message stringlengths 2 433k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that <image>.
Input
The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], ..., a[n] (|a[i]| ≤ 109) — the elements of array a.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Examples
Input
5
1 2 3 0 3
Output
2
Input
4
0 1 -1 0
Output
1
Input
2
4 1
Output
0 | instruction | 0 | 12,013 | 12 | 24,026 |
Tags: binary search, brute force, data structures, dp, two pointers
Correct Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
s = sum(a)
if s % 3 != 0:
print(0)
else:
b = [a[0]]
s = s//3
for i in range(1,n):
b.append(b[i-1]+a[i])
if s==0:
m = b.count(0)
print((m-1) *(m-2) //2 if m>2 else 0)
else:
t = 0
c = [0]
for i in range(n-1,-1,-1):
if b[i] == s*2:
t+=1
elif b[i] == s:
c.append(t+c[-1])
t = 0
print(sum(c))
``` | output | 1 | 12,013 | 12 | 24,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that <image>.
Input
The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], ..., a[n] (|a[i]| ≤ 109) — the elements of array a.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Examples
Input
5
1 2 3 0 3
Output
2
Input
4
0 1 -1 0
Output
1
Input
2
4 1
Output
0 | instruction | 0 | 12,014 | 12 | 24,028 |
Tags: binary search, brute force, data structures, dp, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
s = sum(a)
m = 0
c = 0
ans = 0
if s % 3 == 0:
s //= 3
for i in range(n-1):
m += a[i]
if m == s * 2:
ans += c
if m == s:
c += 1
print(ans)
``` | output | 1 | 12,014 | 12 | 24,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that <image>.
Input
The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], ..., a[n] (|a[i]| ≤ 109) — the elements of array a.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Examples
Input
5
1 2 3 0 3
Output
2
Input
4
0 1 -1 0
Output
1
Input
2
4 1
Output
0 | instruction | 0 | 12,015 | 12 | 24,030 |
Tags: binary search, brute force, data structures, dp, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
ret, s = 0, sum(a)
if s % 3 == 0:
p1 = s // 3
p2 = s - p1
s = c = 0
for i in range(n - 1):
s += a[i]
if s == p2:
ret += c
if s == p1:
c += 1
print(ret)
``` | output | 1 | 12,015 | 12 | 24,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that <image>.
Input
The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], ..., a[n] (|a[i]| ≤ 109) — the elements of array a.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Examples
Input
5
1 2 3 0 3
Output
2
Input
4
0 1 -1 0
Output
1
Input
2
4 1
Output
0 | instruction | 0 | 12,016 | 12 | 24,032 |
Tags: binary search, brute force, data structures, dp, two pointers
Correct Solution:
```
def countways(a, n):
cnt = [0 for i in range(n)]
s = 0
s = sum(a)
if (s % 3 != 0):
return 0
s //= 3
ss = 0
for i in range(n - 1, -1, -1):
ss += a[i]
if (ss == s):
cnt[i] = 1
for i in range(n - 2, -1, -1):
cnt[i] += cnt[i + 1]
ans = 0
ss = 0
for i in range(0, n - 2):
ss += a[i]
if (ss == s):
ans += cnt[i + 2]
return ans
n=int(input())
a=list(map(int,input().split()))
print(countways(a,n))
``` | output | 1 | 12,016 | 12 | 24,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that <image>.
Input
The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], ..., a[n] (|a[i]| ≤ 109) — the elements of array a.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Examples
Input
5
1 2 3 0 3
Output
2
Input
4
0 1 -1 0
Output
1
Input
2
4 1
Output
0
Submitted Solution:
```
from bisect import bisect_left
def bini(a,x):
i=bisect_left(a,x)
return i;
a=int(input())
z=list(map(int,input().split()))
if(sum(z)%3!=0):
print(0)
else:
t=sum(z)//3
total=0
count=0
save=[]
save1=[]
ans=[]
for i in range(len(z)):
total=total+z[i]
if(total==t and i!=len(z)-1):
save.append(i)
if(total==2*t and i!=0 and i!=len(z)-1):
save1.append(i)
for i in range(len(save1)):
ans.append(bini(save,save1[i]))
print(sum(ans))
``` | instruction | 0 | 12,017 | 12 | 24,034 |
Yes | output | 1 | 12,017 | 12 | 24,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that <image>.
