message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.
The person numbered i says:
* "In front of me, exactly A_i people are wearing hats with the same color as mine."
Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.
Since the count can be enormous, compute it modulo 1000000007.
Constraints
* 1 \leq N \leq 100000
* 0 \leq A_i \leq N-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print the number of possible combinations of colors of the N people's hats, modulo 1000000007.
Examples
Input
6
0 1 2 3 4 5
Output
3
Input
3
0 0 0
Output
6
Input
54
0 0 1 0 1 2 1 2 3 2 3 3 4 4 5 4 6 5 7 8 5 6 6 7 7 8 8 9 9 10 10 11 9 12 10 13 14 11 11 12 12 13 13 14 14 15 15 15 16 16 16 17 17 17
Output
115295190
Submitted Solution:
```
N = int(input())
*A, = map(int,input().split())
mod = 10**9+7
ans = 3
if A[0]!=0: ans = 0
dp = [1,0,0]
for a in A[1:]:
ans *= dp.count(a)
ans %= mod
if ans == 0: break
dp[dp.index(a)] += 1
print(ans)
``` | instruction | 0 | 46,869 | 7 | 93,738 |
Yes | output | 1 | 46,869 | 7 | 93,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.
The person numbered i says:
* "In front of me, exactly A_i people are wearing hats with the same color as mine."
Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.
Since the count can be enormous, compute it modulo 1000000007.
Constraints
* 1 \leq N \leq 100000
* 0 \leq A_i \leq N-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print the number of possible combinations of colors of the N people's hats, modulo 1000000007.
Examples
Input
6
0 1 2 3 4 5
Output
3
Input
3
0 0 0
Output
6
Input
54
0 0 1 0 1 2 1 2 3 2 3 3 4 4 5 4 6 5 7 8 5 6 6 7 7 8 8 9 9 10 10 11 9 12 10 13 14 11 11 12 12 13 13 14 14 15 15 15 16 16 16 17 17 17
Output
115295190
Submitted Solution:
```
N = int(input())
A = list(map(int, input().split()))
kMod = 10**9+7
mem = [0, 0, 0]
cnt = 1
for a in A:
if a not in mem:
cnt = 0
break
cnt *= mem.count(a)
cnt %= kMod
mem[mem.index(a)] = a+1
print(cnt)
``` | instruction | 0 | 46,870 | 7 | 93,740 |
Yes | output | 1 | 46,870 | 7 | 93,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.
The person numbered i says:
* "In front of me, exactly A_i people are wearing hats with the same color as mine."
Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.
Since the count can be enormous, compute it modulo 1000000007.
Constraints
* 1 \leq N \leq 100000
* 0 \leq A_i \leq N-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print the number of possible combinations of colors of the N people's hats, modulo 1000000007.
Examples
Input
6
0 1 2 3 4 5
Output
3
Input
3
0 0 0
Output
6
Input
54
0 0 1 0 1 2 1 2 3 2 3 3 4 4 5 4 6 5 7 8 5 6 6 7 7 8 8 9 9 10 10 11 9 12 10 13 14 11 11 12 12 13 13 14 14 15 15 15 16 16 16 17 17 17
Output
115295190
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
l=[3]+[0]*(10**5)
m=10**9+7
x=1
for i in a:
if l[i]:
x=(x*l[i])%m
l[i]-=1
l[i+1]+=1
else:
x=0
break
print(x)
``` | instruction | 0 | 46,871 | 7 | 93,742 |
Yes | output | 1 | 46,871 | 7 | 93,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.
The person numbered i says:
* "In front of me, exactly A_i people are wearing hats with the same color as mine."
Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.
Since the count can be enormous, compute it modulo 1000000007.
Constraints
* 1 \leq N \leq 100000
* 0 \leq A_i \leq N-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print the number of possible combinations of colors of the N people's hats, modulo 1000000007.
Examples
Input
6
0 1 2 3 4 5
Output
3
Input
3
0 0 0
Output
6
Input
54
0 0 1 0 1 2 1 2 3 2 3 3 4 4 5 4 6 5 7 8 5 6 6 7 7 8 8 9 9 10 10 11 9 12 10 13 14 11 11 12 12 13 13 14 14 15 15 15 16 16 16 17 17 17
Output
115295190
Submitted Solution:
```
p = 10**9+7
N = int(input())
A = list(map(int,input().split()))
C={i:0 for i in range(N)}
C[0]=1
ans = 3
for i in range(1,N):
a = A[i]
if a not in C:
C[a]=0
C[a] += 1
if C[a]==1:
ans = (ans*C[a-1])%p
elif a==0 and C[a]==2:
ans = ans*2
elif a>0 and C[a]==2:
ans = (ans*(C[a-1]-1))%p
elif C[a]>3:
ans = 0
print(max(ans,0))
``` | instruction | 0 | 46,872 | 7 | 93,744 |
No | output | 1 | 46,872 | 7 | 93,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.
The person numbered i says:
* "In front of me, exactly A_i people are wearing hats with the same color as mine."
Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.
Since the count can be enormous, compute it modulo 1000000007.
Constraints
* 1 \leq N \leq 100000
* 0 \leq A_i \leq N-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print the number of possible combinations of colors of the N people's hats, modulo 1000000007.
Examples
Input
6
0 1 2 3 4 5
Output
3
Input
3
0 0 0
Output
6
Input
54
0 0 1 0 1 2 1 2 3 2 3 3 4 4 5 4 6 5 7 8 5 6 6 7 7 8 8 9 9 10 10 11 9 12 10 13 14 11 11 12 12 13 13 14 14 15 15 15 16 16 16 17 17 17
Output
115295190
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
x = 0
y = 0
z = 0
ans = 1
for i in range(n):
if a[i] == x:
if x == y == z:
ans *= 3
elif x == y != z:
ans *= 2
x += 1
elif a[i] == y:
if y == z:
ans *= 2
y += 1
else:
z += 1
ans %= 1000000007
print(ans)
``` | instruction | 0 | 46,873 | 7 | 93,746 |
No | output | 1 | 46,873 | 7 | 93,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.
The person numbered i says:
* "In front of me, exactly A_i people are wearing hats with the same color as mine."
Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.
Since the count can be enormous, compute it modulo 1000000007.
Constraints
* 1 \leq N \leq 100000
* 0 \leq A_i \leq N-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print the number of possible combinations of colors of the N people's hats, modulo 1000000007.
Examples
Input
6
0 1 2 3 4 5
Output
3
Input
3
0 0 0
Output
6
Input
54
0 0 1 0 1 2 1 2 3 2 3 3 4 4 5 4 6 5 7 8 5 6 6 7 7 8 8 9 9 10 10 11 9 12 10 13 14 11 11 12 12 13 13 14 14 15 15 15 16 16 16 17 17 17
Output
115295190
Submitted Solution:
```
n = int(input())
lst = list(map(int,input().split()))
mod = 10**9 + 7
col_lst = [0 , 0 , 0]
ans = 1
for i in range(n):
tmp = lst[i]
val = col_lst.count(tmp)
ans = (ans * val) % mod
col_lst[col_lst.index(tmp)] += 1
print(ans % mod)
``` | instruction | 0 | 46,874 | 7 | 93,748 |
No | output | 1 | 46,874 | 7 | 93,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.
The person numbered i says:
* "In front of me, exactly A_i people are wearing hats with the same color as mine."
Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.
Since the count can be enormous, compute it modulo 1000000007.
Constraints
* 1 \leq N \leq 100000
* 0 \leq A_i \leq N-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print the number of possible combinations of colors of the N people's hats, modulo 1000000007.
Examples
Input
6
0 1 2 3 4 5
Output
3
Input
3
0 0 0
Output
6
Input
54
0 0 1 0 1 2 1 2 3 2 3 3 4 4 5 4 6 5 7 8 5 6 6 7 7 8 8 9 9 10 10 11 9 12 10 13 14 11 11 12 12 13 13 14 14 15 15 15 16 16 16 17 17 17
Output
115295190
Submitted Solution:
```
n = int(input())
hats = [0, 0, 0]
count = 1
mod = 1000000007
for a in map(int, input().split()):
count *= hats.count(a)
count %= mod
count[count.index(a)] += 1
print(count)
``` | instruction | 0 | 46,875 | 7 | 93,750 |
No | output | 1 | 46,875 | 7 | 93,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Japan, people make offerings called hina arare, colorful crackers, on March 3.
We have a bag that contains N hina arare. (From here, we call them arare.)
It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow.
We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`.
If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`.
Constraints
* 1 \leq N \leq 100
* S_i is `P`, `W`, `G` or `Y`.
* There always exist i, j and k such that S_i=`P`, S_j=`W` and S_k=`G`.
Input
Input is given from Standard Input in the following format:
N
S_1 S_2 ... S_N
Output
If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`.
Examples
Input
6
G W Y P Y W
Output
Four
Input
9
G W W G P W P G G
Output
Three
Input
8
P Y W G Y W Y Y
Output
Four
Submitted Solution:
```
#89b
n=int(input())
a=list(input())
a=set(a)
c=['P','W','G','Y']
if c-a==0:
print('Four')
else:
print('Three')
``` | instruction | 0 | 46,921 | 7 | 93,842 |
No | output | 1 | 46,921 | 7 | 93,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Japan, people make offerings called hina arare, colorful crackers, on March 3.
We have a bag that contains N hina arare. (From here, we call them arare.)
It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow.
We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`.
If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`.
Constraints
* 1 \leq N \leq 100
* S_i is `P`, `W`, `G` or `Y`.
* There always exist i, j and k such that S_i=`P`, S_j=`W` and S_k=`G`.
Input
Input is given from Standard Input in the following format:
N
S_1 S_2 ... S_N
Output
If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`.
Examples
Input
6
G W Y P Y W
Output
Four
Input
9
G W W G P W P G G
Output
Three
Input
8
P Y W G Y W Y Y
Output
Four
Submitted Solution:
```
N = int(input())
arare = input().split()
print(arare)
y = 0
for i in range(len(arare)):
if arare[i] == "Y":
y += 1
break
if y == 1:
print("Four")
else:
print("Three")
``` | instruction | 0 | 46,922 | 7 | 93,844 |
No | output | 1 | 46,922 | 7 | 93,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Japan, people make offerings called hina arare, colorful crackers, on March 3.
We have a bag that contains N hina arare. (From here, we call them arare.)
It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow.
We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`.
If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`.
Constraints
* 1 \leq N \leq 100
* S_i is `P`, `W`, `G` or `Y`.
* There always exist i, j and k such that S_i=`P`, S_j=`W` and S_k=`G`.
