ in range(n):
params = readLine()
ret.append((params[0],params[1]))
# ret = []
#
# for _ in range(100000):
# ret.append((random.randint(100000,1000000),random.randint(100000,1000000)))
#
# import time
#
# t0 = time.time()
a,b = ask(ret)
print(a,b)
# t1 = time.time()
# total = t1-t0
# print(total)
```
Yes
| 13,492 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Submitted Solution:
```
n=int(input())
a=[]
for i in range(n):
x,y=map(int,input().split())
a.append([x,1])
a.append([y,-1])
a.sort()
k=0
y=0
m=0
for i in range(len(a)):
k+=a[i][1]
if m<k:
m=k
y=a[i][0]
print(y,m)
```
Yes
| 13,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Submitted Solution:
```
n = int(input())
births = []
deaths = []
for i in range(n):
b, d = map(int, input().split())
births.append(b)
deaths.append(d)
births = sorted(births)
deaths = sorted(deaths)
living = 0
i = 0
j = 0
year = 0
current = 0
change = 0
while i < n and j < n:
current = min(births[i], deaths[j])
if births[i] <= current:
change += 1
i += 1
if deaths[j] <= current:
change -= 1
j += 1
if change > living:
living = change
year = current
if i < n:
if change + n - i + 1 > living:
living = change + n - i + 1
year = births[-1]
print(f"{year} {living}")
```
Yes
| 13,494 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Submitted Solution:
```
anos = {}
for _ in range(int(input())):
ini, fim = [int(x) for x in input().split()]
if ini in anos:
anos[ini] += 1
else:
anos[ini] = 1
if fim in anos:
anos[fim] -= 1
else:
anos[fim] = -1
vivos = 0
maior = 0
anoMaior = 0
for ano, saldo in sorted(anos.items()):
vivos += saldo
if maior < vivos:
maior = vivos
anoMaior = ano
print(anoMaior, maior)
```
Yes
| 13,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Submitted Solution:
```
import sys, math, itertools, collections, copy
input = sys.stdin.readline
N = int(input())
events = []
for i in range(N):
begin, end = map(int, input().split())
end -= 1
events.append([begin, end])
events.sort(key = lambda x: (x[0], x[1]))
begin = events[0]
ans = 0
cur = 1
second = events[0][0]
for i in range(1, len(events)):
if max(begin[0], events[i][0]) <= min(begin[1], events[i][1]):
cur += 1
ans = max(ans, cur)
if ans == cur:
second = max(begin[0], events[i][0])
else:
ans = max(ans, cur)
cur = 1
begin = events[i]
print(second, ans)
```
No
| 13,496 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Submitted Solution:
```
from sys import stdout, stdin
input = stdin.readline
print = stdout.write
def main():
mp = lambda: map(int, input().strip().split())
it = int
n = it(input())
a = [list(mp()) for _ in [0] * n]
a.sort(key=lambda x: (x[0], -x[1]))
queue = [a[0]]
m = 0
c = 0
y = 0
for i in range(1, n):
if a[i][0] != a[i - 1][0]:
queue.append(a[i])
if a[i][0] >= queue[0][1]:
if c > m:
m = c
y = a[i - 1][0]
queue.pop(0)
c = 0
else:
c += 1
print(str(y) + ' ' + str(m + 1) + '\n')
main()
```
No
| 13,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Submitted Solution:
```
import sys
from random import *
from bisect import *
from heapq import *
#from collections import deque
pl=1
from math import gcd,sqrt,ceil
from copy import *
sys.setrecursionlimit(10**5)
if pl:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('outpt.txt','w')
def li():
return [int(xxx) for xxx in input().split()]
def fi():
return int(input())
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
t=1
while t>0:
t-=1
n=fi()
maxi=[0]*2
a=[]
for i in range(n):
l,r=mi()
a.append([l,0])
a.append([r-1,1])
a.sort()
c=0
for i in range(n):
if a[i][1]==0:
c+=1
if c>maxi[0]:
maxi=[c,a[i][0]]
else:
c-=1
print(*maxi[::-1])
```
No
| 13,498 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range (1, 10^9)! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer n (1 β€ n β€ 10^5) β the number of people.
Each of the following n lines contain two integers b and d (1 β€ b < d β€ 10^9) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, y β the year with a maximum number of people alive and k β the number of people alive in year y.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from b to d has been born at the beginning of b and died at the beginning of d, and therefore living for d - b years.
Submitted Solution:
```
n = int(input())
arr = []
ans = []
i = 0
j = 1
count = 0
for _ in range(n):
x = list(map(int,input().split()))
arr.append(x)
arr.sort()
for j in range(n):
if arr[i][1]>arr[j][0] and arr[i][0]<=arr[j][0]:
count += 1
ans.append(arr[j][0])
else:
i += 1
j += 1
print(max(ans),count)
```
No
| 13,499 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.