Input
The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], ..., a[n] (|a[i]| ≤ 109) — the elements of array a.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Examples
Input
5
1 2 3 0 3
Output
2
Input
4
0 1 -1 0
Output
1
Input
2
4 1
Output
0
Submitted Solution:
```
n = int(input())
x = input().split()
a = list(int(x[i]) for i in range(len(x)))
way = 0
suma = sum(a[i] for i in range(len(a)))
if (suma % 3 > 0) or (len(a) < 3):
print(0)
else:
sign = 1
target = suma // 3
if target == 0:
segment = 0
sumseg = 0
for i in range(len(a)):
sumseg += a[i]
if sumseg == 0:
segment += 1
way = (segment - 1) * (segment - 2) // 2
else:
sumseg = 0
list0 = []
dict0 = {}
dict1 = {}
i = 1
while i < len(a) - 1:
if a[i] == 0:
i0 = i - 1
while a[i] == 0:
i += 1
dict0[i0] = dict1[i] = i - i0
else:
i += 1
i = 0
while i < len(a) - 2:
sumseg += a[i]
if sumseg == target:
list0.insert(0, i)
if i in dict0.keys():
i += dict0[i]
continue
i += 1
sumseg = 0
list1 = []
i = len(a) - 1
while i > 1:
sumseg += a[i]
if sumseg == target:
list1.insert(0, i)
if i in dict1.keys():
i -= dict1[i]
continue
i -= 1
flag = 0
sumlist1 = 0
sumlist0 = 0
for i in range(len(list1)):
sumlist1 += 1 if list1[i] not in dict1.keys() else dict1[list1[i]]
for i in range(len(list0)):
sumlist0 += 1 if list0[i] not in dict0.keys() else dict0[list0[i]]
minusadd = 0
for i in range(len(list0)):
if list0[i] < list1[0]:
break
else:
for j in range(len(list1)):
if list0[i] < list1[j]:
for k in range(j):
minusadd += (1 if list1[j] not in dict1.keys() else dict1[list1[j]]) * (1 if list0[i] not in dict0.keys() else dict0[list0[i]])
break
way = sumlist1 * sumlist0 - minusadd
print(way)
``` | instruction | 0 | 12,019 | 12 | 24,038 |
Yes | output | 1 | 12,019 | 12 | 24,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that <image>.
Input
The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], ..., a[n] (|a[i]| ≤ 109) — the elements of array a.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Examples
Input
5
1 2 3 0 3
Output
2
Input
4
0 1 -1 0
Output
1
Input
2
4 1
Output
0
Submitted Solution:
```
n = int(input())
li = list(map(int, input().split()))
if(sum(li)%3 != 0):
print(0)
quit()
t = sum(li)//3
for i in range(1, n):
li[i] += li[i-1]
cnt = [0]*n
for i in range(n-2, -1, -1):
cnt[i] = cnt[i+1]
if li[i] == 2*t:
cnt[i] += 1
ans = 0
for i in range(0, n-1):
if li[i] != t:
continue
ans += cnt[i+1]
print(ans)
``` | instruction | 0 | 12,020 | 12 | 24,040 |
Yes | output | 1 | 12,020 | 12 | 24,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that <image>.
Input
The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], ..., a[n] (|a[i]| ≤ 109) — the elements of array a.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Examples
Input
5
1 2 3 0 3
Output
2
Input
4
0 1 -1 0
Output
1
Input
2
4 1
Output
0
Submitted Solution:
```
n, a = int(input()), list(map(int, input().split()))
sums, s = [[0, 0]] + [0] * n, sum(a)
for i in range(n):
sums[i + 1] = [sums[i][0] + a[i], sums[i][1] + int(i and (sums[i][0] + a[i]) * 3 == s * 2)]
print(0 if s % 3 else sum(sums[-1][1] - sums[i][1] for i in range(2, n) if sums[i][0] * 3 == s))
``` | instruction | 0 | 12,022 | 12 | 24,044 |
No | output | 1 | 12,022 | 12 | 24,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that <image>.