Input
Input is given from Standard Input in the following format:
N
S_1 S_2 ... S_N
Output
If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`.
Examples
Input
6
G W Y P Y W
Output
Four
Input
9
G W W G P W P G G
Output
Three
Input
8
P Y W G Y W Y Y
Output
Four
Submitted Solution:
```
ips = []
ips.append(input())
while True:
try:
ips.append(input())
except (EOFError):
break
num = int(ips[0])
hina = ips[1]
v = []
for i in range(num):
if hina[i] not in v:
v.append(hina[i])
if len(v) == 3:
print("three")
elif len(v) == 4:
print("four")
``` | instruction | 0 | 46,923 | 7 | 93,846 |
No | output | 1 | 46,923 | 7 | 93,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching. | instruction | 0 | 47,253 | 7 | 94,506 |
Tags: greedy, sortings, two pointers
Correct Solution:
```
def solve():
n, l, r = map(int, input().split())
c = list(map(int, input().split()))
lc = [0] * (n+1)
rc = [0] * (n+1)
res = 0
for i in range(l):
lc[c[i]] += 1
for i in range(l, n):
rc[c[i]] += 1
for i in range(n):
d = min(lc[i], rc[i])
lc[i] -= d
rc[i] -= d
# print(lc)
# print(rc)
for i in range(n+1):
# print(lc)
# print(rc)
# print()
if lc[i] > rc[i]:
if l > r:
d = (l - r) // 2
give = min((lc[i] - rc[i]) // 2, d)
rc[i] += give
lc[i] -= give
l -= give
r += give
res += give
else:
if r > l:
d = (r - l) // 2
give = min((rc[i] - lc[i]) // 2, d)
r -= give
l += give
rc[i] -= give
lc[i] += give
# print('give', i, give)
res += give
res += abs(r-l) // 2
d = 0
for i in range(n+1):
d += abs(lc[i] - rc[i])
res += d // 2 + int(d % 2 == 1)
print(res)
def main():
for _ in range(int(input())):
solve()
main()
``` | output | 1 | 47,253 | 7 | 94,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching. | instruction | 0 | 47,254 | 7 | 94,508 |
Tags: greedy, sortings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
def map_int():
return map(int, input().split())
def solve():
n, l, r=map_int()
c=list(map_int())
if r>l:
c=c[l:n]+c[:l]
l, r = r, l
color = [0]*(n+1)
for i in range(n):
color[c[i]]+=1 if i < l else -1
ans=rem=n//2-r
bad=0
for cnt in color:
if cnt>1:
used=min(cnt//2, rem)
cnt-=2*used
rem-=used
if cnt>0:
bad+=cnt
print(ans+bad-rem)
for _ in range(int(input())):
solve()
``` | output | 1 | 47,254 | 7 | 94,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching. | instruction | 0 | 47,255 | 7 | 94,510 |
Tags: greedy, sortings, two pointers
Correct Solution:
```
# https://codeforces.com/contest/1515/problem/D
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
# do magic here
t = int(input())
def makeDict(left):
leftMap = {}
for item in left:
if item not in leftMap:
leftMap[item] = 0
leftMap[item] += 1
return leftMap
for _ in range(t):
n, l, r = map(int, input().split())
left = list(map(int, input().split()))
right = []
while len(left) > l:
right.append(left.pop())
leftMap = makeDict(left)
rightMap = makeDict(right)
lkeys = set(leftMap.keys())
rkeys = set(rightMap.keys())
for k in lkeys:
if k in rkeys:
mn = min(leftMap[k], rightMap[k])
leftMap[k] -= mn
if not leftMap[k]:
leftMap.pop(k)
rightMap[k] -= mn
if not rightMap[k]:
rightMap.pop(k)
# print(leftMap, rightMap)
left = []
right = []
a, b = sum(leftMap.values()), sum(rightMap.values())
if a < b:
leftMap, rightMap = rightMap, leftMap
a, b = b, a
elif a == b:
# only change colors
pass
lkeys = leftMap.keys()
gap = abs(a-b)
cost = 0
vals = sorted(leftMap.values(), reverse=True)
for i in range(len(vals)):
val = vals[i]
while gap and val >= 2:
gap -= 2
vals[i] -= 2
val -= 2
cost += 1
X = sum(vals)
while gap:
cost += 2
X -= 2
gap -= 2
cost += X
print(cost)
``` | output | 1 | 47,255 | 7 | 94,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching. | instruction | 0 | 47,256 | 7 | 94,512 |
Tags: greedy, sortings, two pointers
Correct Solution:
```
#!/usr/bin/env python3
import sys, getpass
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline # to read input quickly
# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy
M9 = 10**9 + 7 # 998244353
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)] # hexagonal layout
MAXINT = sys.maxsize
# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False # codechef does not allow getpass
def log(*args):
if OFFLINE_TEST:
print('\033[36m', *args, '\033[0m', file=sys.stderr)
def solve(*args):
# screen input
if OFFLINE_TEST:
log("----- solving ------")
log(*args)
log("----- ------- ------")
return solve_(*args)
def read_matrix(rows):
return [list(map(int,input().split())) for _ in range(rows)]
def read_strings(rows):
return [input().strip() for _ in range(rows)]
# ---------------------------- template ends here ----------------------------
def solve_(arr, brr):
if len(arr) < len(brr):
arr, brr = brr, arr
diff = len(arr) - len(brr)
# your solution here
c1 = Counter(arr)
c2 = Counter(brr)
for k in list(c1.keys()):
comb = min(c1[k], c2[k])
c1[k] -= comb
c2[k] -= comb
if c1[k] == 0:
del c1[k]
if c2[k] == 0:
del c2[k]
log(c1)
log(c2)
asum = sum(c1.values())
bsum = sum(c2.values())
matched = 0
for k in list(c1.keys()):
flip = c1[k]//2
matched += flip
c1[k] -= flip*2
return bsum + max((asum-bsum)//2, asum-bsum-matched)
# for k,v in c1.items():
# res += v
# for k,v in c2.items():
# res += v
return res
# for case_num in [0]: # no loop over test case
# for case_num in range(100): # if the number of test cases is specified
for case_num in range(int(input())):
# read line as an integer
# k = int(input())
# read line as a string
# srr = input().strip()
# read one line and parse each word as a string
# lst = input().split()
# read one line and parse each word as an integer
_,left,right = list(map(int,input().split()))
lst = list(map(int,input().split()))
# read multiple rows
# mrr = read_matrix(k) # and return as a list of list of int
# arr = read_strings(k) # and return as a list of str
arr = lst[:left]
brr = lst[left:]
res = solve(arr, brr) # include input here
# print result
# Google and Facebook - case number required
# print("Case #{}: {}".format(case_num+1, res))
# Other platforms - no case number required
print(res)
# print(len(res))
# print(*res) # print a list with elements
# for r in res: # print each list in a different line
# print(res)
# print(*res)
``` | output | 1 | 47,256 | 7 | 94,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching. | instruction | 0 | 47,257 | 7 | 94,514 |
Tags: greedy, sortings, two pointers
Correct Solution:
```
# template begins
#####################################
from io import BytesIO, IOBase
import sys
import math
import os
from collections import defaultdict
from math import ceil
from bisect import bisect_left, bisect_left
from time import perf_counter
# 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)
def input(): return sys.stdin.readline().rstrip("\r\n")
def mint(): return map(int, input().split())
def mfloat(): return map(float, input().split())
#####################################
# template ends
# Use the recursion snippet if heavy recursion is needed (depth>1000)
def solve():
n, l, r = mint()
*a, = mint()
left_socks={}
right_socks={}
for i in range(l):
left_socks[a[i]]=left_socks.get(a[i], 0)+1
for i in range(l, n):
sock = a[i]
if left_socks.get(sock, 0)>0:
left_socks[sock]-=1
else:
right_socks[sock] = right_socks.get(sock, 0) + 1
# you can convert 2l to 1l 1r
"""
you can pay a dollar to match one left with one right
avoid the ones that are 2 of same type, since you can pair them up later
"""
left = sum(left_socks.values())
right = sum(right_socks.values())
if left<right:
pair_up = right_socks.values()
remaining = right-left
cost=left
else:
pair_up = left_socks.values()
remaining = left-right
cost=right
for i in pair_up:
pair = i//2*2
take = min(pair, remaining-remaining%2)
cost += take//2
remaining -= take
if not remaining:
break
# print(cost, remaining)
cost += remaining
print(cost)
def main():
t = int(input())
# t = 1
for _ in range(t):
solve()
if __name__ == "__main__":
main()
``` | output | 1 | 47,257 | 7 | 94,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching. | instruction | 0 | 47,258 | 7 | 94,516 |
Tags: greedy, sortings, two pointers
Correct Solution:
```
from collections import defaultdict as dd
for _ in range(int(input())):
n,l,r=map(int,input().split())
a=list(map(int,input().split()))
le=dd(int)
ri=dd(int)
k=abs(l-r)
k//=2
for i in range(l):
le[a[i]]+=1
for i in range(l,n):
ri[a[i]]+=1
ans=0
for i in ri.keys():
x=min(ri[i],le[i])
ri[i]-=x
le[i]-=x
if l>r:
for i in le.keys():
m=le[i]//2
m=min(m,k)
k-=m
ans+=m
le[i]-=2*m
else:
for i in ri.keys():
m=ri[i]//2
m=min(m,k)
k-=m
ans+=m
ri[i]-=2*m
k=abs(sum(le.values())-sum(ri.values()))
x=min(sum(le.values()),sum(ri.values()))
ans+=x
ans+=k
print(ans)
``` | output | 1 | 47,258 | 7 | 94,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching. | instruction | 0 | 47,259 | 7 | 94,518 |
Tags: greedy, sortings, two pointers
Correct Solution:
```
from sys import stdin,stdout
stdin.readline
def mp(): return list(map(int, stdin.readline().strip().split()))
def it():return int(stdin.readline().strip())
from collections import defaultdict as dd,Counter as C,deque
from math import ceil,gcd,sqrt,factorial,log2,floor
def fun(dr,temp):
odd = 0
for i in dr.keys():
if dr[i]&1:
odd += 1
if odd >= temp:
for i in dr.keys():
if dr[i]&1 and temp:
dr[i] -= 1
temp -= 1
else:
for i in dr.keys():
if dr[i]&1 and temp:
dr[i] -= 1
temp -= 1
for i in dr.keys():
if temp:
if temp >= dr[i]:
temp -= dr[i]
dr[i] = 0
else:
dr[i] -= temp
temp = 0
else:
break
def fun2(dr):
x = 0
ans = 0
for i in dr.keys():
if dr[i] and not dr[i]&1:
ans += dr[i]//2
elif dr[i] and dr[i]&1:
if x:
ans += dr[i]//2
ans += 2
x = 0
else:
ans += dr[i]//2
x = 1
return ans
for _ in range(it()):
n,l,r = mp()
a = mp()
left = a[:l]
right = a[l:]
dl = C(left)
dr = C(right)
for i in range(1,n+1):
if dr[i] and dl[i]:
temp = min(dl[i],dr[i])
dl[i] -= temp
dr[i] -= temp
ans = min(sum(dl.values()),sum(dr.values()))
temp = ans
if ans == sum(dl.values()):
fun(dr,temp)
print(ans + fun2(dr))
else:
fun(dl,temp)
print(ans+fun2(dl))
``` | output | 1 | 47,259 | 7 | 94,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching. | instruction | 0 | 47,260 | 7 | 94,520 |
Tags: greedy, sortings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
import math
mod=10**9+7
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
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()))
for _ in range(inp()):
n,l,r=invr()
a=inlt()
b=[]
for i in range(l,n):
b.append(a[i])
a=a[:l]
a.sort()
b.sort()
a1=[0 for i in range(n)]
b1 = a1[:]
for i in range (l):
a1[a[i]-1]+=1
for i in range(r):
b1[b[i]-1]+=1
for i in range(n):
m= min(a1[i],b1[i])
a1[i]-=m
b1[i]-=m
a=[]
b=[]
for i in range(n):
for j in range(a1[i]):
a.append(i+1)
for j in range(b1[i]):
b.append(i+1)
l=len(a)
r=len(b)
if l<r:
t=a
a=b
b=t
l,r=r,l
a.sort()
b.sort()
req=int((l+r)/2)-r
c=req
if req!=0:
y=[]
x=[[0,i+1] for i in range(n)]
for i in range(l):
x[a[i]-1][0]+=1
x.sort(key=lambda x:x[0])
x.reverse()
g= n-1
for i in range (n):
if x[i][0]==0:
g= i-1
break
p=0
while req>0:
t=req
req=max(req-(x[p][0]//2),0)
for i in range(t-req):
y.append(x[p][1])
if p== n-1:
for i in range(req):
req-=1
y.append(x[g-i][1])
p+=1
for i in y:
b.append(i)
y.sort()
h=[1 for i in range(l)]
p=0
g=len(y)
for i in range(l):
if a[i]==y[p]:
h[i]=0
if p==g-1:
break
p+=1
t=[]
for i in range(l):
if h[i]==1:
t.append(a[i])
a=t
a.sort()
b.sort()
p=0
x=0
d=[0 for i in range(n)]
n=(l+r)
for i in range(int(n/2)):
d[a[i]-1]+=1
for i in range(int(n/2)):
if d[b[i]-1]!=0:
d[b[i]-1]-=1
print(c+sum(d))
``` | output | 1 | 47,260 | 7 | 94,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching.