Input
The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], ..., a[n] (|a[i]| ≤ 109) — the elements of array a.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Examples
Input
5
1 2 3 0 3
Output
2
Input
4
0 1 -1 0
Output
1
Input
2
4 1
Output
0
Submitted Solution:
```
n=int(input())
ch=str(input())
LLL=ch.split()
L=list()
L1=list()
total=0
for c in LLL:
L.append(int(c))
total+=int(c)
L1.append(total)
if (total%3!=0):
print(0)
elif (total!=0):
c1=0
c2=0
res=0
for i in range(0,n-1):
if L1[i]==total//3:
c1+=1
if (L1[i]==(total//3)*2) and (i!=0) and (i!=n-1):
res+=c1
print(res)
else:
print(1)
``` | instruction | 0 | 12,023 | 12 | 24,046 |
No | output | 1 | 12,023 | 12 | 24,047 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>. | instruction | 0 | 12,218 | 12 | 24,436 |
Tags: constructive algorithms
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=1000000007
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(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 stdin.read().split()
range = xrange # not for python 3.0+
# main code
n,x=in_arr()
if n==2 and x==0:
pr('NO')
exit()
pr('YES\n')
ans=0
if n==1:
pr_num(x)
elif n==2:
pr_arr([x,0])
else:
pw=2**17
for i in range(1,n-2):
pr(str(i)+' ')
ans^=i
if ans==x:
pr_arr([pw,pw*2,pw^(pw*2)])
else:
pr_arr([0,pw,pw^x^ans])
``` | output | 1 | 12,218 | 12 | 24,437 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2 | instruction | 0 | 12,227 | 12 | 24,454 |
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
a = [int(s) for s in input().split(' ')]
e = 0
for i in range(1, n - 1):
if a[i] > max(a[i - 1], a[i + 1]) or a[i] < min(a[i - 1], a[i + 1]):
e += 1
print(e)
``` | output | 1 | 12,227 | 12 | 24,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2 | instruction | 0 | 12,228 | 12 | 24,456 |
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
x = list(map(int,input().split()))
r = 0
if n<=2:
print(0)
else:
for i in range(1,n-1):
if x[i]>x[i-1] and x[i]>x[i+1]:
r+=1
if x[i]<x[i-1] and x[i]<x[i+1]:
r+=1
print(r)
``` | output | 1 | 12,228 | 12 | 24,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2 | instruction | 0 | 12,229 | 12 | 24,458 |
Tags: brute force, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
from math import*
############ ---- Input Functions ---- ############
def inp():
return(int(input().strip()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
n=inp()
li=inlt()
cnt=0
for i in range(1,n-1):
if li[i]< li[i-1] and li[i]<li[i+1]:
cnt+=1
elif li[i]> li[i-1] and li[i]>li[i+1]:
cnt+=1
print(cnt)
``` | output | 1 | 12,229 | 12 | 24,459 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2 | instruction | 0 | 12,230 | 12 | 24,460 |
Tags: brute force, implementation
Correct Solution:
```
number = int(input())
num = 0
lst = [int(i) for i in input().split()][:number]
for i in range(len(lst)):
if not(i == 0 or i == len(lst)-1) and (lst[i] > lst[i-1] and lst[i]>lst[i+1] or lst[i]<lst[i-1] and lst[i]<lst[i+1]):
num+=1
print(num)
``` | output | 1 | 12,230 | 12 | 24,461 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2 | instruction | 0 | 12,231 | 12 | 24,462 |
Tags: brute force, implementation
Correct Solution:
```
def find_extrema(n, lst):
maxima = list()
minima = list()
if n >= 3:
for i in range(1, n - 1):
if lst[i] > lst[i - 1] and lst[i] > lst[i + 1]:
maxima.append(lst[i])
elif lst[i] < lst[i - 1] and lst[i] < lst[i + 1]:
minima.append(lst[i])
return len(minima) + len(maxima)
return 0
m = int(input())
a = [int(j) for j in input().split()]
print(find_extrema(m, a))
``` | output | 1 | 12,231 | 12 | 24,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2 | instruction | 0 | 12,232 | 12 | 24,464 |
Tags: brute force, implementation
Correct Solution:
```
def main():
input()
nums = list(map(int, input().split()))
lastn = nums.pop(0)
mc = 0
while len(nums) > 1:
buff = nums.pop(0)
if (lastn < buff and buff > nums[0]) or (lastn > buff and buff < nums[0]):
mc += 1
lastn = buff
return mc
if __name__ == "__main__":
print(main())
``` | output | 1 | 12,232 | 12 | 24,465 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2 | instruction | 0 | 12,233 | 12 | 24,466 |
Tags: brute force, implementation
Correct Solution:
```
x=input()
x=int(x)
y=input()
y=y.split()
min1=0
max1=0
for i in range(1,x-1):
if int(y[i]) < int(y[i-1]) and int(y[i])<int(y[i+1]):
min1=min1+1
if int(y[i])>int(y[i+1]) and int(y[i])>int(y[i-1]):
max1=max1+1
res=min1+max1
if x==2 or x==1:
print(0)
else:
print(res)
``` | output | 1 | 12,233 | 12 | 24,467 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2 | instruction | 0 | 12,234 | 12 | 24,468 |
Tags: brute force, implementation
Correct Solution:
```
# bsdk idhar kya dekhne ko aaya hai, khud kr!!!
# from math import *
# from itertools import *
# import random
n = int(input())
arr = list(map(int, input().split()))
count_ = 0
for i in range(1, n-1):
if (arr[i] < arr[i-1] and arr[i] < arr[i+1]) or (arr[i] > arr[i-1] and arr[i] > arr[i+1]):
count_ += 1
else:
continue
print(count_)
``` | output | 1 | 12,234 | 12 | 24,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
m=int(input())
a=list(map(int,input().split()))
c=0
for i in range(1,len(a)-1):
if (a[i]>a[i+1] and a[i]>a[i-1]) or (a[i]<a[i-1] and a[i]<a[i+1]) :
c+=1
else :
pass
print(c)
``` | instruction | 0 | 12,235 | 12 | 24,470 |
Yes | output | 1 | 12,235 | 12 | 24,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
n = int(input())
nums = list(map(int, input().split()))
res = 0
for i in range(1, n - 1):
f = -1
if(nums[i] > nums[i - 1]):f = 1
if(nums[i] < nums[i - 1]):f = 2
s = -2
if(nums[i] > nums[i + 1]):s = 1
if(nums[i] < nums[i + 1]):s = 2
if(s == f):
res += 1
print(res)
``` | instruction | 0 | 12,236 | 12 | 24,472 |
Yes | output | 1 | 12,236 | 12 | 24,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
n=int(input())
line=list(input().split(' '))
line=[int(i) for i in line]
s=0
for i in range(1,n-1):
if line[i-1]<line[i]>line[i+1]:
s+=1
elif line[i-1]>line[i]<line[i+1]:
s+=1
print(s)
``` | instruction | 0 | 12,237 | 12 | 24,474 |
Yes | output | 1 | 12,237 | 12 | 24,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
count=0
for i in range(1,n-1):
if l[i]>l[i-1] and l[i]>l[i+1]:
count+=1
elif l[i]<l[i-1] and l[i]<l[i+1]:
count+=1
print(count)
``` | instruction | 0 | 12,238 | 12 | 24,476 |
Yes | output | 1 | 12,238 | 12 | 24,477 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
n=int(input())
f=[int(i) for i in input().split()]
gh=0
b=0
t=0
for i in range(1,n-1):
b=f[i+1]
gh=f[i-1]
if i!=0 and i!=n-1 and (gh>f[i-1]<b or gh<f[i]>b):
t=t+1
print(t)
``` | instruction | 0 | 12,239 | 12 | 24,478 |
No | output | 1 | 12,239 | 12 | 24,479 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
input()
data = [int(x) for x in input().split()]
al=0
for x in range(len(data)):
if x==0 or x==len(data)-1:
continue
if (data[x]<data[x+1] and data[x]<data[x-1]) or (data[x]>data[x+1] and data[x]>data[x-1]):
al+=x-1
print(al)
``` | instruction | 0 | 12,240 | 12 | 24,480 |
No | output | 1 | 12,240 | 12 | 24,481 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
useless = input()
numbers = input().split()
def local_extremums(array):
numbers = list(map(int,array))
i = 1
result = 0
while i < len(numbers) - 1:
if (numbers[i] < numbers[i-1] and numbers[i] < numbers[i + 1]) or (numbers[i] > numbers[i-1] and numbers[i] > numbers[i + 1]):
result += 1
i += 1
return result
print(local_extremums)
``` | instruction | 0 | 12,241 | 12 | 24,482 |
No | output | 1 | 12,241 | 12 | 24,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
input()
a = b = s = 0
for c in map(int, input().split()):
if a and a > b < c or a < b > c: s += 1
a, b = b, c
print(s)
``` | instruction | 0 | 12,242 | 12 | 24,484 |
No | output | 1 | 12,242 | 12 | 24,485 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1 | instruction | 0 | 12,349 | 12 | 24,698 |
"Correct Solution:
```
N, A, B = map(int, input().split())
if N < A + B - 1 :
print(-1)
else :
if A > B :
mode = 0
A -= 1
else :
mode = 1
B -= 1
ret = []
u = N - A + 1
l = B
while len(ret) < N :
if (mode and A == 0) or (not mode and B == 0) :
ret = [-1]
break
if mode :
for i in range(max(u, l - B + 1), u + A) :
ret.append(i)
A -= 1
u -= A
mode ^= 1
else :
for i in range(min(l, u + A - 1), l - B, -1) :
ret.append(i)
B -= 1
l += B
mode ^= 1
print(*ret)