Submitted Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
from collections import defaultdict
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def main():
t = int(input())
for _ in range(t):
n, l, r = map(int, input().split())
l_socks = [0] * n
r_socks = [0] * n
for i, ci in enumerate(map(int, input().split())):
if i < l:
l_socks[ci - 1] += 1
else:
r_socks[ci - 1] += 1
sol = 0
l_rem, r_rem = 0, 0
for i in range(n):
if l > r and l_socks[i] > r_socks[i]:
t = min((l_socks[i] - r_socks[i]) // 2, (l - r) // 2)
l_socks[i] -= t
r_socks[i] += t
l -= t
r += t
sol += t
elif r > l and r_socks[i] > l_socks[i]:
t = min((r_socks[i] - l_socks[i]) // 2, (r - l) // 2)
r_socks[i] -= t
l_socks[i] += t
r -= t
l += t
sol += t
m = min(r_socks[i], l_socks[i])
l_socks[i] -= m
r_socks[i] -= m
l_rem += l_socks[i]
r_rem += r_socks[i]
print(sol + abs(l_rem - r_rem) // 2 + abs(l_rem + r_rem) // 2)
# 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")
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)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 47,261 | 7 | 94,522 |
Yes | output | 1 | 47,261 | 7 | 94,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching.
Submitted Solution:
```
import sys
In = sys.stdin.readline
T = int(In())
for tc in range(1, T + 1):
n, l, r = map(int, In().split())
color = [0 for _ in range(n + 1)]
socks = list(map(int, In().split()))
visited = {}
left, right = {"total": 0, "odd": 0}, {"total": 0, "odd": 0}
answer = 0
for i in range(l):
color[socks[i]] += 1
visited[socks[i]] = True
for i in range(l, n):
color[socks[i]] -= 1
visited[socks[i]] = True
for key in visited.keys():
temp = color[key]
if temp == 0:
continue
if temp > 0:
left["total"] += temp
if temp % 2 != 0:
left["odd"] += 1
else:
temp = -temp
right["total"] += temp
if temp % 2 != 0:
right["odd"] += 1
if left["total"] >= right["total"]:
answer += right["total"]
left["odd"] -= right["total"]
if left["odd"] > 0:
answer += left["odd"]
answer += (left["total"] - right["total"] - left["odd"]) // 2
else:
answer += (left["total"] - right["total"]) // 2
else:
answer += left["total"]
right["odd"] -= left["total"]
if right["odd"] > 0:
answer += right["odd"]
answer += (right["total"] - left["total"] - right["odd"]) // 2
else:
answer += (right["total"] - left["total"]) // 2
print(answer)
``` | instruction | 0 | 47,262 | 7 | 94,524 |
Yes | output | 1 | 47,262 | 7 | 94,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun May 2 17:39:03 2021
@author: edwar
"""
import sys
input = sys.stdin.readline
def inp(): # integer
return(int(input()))
def inlt(): # list
return(list(map(int,input().split())))
def insr(): # string
s = input()
return(list(s[:len(s) - 1]))
def invr(): # space separated integers
return(map(int,input().split()))
tests = inp()
for _ in range(tests):
n,l,r = invr()
c = inlt()
colourCount = [[0,0] for _ in range(n)]
for ii in range(l):
colourCount[c[ii]-1][0] +=1
for ii in range(l,n):
colourCount[c[ii]-1][1] +=1
totalCost = 0
spareLeft = 0
spareRight = 0
for col in colourCount:
matches = min(col)
col[0]-=matches
col[1]-=matches
if l>r:
matches = min(col[0]//2,(l-r)//2)
l-=matches
r+=matches
col[0]-=matches*2
totalCost+=matches
elif r>l:
matches = min(col[1]//2,(r-l)//2)
l+=matches
r-=matches
col[1]-=matches*2
totalCost+=matches
matches = min(col[0],spareRight)
spareRight-=matches
col[0]-=matches
spareLeft+=col[0]
totalCost+=matches
matches = min(col[1],spareLeft)
spareLeft-=matches
col[1]-=matches
spareRight+=col[1]
totalCost+=matches
totalCost+=(spareLeft+spareRight)
print(totalCost)
``` | instruction | 0 | 47,263 | 7 | 94,526 |
Yes | output | 1 | 47,263 | 7 | 94,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching.
Submitted Solution:
```
from collections import Counter
import sys
input=sys.stdin.readline
# "". join(strings)
def ri():
return int(input())
def rl():
return list(map(int, input().split()))
t = ri()
for _ in range(t):
# print("#"*50)
n, l, r = rl()
ss = rl()
# if l > r:
# ss = ss[::-1]
# l, r = r, l
ll = ss[:l]
rr = ss[l:]
L= Counter(ll)
R = Counter(rr)
All = Counter(ss)
solution = 0
left_alone = 0
right_alone = 0
left_pairs = 0
right_pairs = 0
for k in All.keys():
if L[k] > R[k]:
excess = L[k] - R[k]
parity = excess % 2
left_alone += parity
left_pairs += (excess-parity)
elif L[k] < R[k]:
excess = R[k] - L[k]
parity = excess % 2
right_alone += parity
right_pairs += (excess-parity)
# if left_alone > right_alone:
# left_alone, right_alone = right_alone, left_alone
# print(left_pairs, left_alone, right_pairs, right_alone)
if left_alone > right_alone:
left_alone, left_pairs, right_alone, right_pairs = right_alone, right_pairs,left_alone, left_pairs
# print(left_pairs, left_alone, right_pairs, right_alone)
solution += left_alone
right_alone -= left_alone
left_alone = 0
# print("sol", solution)
# print(left_pairs, left_alone, right_pairs, right_alone)
if right_alone > left_pairs:
solution += left_pairs
right_alone -= left_pairs
left_pairs = 0
solution += right_pairs//2 + right_alone
else:
solution += right_alone
left_pairs -= right_alone
solution += right_pairs//2 + left_pairs//2
# if left_alone > right_pairs:
# solution += right_pairs
# left_alone -= right_pairs
# right_pairs = 0
# else:
# solution += left_alone
# right_pairs -= left_alone
# left_alone = 0
# if right_alone > left_pairs:
# solution += left_pairs
# right_alone -= left_pairs
# left_pairs = 0
# else:
# solution += right_alone
# left_pairs -= right_alone
# right_alone = 0
# # print(left_pairs, left_alone, right_pairs, right_alone)
# if left_alone > 0 and right_alone > 0:
# mini = min(left_alone, right_alone)
# maxi = max(left_alone, right_alone)
# solution += maxi
# elif left_alone > 0 and left_pairs > 0:
# left_alone += left_pairs % 2
# left_pairs -= left_pairs % 2
# solution += left_pairs//2 + left_alone
# elif right_alone > 0 and right_pairs > 0:
# right_alone += right_pairs % 2
# right_pairs -= right_pairs % 2
# solution += right_pairs//2 + right_alone
# elif left_pairs > 0 and right_pairs > 0:
# if left_pairs % 2 == 1:
# left_pairs -= 1
# right_pairs -= 1
# solution += 2
# solution += left_pairs//2 + right_pairs//2
print(solution)
# solution += left_pairs //2 + right_pairs//2 + left_alone
# excess = right_alone - left_alone
# solution += excess
# print(solution)
``` | instruction | 0 | 47,264 | 7 | 94,528 |
Yes | output | 1 | 47,264 | 7 | 94,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching.
Submitted Solution:
```
#def main():
# from sys import stdin, stdout
# n, k = stdin.readline().split()
# n = int(n)
# k = int(k)
# cnt = 0
# lines = stdin.readlines()
# for line in lines:
# if int(line) % k == 0:
# cnt += 1
# stdout.write(str(cnt))
from sys import stdin,stdout
def readInt():
return int(stdin.readline())
def readInts():
return map(int, stdin.readline().strip().split())
def readString():
return stdin.readline().strip()
def writeLine(s):
stdout.write(str(s)+'\n')
return
def main():
t=readInt()
for _ in range(t):
n,l,r=readInts()
c=list(readInts())
ld={}
rd={}
for i in range(l):
ld[c[i]]=ld.get(c[i],0)+1
for i in range(l,n):
rd[c[i]]=rd.get(c[i],0)+1
diff={}
for i in range(1,n+1):
diff[i]=ld.get(i,0)-rd.get(i,0)
res=0
sum=0
for key in diff:
sum+=diff[key]
res+=abs(sum)
mc=0
for key in diff:
mc+=diff[key]%2
res+=mc/2
writeLine(res)
if __name__ == "__main__":
main()
``` | instruction | 0 | 47,265 | 7 | 94,530 |
No | output | 1 | 47,265 | 7 | 94,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching.
Submitted Solution:
```
from sys import stdin,stdout
stdin.readline
def mp(): return list(map(int, stdin.readline().strip().split()))
def it():return int(stdin.readline().strip())
from collections import defaultdict as dd,Counter as C,deque
from math import ceil,gcd,sqrt,factorial,log2,floor
def fun(dr,temp):
odd = 0
for i in dr.keys():
if dr[i]&1:
odd += 1
if odd >= temp:
for i in dr.keys():
if dr[i]&1 and temp:
dr[i] -= 1
temp -= 1
else:
for i in dr.keys():
if dr[i]&1 and temp:
dr[i] -= 1
temp -= 1
for i in dr.keys():
if temp:
if temp >= dr[i]:
temp -= dr[i]
dr[i] = 0
else:
dr[i] -= temp
temp = 0
else:
break
for _ in range(it()):
n,l,r = mp()
a = mp()
left = a[:l]
right = a[l:]
dl = C(left)
dr = C(right)
# print(dict(dl),dict(dr))
for i in range(1,n+1):
if dr[i] and dl[i]:
dl[i] -= min(dl[i],dr[i])
dr[i] -= min(dl[i],dr[i])
temp = ans = min(sum(dl.values()),sum(dr.values()))
if ans == sum(dl.values()):
fun(dr,temp)
x = 0
for i in dr.keys():
if dr[i] and not dr[i]&1:
ans += dr[i]//2
elif dr[i] and dr[i]&1:
if x:
ans += dr[i]//2
ans += 2
x = 0
else:
ans += dr[i]//2
x = 1
else:
fun(dl,temp)
x = 0
for i in dl.keys():
if dl[i] and not dl[i]&1:
ans += dl[i]//2
elif dl[i] and dl[i]&1:
if x:
ans += dl[i]//2
ans += 2
x=0
else:
ans += dl[i]//2
x = 1
print(ans)
``` | instruction | 0 | 47,266 | 7 | 94,532 |
No | output | 1 | 47,266 | 7 | 94,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching.
Submitted Solution:
```
def solver(n, l, r, c):
cost = 0
cnew = sorted([f'{color}_left' for color in c[:l]] + [f'{color}_rigth' for color in c[l:]])
cnewnew = [None for i in range(len(cnew))]
counter = 0
for i in range(len(cnew)):
if i > 0 and cnew[i] != cnew[i-1]:
counter = 0
counter += 1
temp = cnew[i].split('_')
cnewnew[i] = f'{temp[0]}_{counter}_{temp[1]}'
cnew = sorted(cnewnew)
#print(cnew)
mask = [0 for i in range(len(cnew))]
for i in range(1, len(cnew)):
color_equal = cnew[i].split('_')[0] == cnew[i-1].split('_')[0]
left_rigth = cnew[i].split('_')[2] != cnew[i-1].split('_')[2]
if color_equal and left_rigth:
mask[i] = 1
mask[i-1] = 1
cnew = [x for x, y in zip(cnew, mask) if y != 1]
mask = [0 for i in range(len(cnew))]
# ΠΈΠ·ΠΌΠ΅Π½ΠΈΡΡ ΡΠΎΠ»ΡΠΊΠΎ ΡΡΠΎΡΠΎΠ½Ρ
for i in range(1, len(cnew)):
if mask[i-1] == 1:
continue
color_equal = cnew[i].split('_')[0] == cnew[i-1].split('_')[0]
if color_equal:
mask[i] = 1
mask[i-1] = 1
cost += 1
continue
left_rigth = cnew[i].split('_')[2] != cnew[i-1].split('_')[2]
if left_rigth:
mask[i] = 1
mask[i-1] = 1
cost += 1
cnew = [x for x, y in zip(cnew, mask) if y != 1]
left_count = len([el for el in cnew if 'left' in el])
rigth_count = len([el for el in cnew if 'rigth' in el])
if left_count < rigth_count:
left_count, rigth_count = rigth_count, left_count
min_possible = min(left_count,rigth_count)
cost += min_possible
left_count -= min_possible
if left_count > 0:
cost += left_count
#print(left_count, rigth_count, cnew, cost,min_possible)
return str(cost)
t = int(input())
answers = []
for i in range(t):
n, l, r = list(map(int, input().split()))
c = list(map(int, input().split()))
ans = solver(n, l, r, c)
answers.append(ans)
print('\n'.join(answers))
``` | instruction | 0 | 47,267 | 7 | 94,534 |
No | output | 1 | 47,267 | 7 | 94,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color c_i and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
* recolor a sock to any color c' (1 β€ c' β€ n)
* turn a left sock into a right sock
* turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn't change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains three integers n, l, and r (2 β€ n β€ 2 β
10^5; n is even; 0 β€ l, r β€ n; l+r=n) β the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers c_i (1 β€ c_i β€ n) β the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2 β
10^5.
Output
For each test case, print one integer β the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
Input
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
Output
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
* recolor sock 1 to color 2
* recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1, 4), (2, 5), and (3, 6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
* turn sock 6 from a right sock to a left sock
* recolor sock 3 to color 1
* recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1, 3), (2, 4), and (5, 6) are matching.
Submitted Solution:
```
import math
from collections import defaultdict
from heapq import heappush, heappop
DEBUG = True
def log(*args, **kwargs):
if DEBUG:
print(*args, **kwargs)
def ri():
return int(input())
def rl(f=int):
return list(map(f, input().split()))
def rs():
return input()
class Solution:
def __init__(self):
pass
def run(self):
n, l, r = rl()
c = rl()
left = defaultdict(int)
right = defaultdict(int)
colors = set()
for i in range(l):
left[c[i]] += 1
colors.add(c[i])
for i in range(l, n):
right[c[i]] += 1
colors.add(c[i])
#print(left)
#print(right)
#print('---')
exleft = defaultdict(int)
exright = defaultdict(int)
result = 0
totalleft = 0
totalright = 0
for co in colors:
diff = left[co] - right[co]
if diff > 0:
exleft[co] += diff
totalleft += diff
elif diff < 0:
exright[co] += (-diff)
totalright += (-diff)
#print(exleft)
#print(exright)
#print('---')
if totalleft == totalright:
return totalleft
largest = defaultdict(int)
smallesttotal = min(totalleft, totalright)
if totalleft > totalright:
largest = exleft
else:
largest = exright
#print(smallesttotal, largest)
changed = 0
for k, v in sorted(largest.items(), key = lambda x: (int(not (x[1] % 2)), x[1])):
#print(k, v, changed, min(v, smallesttotal - changed))
largest[k] -= min(v, smallesttotal - changed)
changed += min(v, smallesttotal - changed)
#print(largest[k])
#print(exleft)
#print(exright)
#print(changed, largest)
#print('---')
result += smallesttotal
for k,v in largest.items():
result += v // 2
result += v % 2
return result
if __name__ == '__main__':
t = int(input())
s = Solution()
#print(s.run())
#s.run()
for i in range(t):
# s.run()
print(s.run())
``` | instruction | 0 | 47,268 | 7 | 94,536 |
No | output | 1 | 47,268 | 7 | 94,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3 | instruction | 0 | 47,374 | 7 | 94,748 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n,m=map(int,input().split())
ans=[-1]*(n+1)
for k in range(m):
dancers=list(map(int,input().split()))
need=[1,2,3]
for i in dancers:
if(ans[i]!=-1):
need.remove(ans[i])
for i in dancers:
if(ans[i]==-1):
ans[i]=need.pop(-1)
print(*ans[1:])
``` | output | 1 | 47,374 | 7 | 94,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3 | instruction | 0 | 47,375 | 7 | 94,750 |
Tags: constructive algorithms, implementation
Correct Solution:
```
from sys import stdin,stdout
stdin.readline
def mp(): return list(map(int, stdin.readline().strip().split()))
def it():return int(stdin.readline().strip())
from collections import defaultdict as dd,Counter as C,deque
from math import ceil,gcd,sqrt,factorial,log2,floor
from bisect import bisect_right as br,bisect_left as bl
import heapq
n,m = mp()
d = dd(lambda:0)
for _ in range(m):
l = mp()
s = set()
s.add(1)
s.add(2)
s.add(3)
for i in l:
if d[i]:
s.remove(d[i])
for i in l:
if s and not d[i]:
d[i] = s.pop()
for i in sorted(d.keys()):
print(d[i],end=' ')
``` | output | 1 | 47,375 | 7 | 94,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3 | instruction | 0 | 47,376 | 7 | 94,752 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n,m=map(int,input().split())
arr=[0 for i in range(n+1)]
for i in range(m):
x,y,z=map(int,input().split())
if([arr[x],arr[y],arr[z]]==[0,0,0]):
arr[x],arr[y],arr[z]=1,2,3
else:
ans=[1,2,3]
if(arr[x]!=0):
ans.remove(arr[x])
elif(arr[y]!=0):
ans.remove(arr[y])
elif(arr[z]!=0):
ans.remove(arr[z])
if(arr[x]==0):
arr[x]=ans[0]
ans.remove(ans[0])
if(arr[y]==0):
arr[y]=ans[0]
ans.remove(ans[0])
if(arr[z]==0):
arr[z]=ans[0]
ans.remove(ans[0])
print(*arr[1:])
``` | output | 1 | 47,376 | 7 | 94,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3 | instruction | 0 | 47,377 | 7 | 94,754 |
Tags: constructive algorithms, implementation
Correct Solution:
```
def main():
n, m = [int(i) for i in input().split()]
colors = [0 for i in range(n + 1)]
for i in range(m):
arr = [int(i) for i in input().split()]
early = colors[arr[0]] or colors[arr[1]] or colors[arr[2]]
l = [1, 2, 3]
if early: l.remove(early)
for i in arr:
if colors[i] == 0:
colors[i] = l.pop()
print(' '.join(str(colors[i]) for i in range(1, n + 1)))
main()
``` | output | 1 | 47,377 | 7 | 94,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3 | instruction | 0 | 47,378 | 7 | 94,756 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n,m = map(int,input().split(' '))
result,participated = [None]*n,{} #dancer : color
for x in range(m):
dancers = [int(i) for i in input().split(' ')]
colors,temp = set([1,2,3]),-1
for d in range(3):
if dancers[d] in participated:
colors.remove(participated[dancers[d]])
temp = d
it = iter(colors)
for i in range(3):
t = dancers[i]
if i != temp:
result[dancers[i] - 1] = next(it)
participated[t] = result[dancers[i] - 1]
print(*result)
'''
KaΕΌdy taniec musi siΔ skΕadaΔ z kolorΓ³w 1,2,3.