``` | output | 1 | 12,349 | 12 | 24,699 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1 | instruction | 0 | 12,350 | 12 | 24,700 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
"""
n-増大列、m-減少列まで -> 長さ nm 以下であることを示す。帰納法。
(n+1)m + 1 項があるとする。(n+2増大 or m+1減少)の存在をいう。
A：左 nm 項 B：右 m+1 項
Aに1項加えると、(n+1,m+1)のどちらかができる。(n+1)-増大ができるとしてよい。
Bの各項 b に対して、bで終わる(n+1)-増大列が存在する。
Bの中に2-増大列があれば(n+2)増大列ができる。そうでなければBが(m+1)-減少列なのでよい
"""
N,A,B = map(int,input().split())
if A+B-1 > N:
print(-1)
exit()
if A*B < N:
print(-1)
exit()
# 減少列をA個並べる
if B == 1:
size = [1] * A
else:
q,r = divmod(N-A,B-1)
if q < A:
size = [B] * q + [1+r] + [1] * (A-q-1)
else:
size = [B] * A
answer = []
start = 1
for s in size:
end = start + s
answer += list(range(end-1, start-1, -1))
start = end
print(*answer)
``` | output | 1 | 12,350 | 12 | 24,701 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1 | instruction | 0 | 12,351 | 12 | 24,702 |
"Correct Solution:
```
N, A, B = map(int, input().split())
if A * B < N:
print(-1)
elif A + B - 1 > N:
print(-1)
else:
P = [0] * N
b = (N - A) // (B - 1) if B > 1 else 0
r = (N - A) % (B - 1) + 1 if B > 1 else 1
i = 1
pos = 0
while i <= N:
if b:
for j in range(B):
P[pos + B - j - 1] = i
i += 1
pos += B
b -= 1
elif r:
for j in range(r):
P[pos + r - j - 1] = i
i += 1
pos += r
r = 0
else:
P[pos] = i
i += 1
pos += 1
for p in P:
print(p, end=' ')
``` | output | 1 | 12,351 | 12 | 24,703 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1 | instruction | 0 | 12,352 | 12 | 24,704 |
"Correct Solution:
```
N, A, B = map(int, input().split())
ans = list(range(A))
m = 0
rest = N - A
B -= 1
if rest < B:
print(-1)
exit()
if rest / A > B:
print(-1)
exit()
while rest > B:
s = min(A, rest - B + 1)
ans += list(range(m - s, m))
m -= s
rest -= s
B -= 1
ans += list(range(m - B, m))[::-1]
m -= B
print(" ".join([str(x - m + 1) for x in ans]))
``` | output | 1 | 12,352 | 12 | 24,705 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1 | instruction | 0 | 12,353 | 12 | 24,706 |
"Correct Solution:
```
import sys
def input(): return sys.stdin.readline().strip()
def mapint(): return map(int, input().split())
sys.setrecursionlimit(10**9)
N, A, B = mapint()
from collections import deque
if A+B-1>N or N>A*B:
print(-1)
else:
blocks = []
blocks.append(' '.join(map(str, range(B, 0, -1))))
now = B+1
rest = N-A+2
while now:
if now==rest:
break
if now+B>rest:
blocks.append(' '.join(map(str, range(rest, now-1, -1))))
rest += 1
break
blocks.append(' '.join(map(str, range(now+B-1, now-1, -1))))
now = now+B
rest += 1
blocks.append(' '.join(map(str, range(rest, N+1))))
print(' '.join(blocks))
``` | output | 1 | 12,353 | 12 | 24,707 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1 | instruction | 0 | 12,354 | 12 | 24,708 |
"Correct Solution:
```
n,a,b=map(int,input().split())
if a+b-1>n:exit(print(-1))
if a*b<n:exit(print(-1))
ans=[]
c=0
nn=n
for i in range(a):
ans.append([])
t=0--nn//(a-i)
if i==0:
t=b
nn-=min(t,b)
for j in range(min(b,t)):
ans[-1].append((i+1)*b-j)
c+=1
if c==n:break
if c==n:break
anss=[]
for i in ans:anss+=i
def position_zip(a,flag):
j=1
d={}
for i in sorted(a):
if i in d:continue
d[i]=j
j+=1
if flag==1:return d
return [d[i] for i in a]
print(*position_zip(anss,0))
``` | output | 1 | 12,354 | 12 | 24,709 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1 | instruction | 0 | 12,355 | 12 | 24,710 |
"Correct Solution:
```
r,p=range,print
n,a,b=map(int,input().split())
if a+b>n+1or a*b<n:exit(p(-1))
l=[[]for i in r(b-1)]+[list(r(1,a+1))]
for i in r(a,n):l[-2-(i-a)%(b-1)]+=[i+1]
for i in l:p(*i,end=" ")
``` | output | 1 | 12,355 | 12 | 24,711 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1 | instruction | 0 | 12,356 | 12 | 24,712 |
"Correct Solution:
```
# でつoO(YOU PLAY WITH THE CARDS YOU'RE DEALT..)
import sys
def main(N, A, B):
if A + B - 1 > N or A * B < N:
print(-1)
return
P = []
r = A * B - N
for b in range(B):
for a in range(A):
if b >= 1 and a >= 1 and r > 0:
if r >= A - 1:
r -= A - 1
break
r -= 1
continue
P.append((B - b) * A + a + 1)
s = sorted([(p, i) for i, p in enumerate(P)], key=lambda x: x[0])
s = sorted([(i, j) for j, (p, i) in enumerate(s)], key=lambda x: x[0])
s = [j + 1 for i, j in s]
print(*s)
if __name__ == '__main__':
input = sys.stdin.readline
N, A, B = map(int, input().split())
main(N, A, B)
``` | output | 1 | 12,356 | 12 | 24,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1
Submitted Solution:
```
N, A, B = map(int, input().split())
if A*B < N or A+B-1 > N:
print(-1)
else:
array = [i for i in reversed(range(1, N+1))]
if A > 1:
f = array[:B]
r = array[B:]
L = [f]
span = len(r)//(A-1)
rem = len(r)%(A-1)
i = 0
for _ in range(A-1):
if rem > 0:
L.append(r[i:i+span+1])
rem -= 1
i += span + 1
else:
L.append(r[i:i+span])
i += span
array = []
for l in reversed(L):
array += l
print(' '.join([str(a) for a in array]))
``` | instruction | 0 | 12,358 | 12 | 24,716 |
Yes | output | 1 | 12,358 | 12 | 24,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1
Submitted Solution:
```
#!/usr/bin/env python
import numpy as np
def compress(array_like, num):
tmp = sorted(array_like)
m = {}
for i, v in enumerate(tmp):
if i == num:
break
m[v] = i + 1
ret = []
for a in array_like:
if a in m:
ret.append(m[a])
return ret
def main():
N, A, B = map(int, input().split())
# N, A, B = 300000, 1000, 1000
# N, A, B = 10, 4, 5
if A + B - 1 > N:
return -1
if A * B < N:
return -1
ans = np.arange(
1, A * B + 1, dtype="int32").reshape(A, B).T[::-1, :].reshape(A * B)
ans = list(ans)
j = 0
i = 0
for _ in range(A * B - N):
ans[j * A + i] = 1000000
i += 1
if i == A - j - 1:
i += 1
if i >= A:
i = 0
j += 1
if i == A - j - 1:
i += 1
# print(ans)
# ans = np.array(ans).reshape(A * B)
# ans = ans[ans != -1]
# return
return ' '.join(map(str, compress(ans, N)))
if __name__ == '__main__':
print(main())
``` | instruction | 0 | 12,361 | 12 | 24,722 |
No | output | 1 | 12,361 | 12 | 24,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:
* The length of its longest increasing subsequence is A.
* The length of its longest decreasing subsequence is B.
If it exists, construct one such sequence.
Constraints
* 1 \leq N,A,B \leq 3\times 10^5
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there are no sequences that satisfy the conditions, print `-1`.
Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.