Taniec - 3 osoby, m taΕcΓ³w.
KaΕΌdy taniec moΕΌe mieΔ max 1 taΕcerza z poprzednich taΕcΓ³w.
Na podstawie tych warunkΓ³w musimy ubraΔ wyznaczonych taΕcerzy do taΕcΓ³w.
jeΕΌeli n = 3m tzn ΕΌe moΕΌemy kaΕΌdΔ
oddzielnΔ
trΓ³jkΔ wyznaczyΔ do innego taΕca i
ubraΔ ich wszystkich wyznaczony patern (1,2,3) od 1 do n-tego tancerza.
join('1,2,3') * 3 * m.
solution: O(m)
Zawsze w trΓ³jce bΔdzie maksymalnie 1 powtarzajΔ
cy siΔ a tzn ΕΌe musimy
w takim przypadku dostoswaΔ dwΓ³ch pozostaΕych w trΓ³jce pod niego.
W przypadku unikalnej trΓ³jki, po prostu 1,2,3.
implementacja:
lista ludzi ktΓ³rzy juΕΌ taΕczyli = hashmap{[1-n]:[1-3]}
WczytujΔ
c trΓ³jkΔ:
najpierw staramy siΔ wykryΔ potencjalnie jednego uczestnika ktΓ³ry juΕΌ braΕ udziaΕ w poprzednim taΕcu.
JeΕΌeli go wykryjemy, spr jego strΓ³j i dodajemy odpowiedniΔ
kombinacje do rezultatu: .join(wykryty,dwa pozostaΕe)
JeΕΌeli nie: dodajemy do rezultatu (1,2,3).
Dodajemy do zbioru uczestnikΓ³w i ich stroje.
'''
``` | output | 1 | 47,378 | 7 | 94,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3 | instruction | 0 | 47,379 | 7 | 94,758 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n,m=map(int,input().split())
ans=[-1]*(n+1)
for _ in range(m):
dancers=list(map(int,input().split()))
need=[1,2,3]
for i in dancers:
if(ans[i]!=-1):
need.remove(ans[i])
for i in dancers:
if(ans[i]==-1):
ans[i]=need.pop(-1)
print(*ans[1:])
``` | output | 1 | 47,379 | 7 | 94,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3 | instruction | 0 | 47,380 | 7 | 94,760 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n,m=map(int,input().split())
li=[]
for i in range(m):
li.append([int(x) for x in input().split()])
d={}
for i in range(m):
x=li[i]
use=[True]*3
l=[]
for ele in x:
if d.get(ele,0)!=0:
use[d.get(ele)-1]=False
for i in range(3):
if use[i]==True:
l.append(i)
for ele in x:
if d.get(ele,0)==0:
d[ele]=l.pop()+1
for ele in sorted(d):
print(d[ele],end=" ")
``` | output | 1 | 47,380 | 7 | 94,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3 | instruction | 0 | 47,381 | 7 | 94,762 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n, m = map(int, input().split())
l = [-1]*n
for i in range(m):
a, b, c = map(int, input().split())
if(l[a-1] == -1 and l[b-1] == -1 and l[c-1] == -1):
l[a-1] = 1
l[b-1] = 2
l[c-1] = 3
elif(l[a-1] == -1 and l[b-1] == -1):
k = [1, 2, 3]
k.remove(l[c-1])
l[a-1] = k[0]
l[b-1] = k[1]
elif(l[c-1] == -1 and l[b-1] == -1):
k = [1, 2, 3]
k.remove(l[a-1])
l[c-1] = k[0]
l[b-1] = k[1]
elif(l[a-1] == -1 and l[c-1] == -1):
k = [1, 2, 3]
k.remove(l[b-1])
l[a-1] = k[0]
l[c-1] = k[1]
print(*l)
``` | output | 1 | 47,381 | 7 | 94,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3
Submitted Solution:
```
n, m = map(int, input().split())
res = [0] * (n + 1)
for _ in range(m):
arr = list(map(int, input().split()))
for i in range(3):
if res[arr[i]] == 1:
res[arr[(i + 1) % 3]] = 2
res[arr[(i + 2) % 3]] = 3
break
elif res[arr[i]] == 2:
res[arr[(i + 1) % 3]] = 1
res[arr[(i + 2) % 3]] = 3
break
elif res[arr[i]] == 3:
res[arr[(i + 1) % 3]] = 1
res[arr[(i + 2) % 3]] = 2
break
if not res[arr[0]]:
res[arr[0]], res[arr[1]], res[arr[2]] = 1, 2, 3
print(*res[1:])
``` | instruction | 0 | 47,382 | 7 | 94,764 |
Yes | output | 1 | 47,382 | 7 | 94,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3
Submitted Solution:
```
import sys;
nbDancers, nbDances = map(int, sys.stdin.readline().split(' '));
dancers = [0]*(nbDancers+1);
for line in sys.stdin:
taken = [False, False, False, False];
ds = list(map(int, line.split(' ')));
for d in ds:
taken[dancers[d]] = True;
for d in ds:
if dancers[d] == 0:
if not taken[1]:
dancers[d] = 1;
taken[1] = True;
elif not taken[2]:
dancers[d] = 2;
taken[2] = True;
elif not taken[3]:
dancers[d] = 3;
taken[3] = True;
print(' '.join(map(str, dancers[1:])));
``` | instruction | 0 | 47,383 | 7 | 94,766 |
Yes | output | 1 | 47,383 | 7 | 94,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3
Submitted Solution:
```
import sys;
nbDancers, nbDances = map(int, sys.stdin.readline().split(' '));
dancers = [0]*(nbDancers+1);
for line in sys.stdin:
taken = [False, False, False, False];
ds = list(map(int, line.split(' ')));
for d in ds:
taken[dancers[d]] = True;
for d in ds:
if dancers[d] == 0:
iCurr = 1;
while iCurr <= 3 and taken[iCurr]:
iCurr += 1;
dancers[d] = iCurr;
taken[iCurr] = True;
print(' '.join(map(str, dancers[1:])));
``` | instruction | 0 | 47,384 | 7 | 94,768 |
Yes | output | 1 | 47,384 | 7 | 94,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3
Submitted Solution:
```
from math import gcd
def take_input(s): #for integer inputs
if s == 1: return int(input())
return map(int, input().split())
n,m = take_input(2)
grid = []
d = {}
for i in range(m):
grid += list(take_input(3))
for i in range(0,m*3,3):
if d.get(grid[i],0) > 0:
if d.get(grid[i],0) == 1:
d[grid[i+1]] = 2
d[grid[i+2]] = 3
elif d.get(grid[i],0) == 2:
d[grid[i+1]] = 1
d[grid[i+2]] = 3
else:
d[grid[i+1]] = 1
d[grid[i+2]] = 2
elif d.get(grid[i+1],0) > 0:
if d.get(grid[i+1],0) == 1:
d[grid[i]] = 2
d[grid[i+2]] = 3
elif d.get(grid[i+1],0) == 2:
d[grid[i]] = 1
d[grid[i+2]] = 3
else:
d[grid[i]] = 1
d[grid[i+2]] = 2
elif d.get(grid[i+2],0) > 0:
if d.get(grid[i+2],0) == 1:
d[grid[i]] = 2
d[grid[i+1]] = 3
elif d.get(grid[i+2],0) == 2:
d[grid[i]] = 1
d[grid[i+1]] = 3
else:
d[grid[i]] = 1
d[grid[i+1]] = 2
else:
d[grid[i]] = 1
d[grid[i+1]] = 2
d[grid[i+2]] = 3
for i in range(n):
print(d[i+1], end = " ")
``` | instruction | 0 | 47,385 | 7 | 94,770 |
Yes | output | 1 | 47,385 | 7 | 94,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3
Submitted Solution:
```
n,m = map(int,input().split(' '))
result,participated = [],{} #dancer : color
for x in range(m):
dancers = [int(i) for i in input().split(' ')]
colors,temp = set([1,2,3]),-1
for d in range(3):
if dancers[d] in participated:
dancers[d] = participated[dancers[d]]
colors.remove(dancers[d])
temp = d
it = iter(colors)
for i in range(3):
t = dancers[i]
if i != temp:
dancers[i] = next(it)
participated[t] = dancers[i]
result.extend(dancers)
print(*result)
'''
KaΕΌdy taniec musi siΔ skΕadaΔ z kolorΓ³w 1,2,3.
Taniec - 3 osoby, m taΕcΓ³w.
KaΕΌdy taniec moΕΌe mieΔ max 1 taΕcerza z poprzednich taΕcΓ³w.
Na podstawie tych warunkΓ³w musimy ubraΔ wyznaczonych taΕcerzy do taΕcΓ³w.
jeΕΌeli n = 3m tzn ΕΌe moΕΌemy kaΕΌdΔ
oddzielnΔ
trΓ³jkΔ wyznaczyΔ do innego taΕca i
ubraΔ ich wszystkich wyznaczony patern (1,2,3) od 1 do n-tego tancerza.
join('1,2,3') * 3 * m.
solution: O(m)
Zawsze w trΓ³jce bΔdzie maksymalnie 1 powtarzajΔ
cy siΔ a tzn ΕΌe musimy
w takim przypadku dostoswaΔ dwΓ³ch pozostaΕych w trΓ³jce pod niego.
W przypadku unikalnej trΓ³jki, po prostu 1,2,3.
implementacja:
lista ludzi ktΓ³rzy juΕΌ taΕczyli = hashmap{[1-n]:[1-3]}
WczytujΔ
c trΓ³jkΔ:
najpierw staramy siΔ wykryΔ potencjalnie jednego uczestnika ktΓ³ry juΕΌ braΕ udziaΕ w poprzednim taΕcu.
JeΕΌeli go wykryjemy, spr jego strΓ³j i dodajemy odpowiedniΔ
kombinacje do rezultatu: .join(wykryty,dwa pozostaΕe)
JeΕΌeli nie: dodajemy do rezultatu (1,2,3).
Dodajemy do zbioru uczestnikΓ³w i ich stroje.
'''
``` | instruction | 0 | 47,386 | 7 | 94,772 |
No | output | 1 | 47,386 | 7 | 94,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3
Submitted Solution:
```
n,m=list(map(int,input().split()))
l=[]
for i in range(m):
l.append(list(map(int,input().split())))
d={}
for i in range(m):
for j in range(3):
d[l[i][j]]=None
for i in range(m):
for j in range(3):
if not d[l[i][j]]:
d[l[i][j]]=j+1
for i in range(1,n):
print(d[i],end=' ')
print(d[n])
``` | instruction | 0 | 47,387 | 7 | 94,774 |
No | output | 1 | 47,387 | 7 | 94,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3
Submitted Solution:
```
n, m = list(map(int, input().split()))
met = {}
for _ in range(m):
arr = list(map(int, input().split()))
if _ == 0:
met[arr[0]] = 1
met[arr[1]] = 2
met[arr[2]] = 3
else:
arr2 = [1, 2, 3]
xxx = 0
for k in range(3):
if arr[k] in met:
arr2.remove(met[arr[k]])
else:
met[arr[k]] = arr2[xxx]
xxx += 1
for kkk in range(1, n+1):
print(met[kkk], end=' ')
``` | instruction | 0 | 47,388 | 7 | 94,776 |
No | output | 1 | 47,388 | 7 | 94,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland, there is the national holiday coming β the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
* overall, there must be m dances;
* exactly three people must take part in each dance;
* each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has n dancers, and their number can be less than 3m. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the m dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the n dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input
The first line contains two space-separated integers n (3 β€ n β€ 105) and m (1 β€ m β€ 105) β the number of dancers and the number of dances, correspondingly. Then m lines follow, describing the dances in the order of dancing them. The i-th line contains three distinct integers β the numbers of the dancers that take part in the i-th dance. The dancers are numbered from 1 to n. Each dancer takes part in at least one dance.
Output
Print n space-separated integers: the i-th number must represent the color of the i-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Examples
Input
7 3
1 2 3
1 4 5
4 6 7
Output
1 2 3 3 2 2 1
Input
9 3
3 6 9
2 5 8
1 4 7
Output
1 1 1 2 2 2 3 3 3
Input
5 2
4 1 5
3 1 2
Output
2 3 1 1 3
Submitted Solution:
```
def next_color(color):
return color % 3 + 1
n, m = list(map(int, input().split()))
dances = [list(map(int, input().split())) for _ in range(m)]
colors = [None for _ in range(n+1)]
color = 1
for dance in dances:
for dancer in dance:
if colors[dancer] is None:
colors[dancer] = color
color = next_color(color)
else:
color = next_color(colors[dancer])
for color in colors[1:]:
print(color, end=' ')
``` | instruction | 0 | 47,389 | 7 | 94,778 |
No | output | 1 | 47,389 | 7 | 94,779 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. A of the Aizu Institute of Biological Research discovered a mysterious insect on a certain southern island. The shape is elongated like a hornworm, but since one segment is shaped like a ball, it looks like a beaded ball connected by a thread. What was strange was that there were many variations in body color, and some insects changed their body color over time. It seems that the color of the somites of all insects is limited to either red, green or blue, but the color of the somites changes every second, and finally all the somites become the same color and settle down. In some cases, the color seemed to keep changing no matter how long I waited.
<image>
As I investigated, I found that all the body segments usually have the same color, but after being surprised or excited by something, the color of the body segments changes without permission. It turns out that once the color of the segment changes, it will continue to change until all the segment are the same color again.
Dr. A was curiously observing how the color changed when he caught and excited many of these insects, but the way the color changed during the color change is as follows. I noticed that there is regularity.
* The color changes only in one pair of two adjacent somites of different colors, and the colors of the other somites do not change. However, when there are multiple such pairs, it is not possible to predict in advance which pair will change color.
* Such a pair will change to a color that is neither of the colors of the two segmentes at the same time (for example, if the green and red segment are adjacent, they will change to blue at the same time).
<image>
The figure above shows all the changes in the color of the insects up to 2 seconds later. Suppose you have an insect that has the color shown in the upper part of the figure. At this time, there are 3 pairs of adjacent somites of different colors, so after 1 second, it will change to one of the 3 colors drawn side by side in the middle row. After 1 second, all the segments can turn green after 2 seconds (second from the left in the lower row of the figure) when they change to the two on the left side of the middle row. On the other hand, when it changes like the one on the far right in the middle row after 1 second, all the segments do not change to the same color after 2 seconds.
He decided to predict if all the insect segments in front of him could be the same color, and if so, at the earliest, how many seconds later.
Create a program that takes the color sequence of the insect body segments in front of you as input and outputs the shortest time required for all the insect body segments to have the same color in seconds. However, if there is no possibility that the colors will be the same, output "NA (half-width uppercase letters)". In addition, the color sequence of the insect body segment is represented by a character string consisting of r (red), g (green), and b (blue) of 2 or more and 10 or less.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. For each dataset, one string is given on one line that represents information about the insect's somites.
The number of datasets does not exceed 100.
Output
For each dataset, the minimum time (integer in seconds) or NA required for all segment colors to be the same is output on one line.
Example
Input
rbgrg
rbbgbbr
bgr
bgrbrgbr
bggrgbgrr
gbrggrbggr
rrrrr
bgbr
0
Output
5
7
1
6
NA
8
0
4 | instruction | 0 | 47,854 | 7 | 95,708 |
"Correct Solution:
```
from collections import deque
def new_color(s, i, color):
return s[:i] + color * 2 + s[i + 2:]
def solve(s):
length = len(s)
monos = ["r" * length, "g" * length, "b" * length]
if s in monos:
print(0)
return
another = {("r", "g"):"b", ("g", "r"):"b",
("r", "b"):"g", ("b", "r"):"g",
("g", "b"):"r", ("b", "g"):"r"}
visited = {s:0}
que = deque()
app = que.append
pop = que.popleft
que.append((s, 0))
while que:
colors, score = pop()
score += 1
temp = colors[0]
for i in range(1, length):
ci = colors[i]
if ci != temp:
new = new_color(colors, i - 1, another[(ci, temp)])
if new in monos:
print(score)
return
if new not in visited:
visited[new] = score
app((new, score))
temp = ci
print("NA")
def main():
while True:
s = input()
if s == "0":
break
solve(s)
main()
``` | output | 1 | 47,854 | 7 | 95,709 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. A of the Aizu Institute of Biological Research discovered a mysterious insect on a certain southern island. The shape is elongated like a hornworm, but since one segment is shaped like a ball, it looks like a beaded ball connected by a thread. What was strange was that there were many variations in body color, and some insects changed their body color over time. It seems that the color of the somites of all insects is limited to either red, green or blue, but the color of the somites changes every second, and finally all the somites become the same color and settle down. In some cases, the color seemed to keep changing no matter how long I waited.
<image>
As I investigated, I found that all the body segments usually have the same color, but after being surprised or excited by something, the color of the body segments changes without permission. It turns out that once the color of the segment changes, it will continue to change until all the segment are the same color again.
Dr. A was curiously observing how the color changed when he caught and excited many of these insects, but the way the color changed during the color change is as follows. I noticed that there is regularity.
* The color changes only in one pair of two adjacent somites of different colors, and the colors of the other somites do not change. However, when there are multiple such pairs, it is not possible to predict in advance which pair will change color.
* Such a pair will change to a color that is neither of the colors of the two segmentes at the same time (for example, if the green and red segment are adjacent, they will change to blue at the same time).
<image>
The figure above shows all the changes in the color of the insects up to 2 seconds later. Suppose you have an insect that has the color shown in the upper part of the figure. At this time, there are 3 pairs of adjacent somites of different colors, so after 1 second, it will change to one of the 3 colors drawn side by side in the middle row. After 1 second, all the segments can turn green after 2 seconds (second from the left in the lower row of the figure) when they change to the two on the left side of the middle row. On the other hand, when it changes like the one on the far right in the middle row after 1 second, all the segments do not change to the same color after 2 seconds.
He decided to predict if all the insect segments in front of him could be the same color, and if so, at the earliest, how many seconds later.
Create a program that takes the color sequence of the insect body segments in front of you as input and outputs the shortest time required for all the insect body segments to have the same color in seconds. However, if there is no possibility that the colors will be the same, output "NA (half-width uppercase letters)". In addition, the color sequence of the insect body segment is represented by a character string consisting of r (red), g (green), and b (blue) of 2 or more and 10 or less.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. For each dataset, one string is given on one line that represents information about the insect's somites.
The number of datasets does not exceed 100.
Output
For each dataset, the minimum time (integer in seconds) or NA required for all segment colors to be the same is output on one line.
Example
Input
rbgrg
rbbgbbr
bgr
bgrbrgbr
bggrgbgrr
gbrggrbggr
rrrrr
bgbr
0
Output
5
7
1
6
NA
8
0
4 | instruction | 0 | 47,855 | 7 | 95,710 |
"Correct Solution:
```
from collections import deque
def only_color(s, monos):
if s in monos:
return True
else:
return False
def new_color(s, i, rgb):
for color in rgb:
if color != s[i] and color != s[i + 1]:
break
return s[:i] + color * 2 + s[i + 2:]
def solve(s):
length = len(s)
monos = ["r" * length, "g" * length, "b" * length]
if s in monos:
print(0)
return
dic = {s:0}
rgb = "rgb"
que = deque()
que.append((s, 0))
while que:
colors, score = que.popleft()
score += 1
temp = colors[0]
for i in range(1, length):
ci = colors[i]
if ci != temp:
new = new_color(colors, i - 1, rgb)
if new in monos:
print(score)
return
if new not in dic:
dic[new] = score
que.append((new, score))
temp = ci
print("NA")
def main():
while True:
s = input()
if s == "0":
break
solve(s)
main()
``` | output | 1 | 47,855 | 7 | 95,711 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. A of the Aizu Institute of Biological Research discovered a mysterious insect on a certain southern island. The shape is elongated like a hornworm, but since one segment is shaped like a ball, it looks like a beaded ball connected by a thread. What was strange was that there were many variations in body color, and some insects changed their body color over time. It seems that the color of the somites of all insects is limited to either red, green or blue, but the color of the somites changes every second, and finally all the somites become the same color and settle down. In some cases, the color seemed to keep changing no matter how long I waited.