Examples
Input
5 3 2
Output
2 4 1 5 3
Input
7 7 1
Output
1 2 3 4 5 6 7
Input
300000 300000 300000
Output
-1
Submitted Solution:
```
#!/usr/bin/env python
import numpy as np
def compress(array_like):
tmp = sorted(array_like)
m = {}
for i, v in enumerate(tmp):
m[v] = i + 1
ret = []
for a in array_like:
ret.append(m[a])
return ret
def main():
N, A, B = map(int, input().split())
# # N, A, B = 300000, 4, 100000
# N, A, B = 10, 4, 5
if A + B - 1 > N:
return -1
if A * B < N:
return -1
ans = np.arange(1, A * B + 1, dtype="int32").reshape(A, B).T[::-1, :]
j = 0
i = 0
for _ in range(A * B - N):
ans[j, i] = -1
i += 1
if i == A - j - 1:
i += 1
if i >= A:
i = 0
j += 1
if i == A - j - 1:
i += 1
# print(ans)
ans = ans.reshape(A * B)
ans = ans[ans != -1]
# return
return ' '.join(map(str, compress(list(ans))))
if __name__ == '__main__':
print(main())
``` | instruction | 0 | 12,364 | 12 | 24,728 |
No | output | 1 | 12,364 | 12 | 24,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10 | instruction | 0 | 12,698 | 12 | 25,396 |
Tags: data structures, hashing, math
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
d = {}
for i in range(n):
x = a[i]-i
if x in d:
d[x] += 1
else:
d[x] = 1
count = 0
for i in d:
count += ((d[i]*(d[i]-1))//2)
print(count)
``` | output | 1 | 12,698 | 12 | 25,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10 | instruction | 0 | 12,699 | 12 | 25,398 |
Tags: data structures, hashing, math
Correct Solution:
```
cases = int(input())
for _ in range(cases):
n = int(input())
ints = [int(n) for n in input().split(" ")]
counts = {}
for i in range(n):
if ints[i]-i not in counts.keys():
counts[ints[i]-i] = 0
counts[ints[i]-i] += 1
count = 0
for c in counts.values():
count += (c*(c-1))//2
print(count)
``` | output | 1 | 12,699 | 12 | 25,399 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10 | instruction | 0 | 12,700 | 12 | 25,400 |
Tags: data structures, hashing, math
Correct Solution:
```
def main():
n=int(input())
a=list(map(int,input().split()))
dictt={}
for i in range(n):
a[i]-=i
# print(a)
for x in a:
if x in dictt:
dictt[x]=dictt[x]+1
else:
dictt[x]=1
ans=0
for x in dictt:
cnt=dictt[x]
ans+=cnt*(cnt-1)/2
# print("ans : "+str(int(ans)))
print(int(ans))
t=int(input())
while t>0:
main()
t-=1
``` | output | 1 | 12,700 | 12 | 25,401 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10 | instruction | 0 | 12,701 | 12 | 25,402 |
Tags: data structures, hashing, math
Correct Solution:
```
def solve(l,n):
t={}
for i in range(n):
if (l[i]-i) in t:
t[l[i]-i]+=1
else:
t[l[i]-i]=1
ans=0
for i in t.values():
ans=ans+(i*(i-1))//2
return ans
for _ in range(int(input())):
n = int(input())
l = list(map(int,input().split( )))
print(solve(l,n),end=' ')
``` | output | 1 | 12,701 | 12 | 25,403 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10 | instruction | 0 | 12,702 | 12 | 25,404 |
Tags: data structures, hashing, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int,input().split()))
s = 0
k = {}
for i in range(n):
k[a[i]-i] = k.get(a[i]-i,0) + 1
for u in k.keys():
s += (k[u]*(k[u]-1))//2
print(s)
print()
``` | output | 1 | 12,702 | 12 | 25,405 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10 | instruction | 0 | 12,703 | 12 | 25,406 |
Tags: data structures, hashing, math
Correct Solution:
```
from collections import defaultdict
t=int(input())
for i in range(t):
n=int(input())
ar=list(map(int,input().split()))
d=defaultdict(int)
ans=0
for i in range(n):
ans+=d[ar[i]-i]
d[ar[i]-i]+=1
print(ans)
``` | output | 1 | 12,703 | 12 | 25,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10 | instruction | 0 | 12,704 | 12 | 25,408 |
Tags: data structures, hashing, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().strip().split()))
d={}
for i in range(n):
l[i]=l[i]-i
d[l[i]]=d.get(l[i],0)+1
c=0
for i in d:
if(d[i]>=2):
c=c+(d[i]-1)*d[i]//2
print(c)
``` | output | 1 | 12,704 | 12 | 25,409 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10 | instruction | 0 | 12,705 | 12 | 25,410 |
Tags: data structures, hashing, math
Correct Solution:
```
#dt = {} for i in x: dt[i] = dt.get(i,0)+1
import sys;input = sys.stdin.readline
#import io,os; input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline #for pypy
inp,ip = lambda :int(input()),lambda :[int(w) for w in input().split()]
for _ in range(1,inp()+1):
n = inp()
x = ip()
y = [x[i]-i for i in range(n)]
dt = {}
for i in y: dt[i] = dt.get(i,0)+1
ans = 0
for i in dt:
ans += dt[i]*(dt[i]-1)//2
print(ans)
``` | output | 1 | 12,705 | 12 | 25,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
# def countPairs(arr, n):
# # To store the frequencies
# # of (arr[i] - i)
# map = dict()
# for i in range(n):
# map[arr[i] - i] = map.get(arr[i] - i, 0) + 1
# # To store the required count
# res = 0
# for x in map:
# cnt = map[x]
# # If cnt is the number of elements
# # whose differecne with their index
# # is same then ((cnt * (cnt - 1)) / 2)
# # such pairs are possible
# res += ((cnt * (cnt - 1)) // 2)
# return res
# # Driver code
# arr = [1, 5, 6, 7, 9]
# n = len(arr)
# print(countPairs(arr, n))
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
d={}
for i in range(n):
d[arr[i]-i]=d.get(arr[i]-i,0)+1
ans=0
for i in d:
n_=d[i]
ans+=((n_*(n_-1))//2)
print(ans)
# d={}
``` | instruction | 0 | 12,708 | 12 | 25,416 |
Yes | output | 1 | 12,708 | 12 | 25,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.