<image>
As I investigated, I found that all the body segments usually have the same color, but after being surprised or excited by something, the color of the body segments changes without permission. It turns out that once the color of the segment changes, it will continue to change until all the segment are the same color again.
Dr. A was curiously observing how the color changed when he caught and excited many of these insects, but the way the color changed during the color change is as follows. I noticed that there is regularity.
* The color changes only in one pair of two adjacent somites of different colors, and the colors of the other somites do not change. However, when there are multiple such pairs, it is not possible to predict in advance which pair will change color.
* Such a pair will change to a color that is neither of the colors of the two segmentes at the same time (for example, if the green and red segment are adjacent, they will change to blue at the same time).
<image>
The figure above shows all the changes in the color of the insects up to 2 seconds later. Suppose you have an insect that has the color shown in the upper part of the figure. At this time, there are 3 pairs of adjacent somites of different colors, so after 1 second, it will change to one of the 3 colors drawn side by side in the middle row. After 1 second, all the segments can turn green after 2 seconds (second from the left in the lower row of the figure) when they change to the two on the left side of the middle row. On the other hand, when it changes like the one on the far right in the middle row after 1 second, all the segments do not change to the same color after 2 seconds.
He decided to predict if all the insect segments in front of him could be the same color, and if so, at the earliest, how many seconds later.
Create a program that takes the color sequence of the insect body segments in front of you as input and outputs the shortest time required for all the insect body segments to have the same color in seconds. However, if there is no possibility that the colors will be the same, output "NA (half-width uppercase letters)". In addition, the color sequence of the insect body segment is represented by a character string consisting of r (red), g (green), and b (blue) of 2 or more and 10 or less.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. For each dataset, one string is given on one line that represents information about the insect's somites.
The number of datasets does not exceed 100.
Output
For each dataset, the minimum time (integer in seconds) or NA required for all segment colors to be the same is output on one line.
Example
Input
rbgrg
rbbgbbr
bgr
bgrbrgbr
bggrgbgrr
gbrggrbggr
rrrrr
bgbr
0
Output
5
7
1
6
NA
8
0
4 | instruction | 0 | 47,856 | 7 | 95,712 |
"Correct Solution:
```
from collections import deque
def change(x,y):
if [x,y] == ["r","g"] or [y,x] == ["r","g"]:
return "b"
elif [x,y] == ["g","b"] or [y,x] == ["g","b"]:
return "r"
elif [x,y] == ["r","b"] or [y,x] == ["r","b"]:
return "g"
rgb = ['r','g','b']
while True:
time = 0
W_dic = {}
W_que = deque([])
worm = tuple(str(input()))
if worm == ('0',):
break
else:
W_que.append([time,worm])
W_dic[worm] = time
while True:
now_list = W_que.popleft()
now_time = now_list[0]
now_worm = now_list[1]
if now_worm.count('r') == len(now_worm) or now_worm.count('g') == len(now_worm) or now_worm.count('b') == len(now_worm):
print(now_time)
break
a_0 = now_worm[0]
for i in range(1,len(now_worm)):
a_1 = now_worm[i]
rgb_c = rgb
worm_c = list(now_worm)
if a_0 != a_1:
coll = change(a_0,a_1)
worm_c[i - 1] = coll
worm_c[i] = coll
if not tuple(worm_c) in W_dic:
W_dic[tuple(worm_c)] = now_time + 1
W_que.append([now_time + 1,worm_c])
a_0 = a_1
if len(W_que) == 0:
print("NA")
break
``` | output | 1 | 47,856 | 7 | 95,713 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. A of the Aizu Institute of Biological Research discovered a mysterious insect on a certain southern island. The shape is elongated like a hornworm, but since one segment is shaped like a ball, it looks like a beaded ball connected by a thread. What was strange was that there were many variations in body color, and some insects changed their body color over time. It seems that the color of the somites of all insects is limited to either red, green or blue, but the color of the somites changes every second, and finally all the somites become the same color and settle down. In some cases, the color seemed to keep changing no matter how long I waited.
<image>
As I investigated, I found that all the body segments usually have the same color, but after being surprised or excited by something, the color of the body segments changes without permission. It turns out that once the color of the segment changes, it will continue to change until all the segment are the same color again.
Dr. A was curiously observing how the color changed when he caught and excited many of these insects, but the way the color changed during the color change is as follows. I noticed that there is regularity.
* The color changes only in one pair of two adjacent somites of different colors, and the colors of the other somites do not change. However, when there are multiple such pairs, it is not possible to predict in advance which pair will change color.
* Such a pair will change to a color that is neither of the colors of the two segmentes at the same time (for example, if the green and red segment are adjacent, they will change to blue at the same time).
<image>
The figure above shows all the changes in the color of the insects up to 2 seconds later. Suppose you have an insect that has the color shown in the upper part of the figure. At this time, there are 3 pairs of adjacent somites of different colors, so after 1 second, it will change to one of the 3 colors drawn side by side in the middle row. After 1 second, all the segments can turn green after 2 seconds (second from the left in the lower row of the figure) when they change to the two on the left side of the middle row. On the other hand, when it changes like the one on the far right in the middle row after 1 second, all the segments do not change to the same color after 2 seconds.
He decided to predict if all the insect segments in front of him could be the same color, and if so, at the earliest, how many seconds later.
Create a program that takes the color sequence of the insect body segments in front of you as input and outputs the shortest time required for all the insect body segments to have the same color in seconds. However, if there is no possibility that the colors will be the same, output "NA (half-width uppercase letters)". In addition, the color sequence of the insect body segment is represented by a character string consisting of r (red), g (green), and b (blue) of 2 or more and 10 or less.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. For each dataset, one string is given on one line that represents information about the insect's somites.
The number of datasets does not exceed 100.
Output
For each dataset, the minimum time (integer in seconds) or NA required for all segment colors to be the same is output on one line.
Example
Input
rbgrg
rbbgbbr
bgr
bgrbrgbr
bggrgbgrr
gbrggrbggr
rrrrr
bgbr
0
Output
5
7
1
6
NA
8
0
4 | instruction | 0 | 47,857 | 7 | 95,714 |
"Correct Solution:
```
from collections import deque
def new_color(s, i, rgb):
for color in rgb:
if color != s[i] and color != s[i + 1]:
break
return s[:i] + color * 2 + s[i + 2:]
def solve(s):
length = len(s)
monos = ["r" * length, "g" * length, "b" * length]
if s in monos:
print(0)
return
dic = {s:0}
rgb = "rgb"
que = deque()
app = que.append
pop = que.popleft
que.append((s, 0))
while que:
colors, score = pop()
score += 1
temp = colors[0]
for i in range(1, length):
ci = colors[i]
if ci != temp:
new = new_color(colors, i - 1, rgb)
if new in monos:
print(score)
return
if new not in dic:
dic[new] = score
app((new, score))
temp = ci
print("NA")
def main():
while True:
s = input()
if s == "0":
break
solve(s)
main()
``` | output | 1 | 47,857 | 7 | 95,715 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. A of the Aizu Institute of Biological Research discovered a mysterious insect on a certain southern island. The shape is elongated like a hornworm, but since one segment is shaped like a ball, it looks like a beaded ball connected by a thread. What was strange was that there were many variations in body color, and some insects changed their body color over time. It seems that the color of the somites of all insects is limited to either red, green or blue, but the color of the somites changes every second, and finally all the somites become the same color and settle down. In some cases, the color seemed to keep changing no matter how long I waited.
<image>
As I investigated, I found that all the body segments usually have the same color, but after being surprised or excited by something, the color of the body segments changes without permission. It turns out that once the color of the segment changes, it will continue to change until all the segment are the same color again.
Dr. A was curiously observing how the color changed when he caught and excited many of these insects, but the way the color changed during the color change is as follows. I noticed that there is regularity.
* The color changes only in one pair of two adjacent somites of different colors, and the colors of the other somites do not change. However, when there are multiple such pairs, it is not possible to predict in advance which pair will change color.
* Such a pair will change to a color that is neither of the colors of the two segmentes at the same time (for example, if the green and red segment are adjacent, they will change to blue at the same time).
<image>
The figure above shows all the changes in the color of the insects up to 2 seconds later. Suppose you have an insect that has the color shown in the upper part of the figure. At this time, there are 3 pairs of adjacent somites of different colors, so after 1 second, it will change to one of the 3 colors drawn side by side in the middle row. After 1 second, all the segments can turn green after 2 seconds (second from the left in the lower row of the figure) when they change to the two on the left side of the middle row. On the other hand, when it changes like the one on the far right in the middle row after 1 second, all the segments do not change to the same color after 2 seconds.
He decided to predict if all the insect segments in front of him could be the same color, and if so, at the earliest, how many seconds later.
Create a program that takes the color sequence of the insect body segments in front of you as input and outputs the shortest time required for all the insect body segments to have the same color in seconds. However, if there is no possibility that the colors will be the same, output "NA (half-width uppercase letters)". In addition, the color sequence of the insect body segment is represented by a character string consisting of r (red), g (green), and b (blue) of 2 or more and 10 or less.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. For each dataset, one string is given on one line that represents information about the insect's somites.
The number of datasets does not exceed 100.
Output
For each dataset, the minimum time (integer in seconds) or NA required for all segment colors to be the same is output on one line.
Example
Input
rbgrg
rbbgbbr
bgr
bgrbrgbr
bggrgbgrr
gbrggrbggr
rrrrr
bgbr
0
Output
5
7
1
6
NA
8
0
4 | instruction | 0 | 47,858 | 7 | 95,716 |
"Correct Solution:
```
from collections import deque
def isOneColor(w):
sw = sorted(w)
if sw[0] == sw[-1]:
return True
else:
return False
def nextColor(c1, c2):
cs = ['r', 'g', 'b']
cs.remove(c1)
cs.remove(c2)
return cs[0]
def bfs(iw):
q = deque()
s = set()
s.add(str(iw))
q.append([iw, 0])
while q:
w, t = q.popleft()
if isOneColor(w):
return t
else:
for i in range(len(w)-1):
if w[i] != w[i+1]:
nw = w.copy()
nc = nextColor(nw[i], nw[i+1])
nw[i], nw[i+1] = nc, nc
if str(nw) not in s:
s.add(str(nw))
q.append([nw, t+1])
return -1
if __name__ == '__main__':
while True:
w = list(input())
if w == ['0']:
break
t = bfs(w)
if t == -1:
print('NA')
else:
print(t)
``` | output | 1 | 47,858 | 7 | 95,717 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. A of the Aizu Institute of Biological Research discovered a mysterious insect on a certain southern island. The shape is elongated like a hornworm, but since one segment is shaped like a ball, it looks like a beaded ball connected by a thread. What was strange was that there were many variations in body color, and some insects changed their body color over time. It seems that the color of the somites of all insects is limited to either red, green or blue, but the color of the somites changes every second, and finally all the somites become the same color and settle down. In some cases, the color seemed to keep changing no matter how long I waited.