The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).
The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).
Output
Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.
Examples
Input
5 4
1 3 5 7 9
6 4 2 8
Output
3 2 1 4
Input
5 5
1 2 1 2 5
3 1 4 1 5
Output
4 2 4 2 5 | instruction | 0 | 12,907 | 12 | 25,814 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
import bisect
if __name__ == "__main__":
n , m = [int(x) for x in input().strip().split()]
a = sorted([int(x) for x in input().strip().split()])
b = [int(x) for x in input().strip().split()]
res = []
for bi in b:
pos = bisect.bisect_left( a , bi)
if pos < n and a[pos] == bi:
res.append( bisect.bisect_right( a , bi ) )
else:
res.append( pos )
print( " ".join([str(x) for x in res]) )
``` | output | 1 | 12,907 | 12 | 25,815 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.
The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).
The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).
Output
Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.
Examples
Input
5 4
1 3 5 7 9
6 4 2 8
Output
3 2 1 4
Input
5 5
1 2 1 2 5
3 1 4 1 5
Output
4 2 4 2 5 | instruction | 0 | 12,908 | 12 | 25,816 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
#editorial approch
from bisect import bisect_right
x, y = map(int, input().split())
*li1, = map(int, input().split())
*li2, = map(int, input().split())
li1.sort()
li3 = []
for i in li2:
li3.append(bisect_right(li1,i))
print(*li3)
``` | output | 1 | 12,908 | 12 | 25,817 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.
The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).
The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).
Output
Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.
Examples
Input
5 4
1 3 5 7 9
6 4 2 8
Output
3 2 1 4
Input
5 5
1 2 1 2 5
3 1 4 1 5
Output
4 2 4 2 5 | instruction | 0 | 12,909 | 12 | 25,818 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
# @oj: codeforces
# @id: hitwanyang
# @email: 296866643@qq.com
# @date: 2020-11-25 17:08
# @url:https://codeforc.es/contest/600/problem/B
import sys,os
from io import BytesIO, IOBase
import collections,itertools,bisect,heapq,math,string
from decimal import *
# region fastio
BUFSIZE = 8192
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")
# ------------------------------
## 注意嵌套括号!!!!!!
## 先有思路,再写代码,别着急!!!
## 先有朴素解法,不要有思维定式,试着换思路解决
## 精度 print("%.10f" % ans)
## sqrt:int(math.sqrt(n))+1
## 字符串拼接不要用+操作，会超时
## 二进制转换:bin(1)[2:].rjust(32,'0')
## array copy:cur=array[::]
## oeis:example 1, 3, _, 1260, _, _, _, _, _, 12164510040883200
def main():
n,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
a.sort()
ans=[]
for i in range(m):
cnt=bisect.bisect_right(a,b[i])
ans.append(cnt)
# print (a,b)
print (" ".join([str(x) for x in ans]))
if __name__ == "__main__":
main()
``` | output | 1 | 12,909 | 12 | 25,819 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.
The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).
The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).
Output
Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.
Examples
Input
5 4
1 3 5 7 9
6 4 2 8
Output
3 2 1 4
Input
5 5
1 2 1 2 5
3 1 4 1 5
Output
4 2 4 2 5 | instruction | 0 | 12,911 | 12 | 25,822 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
import bisect
def less_than_or_equal(x, y):
x.sort()
queries = {}
res = []
for num in y:
if num in queries:
res.append(queries[num])
else:
ind = bisect.bisect(x, num)
res.append(ind)
queries[num] = ind
#print(res)
return ' '.join(str(x) for x in res)
_ = input()
x = [int(x) for x in input().split()]
y = [int(y) for y in input().split()]
print(less_than_or_equal(x, y))
``` | output | 1 | 12,911 | 12 | 25,823 |
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