<image>
As I investigated, I found that all the body segments usually have the same color, but after being surprised or excited by something, the color of the body segments changes without permission. It turns out that once the color of the segment changes, it will continue to change until all the segment are the same color again.
Dr. A was curiously observing how the color changed when he caught and excited many of these insects, but the way the color changed during the color change is as follows. I noticed that there is regularity.
* The color changes only in one pair of two adjacent somites of different colors, and the colors of the other somites do not change. However, when there are multiple such pairs, it is not possible to predict in advance which pair will change color.
* Such a pair will change to a color that is neither of the colors of the two segmentes at the same time (for example, if the green and red segment are adjacent, they will change to blue at the same time).
<image>
The figure above shows all the changes in the color of the insects up to 2 seconds later. Suppose you have an insect that has the color shown in the upper part of the figure. At this time, there are 3 pairs of adjacent somites of different colors, so after 1 second, it will change to one of the 3 colors drawn side by side in the middle row. After 1 second, all the segments can turn green after 2 seconds (second from the left in the lower row of the figure) when they change to the two on the left side of the middle row. On the other hand, when it changes like the one on the far right in the middle row after 1 second, all the segments do not change to the same color after 2 seconds.
He decided to predict if all the insect segments in front of him could be the same color, and if so, at the earliest, how many seconds later.
Create a program that takes the color sequence of the insect body segments in front of you as input and outputs the shortest time required for all the insect body segments to have the same color in seconds. However, if there is no possibility that the colors will be the same, output "NA (half-width uppercase letters)". In addition, the color sequence of the insect body segment is represented by a character string consisting of r (red), g (green), and b (blue) of 2 or more and 10 or less.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. For each dataset, one string is given on one line that represents information about the insect's somites.
The number of datasets does not exceed 100.
Output
For each dataset, the minimum time (integer in seconds) or NA required for all segment colors to be the same is output on one line.
Example
Input
rbgrg
rbbgbbr
bgr
bgrbrgbr
bggrgbgrr
gbrggrbggr
rrrrr
bgbr
0
Output
5
7
1
6
NA
8
0
4 | instruction | 0 | 47,859 | 7 | 95,718 |
"Correct Solution:
```
def solve():
while True:
worm = input()
if worm == '0':
break
q = [worm]
m = {('r', 'g'): 'bb', ('g', 'r'): 'bb', ('r', 'b'): 'gg',
('b', 'r'): 'gg', ('g', 'b'): 'rr', ('b', 'g'): 'rr'}
d = {worm: True}
ans = 0
while q:
nq = []
for w in q:
flag = True
for i, t in enumerate(zip(w, w[1:])):
s1, s2 = t
if s1 != s2:
flag = False
nw = w[:i] + m[t] + w[i+2:]
if nw not in d:
d[nw] = True
nq.append(nw)
if flag:
print(ans)
break
else:
q = nq[:]
ans += 1
continue
break
else:
print('NA')
solve()
``` | output | 1 | 47,859 | 7 | 95,719 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. A of the Aizu Institute of Biological Research discovered a mysterious insect on a certain southern island. The shape is elongated like a hornworm, but since one segment is shaped like a ball, it looks like a beaded ball connected by a thread. What was strange was that there were many variations in body color, and some insects changed their body color over time. It seems that the color of the somites of all insects is limited to either red, green or blue, but the color of the somites changes every second, and finally all the somites become the same color and settle down. In some cases, the color seemed to keep changing no matter how long I waited.
<image>
As I investigated, I found that all the body segments usually have the same color, but after being surprised or excited by something, the color of the body segments changes without permission. It turns out that once the color of the segment changes, it will continue to change until all the segment are the same color again.
Dr. A was curiously observing how the color changed when he caught and excited many of these insects, but the way the color changed during the color change is as follows. I noticed that there is regularity.
* The color changes only in one pair of two adjacent somites of different colors, and the colors of the other somites do not change. However, when there are multiple such pairs, it is not possible to predict in advance which pair will change color.
* Such a pair will change to a color that is neither of the colors of the two segmentes at the same time (for example, if the green and red segment are adjacent, they will change to blue at the same time).
<image>
The figure above shows all the changes in the color of the insects up to 2 seconds later. Suppose you have an insect that has the color shown in the upper part of the figure. At this time, there are 3 pairs of adjacent somites of different colors, so after 1 second, it will change to one of the 3 colors drawn side by side in the middle row. After 1 second, all the segments can turn green after 2 seconds (second from the left in the lower row of the figure) when they change to the two on the left side of the middle row. On the other hand, when it changes like the one on the far right in the middle row after 1 second, all the segments do not change to the same color after 2 seconds.
He decided to predict if all the insect segments in front of him could be the same color, and if so, at the earliest, how many seconds later.
Create a program that takes the color sequence of the insect body segments in front of you as input and outputs the shortest time required for all the insect body segments to have the same color in seconds. However, if there is no possibility that the colors will be the same, output "NA (half-width uppercase letters)". In addition, the color sequence of the insect body segment is represented by a character string consisting of r (red), g (green), and b (blue) of 2 or more and 10 or less.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. For each dataset, one string is given on one line that represents information about the insect's somites.
The number of datasets does not exceed 100.
Output
For each dataset, the minimum time (integer in seconds) or NA required for all segment colors to be the same is output on one line.
Example
Input
rbgrg
rbbgbbr
bgr
bgrbrgbr
bggrgbgrr
gbrggrbggr
rrrrr
bgbr
0
Output
5
7
1
6
NA
8
0
4 | instruction | 0 | 47,860 | 7 | 95,720 |
"Correct Solution:
```
from collections import deque
def new_color(s, i, color):
return s[:i] + color * 2 + s[i + 2:]
def solve(s):
length = len(s)
monos = ["r" * length, "g" * length, "b" * length]
if s in monos:
print(0)
return
rgb = "rgb"
another = {("r", "g"):"b", ("g", "r"):"b",
("r", "b"):"g", ("b", "r"):"g",
("g", "b"):"r", ("b", "g"):"r"}
dic = {s:0}
que = deque()
app = que.append
pop = que.popleft
que.append((s, 0))
while que:
colors, score = pop()
score += 1
temp = colors[0]
for i in range(1, length):
ci = colors[i]
if ci != temp:
new = new_color(colors, i - 1, another[(ci, temp)])
if new in monos:
print(score)
return
if new not in dic:
dic[new] = score
app((new, score))
temp = ci
print("NA")
def main():
while True:
s = input()
if s == "0":
break
solve(s)
main()
``` | output | 1 | 47,860 | 7 | 95,721 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. A of the Aizu Institute of Biological Research discovered a mysterious insect on a certain southern island. The shape is elongated like a hornworm, but since one segment is shaped like a ball, it looks like a beaded ball connected by a thread. What was strange was that there were many variations in body color, and some insects changed their body color over time. It seems that the color of the somites of all insects is limited to either red, green or blue, but the color of the somites changes every second, and finally all the somites become the same color and settle down. In some cases, the color seemed to keep changing no matter how long I waited.
<image>
As I investigated, I found that all the body segments usually have the same color, but after being surprised or excited by something, the color of the body segments changes without permission. It turns out that once the color of the segment changes, it will continue to change until all the segment are the same color again.
Dr. A was curiously observing how the color changed when he caught and excited many of these insects, but the way the color changed during the color change is as follows. I noticed that there is regularity.
* The color changes only in one pair of two adjacent somites of different colors, and the colors of the other somites do not change. However, when there are multiple such pairs, it is not possible to predict in advance which pair will change color.
* Such a pair will change to a color that is neither of the colors of the two segmentes at the same time (for example, if the green and red segment are adjacent, they will change to blue at the same time).
<image>
The figure above shows all the changes in the color of the insects up to 2 seconds later. Suppose you have an insect that has the color shown in the upper part of the figure. At this time, there are 3 pairs of adjacent somites of different colors, so after 1 second, it will change to one of the 3 colors drawn side by side in the middle row. After 1 second, all the segments can turn green after 2 seconds (second from the left in the lower row of the figure) when they change to the two on the left side of the middle row. On the other hand, when it changes like the one on the far right in the middle row after 1 second, all the segments do not change to the same color after 2 seconds.
He decided to predict if all the insect segments in front of him could be the same color, and if so, at the earliest, how many seconds later.
Create a program that takes the color sequence of the insect body segments in front of you as input and outputs the shortest time required for all the insect body segments to have the same color in seconds. However, if there is no possibility that the colors will be the same, output "NA (half-width uppercase letters)". In addition, the color sequence of the insect body segment is represented by a character string consisting of r (red), g (green), and b (blue) of 2 or more and 10 or less.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. For each dataset, one string is given on one line that represents information about the insect's somites.
The number of datasets does not exceed 100.
Output
For each dataset, the minimum time (integer in seconds) or NA required for all segment colors to be the same is output on one line.
Example
Input
rbgrg
rbbgbbr
bgr
bgrbrgbr
bggrgbgrr
gbrggrbggr
rrrrr
bgbr
0
Output
5
7
1
6
NA
8
0
4 | instruction | 0 | 47,861 | 7 | 95,722 |
"Correct Solution:
```
from collections import deque
def new_color(s, i, color):
return s[:i - 1] + color * 2 + s[i + 1:]
def solve(s):
length = len(s)
monos = ["r" * length, "g" * length, "b" * length]
if s in monos:
print(0)
return
another = {("r", "g"):"b", ("g", "r"):"b",
("r", "b"):"g", ("b", "r"):"g",
("g", "b"):"r", ("b", "g"):"r"}
visited = {s:True}
que = deque()
app = que.append
pop = que.popleft
que.append((s, 0))
while que:
colors, score = pop()
score += 1
temp = colors[0]
for i in range(1, length):
ci = colors[i]
if ci != temp:
new = new_color(colors, i, another[(ci, temp)])
if new in monos:
print(score)
return
if new not in visited:
visited[new] = True
app((new, score))
temp = ci
print("NA")
def main():
while True:
s = input()
if s == "0":
break
solve(s)
main()
``` | output | 1 | 47,861 | 7 | 95,723 |
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