message stringlengths 2 433k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary matrix a of size n Γ m. A binary matrix is a matrix where each element is either 0 or 1.
You may perform some (possibly zero) operations with this matrix. During each operation you can inverse the row of this matrix or a column of this matrix. Formally, inverting a row is changing all values in this row to the opposite (0 to 1, 1 to 0). Inverting a column is changing all values in this column to the opposite.
Your task is to sort the initial matrix by some sequence of such operations. The matrix is considered sorted if the array [a_{1, 1}, a_{1, 2}, ..., a_{1, m}, a_{2, 1}, a_{2, 2}, ..., a_{2, m}, ..., a_{n, m - 1}, a_{n, m}] is sorted in non-descending order.
Input
The first line of the input contains two integers n and m (1 β€ n, m β€ 200) β the number of rows and the number of columns in the matrix.
The next n lines contain m integers each. The j-th element in the i-th line is a_{i, j} (0 β€ a_{i, j} β€ 1) β the element of a at position (i, j).
Output
If it is impossible to obtain a sorted matrix, print "NO" in the first line.
Otherwise print "YES" in the first line. In the second line print a string r of length n. The i-th character r_i of this string should be '1' if the i-th row of the matrix is inverted and '0' otherwise. In the third line print a string c of length m. The j-th character c_j of this string should be '1' if the j-th column of the matrix is inverted and '0' otherwise. If there are multiple answers, you can print any.
Examples
Input
2 2
1 1
0 1
Output
YES
00
10
Input
3 4
0 0 0 1
0 0 0 0
1 1 1 1
Output
YES
010
0000
Input
3 3
0 0 0
1 0 1
1 1 0
Output
NO | instruction | 0 | 14,218 | 12 | 28,436 |
Tags: brute force, constructive algorithms
Correct Solution:
```
rint = lambda: int(input())
rmint = lambda: map(int, input().split())
rlist = lambda: list(rmint())
n, m = rmint()
a = []
for i in range(n): a.append(rlist())
ax = []
for i in range(n): ax.append(0)
ay = []
for i in range(m): ay.append(0)
def inv(x,y): a[x][y] = 1 - a[x][y]
def invx(x):
for y in range(m): inv(x,y)
ax[x] = 1 - ax[x]
def invy(y):
for x in range(n): inv(x,y)
ay[y] = 1 - ay[y]
def ex():
print("YES")
for i in ax: print(i,end='')
print()
for i in ay: print(i,end='')
exit(0)
if(n == 1):
ay = a[0]
ex()
def D(x,v):
for y in range(m):
if a[x][y] != v: invy(y)
fx = -1; fy = -1
for i in range(n):
for j in range(m-1):
if a[i][j] != a[i][j+1]:
fx = i
fy = j
if fx < 0:
for i in range(n):
if a[i][0]: invx(i)
else:
if a[fx][fy] > a[fx][fy+1]: invx(fx)
for i in range(fx):
if a[i][0]: invx(i)
for i in range(fx+1,n):
if not a[i][0]: invx(i)
srt = 1
for i in range(n):
for j in range(m):
if i+1 == n and j+1 == m: break
t = i*m + j + 1
x = t // m
y = t % m
if a[x][y] < a[i][j]: srt = 0
if srt: ex()
D(0, 0)
D(n-1, 1)
print("NO")
``` | output | 1 | 14,218 | 12 | 28,437 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary matrix a of size n Γ m. A binary matrix is a matrix where each element is either 0 or 1.
You may perform some (possibly zero) operations with this matrix. During each operation you can inverse the row of this matrix or a column of this matrix. Formally, inverting a row is changing all values in this row to the opposite (0 to 1, 1 to 0). Inverting a column is changing all values in this column to the opposite.
Your task is to sort the initial matrix by some sequence of such operations. The matrix is considered sorted if the array [a_{1, 1}, a_{1, 2}, ..., a_{1, m}, a_{2, 1}, a_{2, 2}, ..., a_{2, m}, ..., a_{n, m - 1}, a_{n, m}] is sorted in non-descending order.
Input
The first line of the input contains two integers n and m (1 β€ n, m β€ 200) β the number of rows and the number of columns in the matrix.
The next n lines contain m integers each. The j-th element in the i-th line is a_{i, j} (0 β€ a_{i, j} β€ 1) β the element of a at position (i, j).
Output
If it is impossible to obtain a sorted matrix, print "NO" in the first line.
Otherwise print "YES" in the first line. In the second line print a string r of length n. The i-th character r_i of this string should be '1' if the i-th row of the matrix is inverted and '0' otherwise. In the third line print a string c of length m. The j-th character c_j of this string should be '1' if the j-th column of the matrix is inverted and '0' otherwise. If there are multiple answers, you can print any.
Examples
Input
2 2
1 1
0 1
Output
YES
00
10
Input
3 4
0 0 0 1
0 0 0 0
1 1 1 1
Output
YES
010
0000
Input
3 3
0 0 0
1 0 1
1 1 0
Output
NO | instruction | 0 | 14,219 | 12 | 28,438 |
Tags: brute force, constructive algorithms
Correct Solution:
```
import sys
n=0
m=0
sq=[[0]*(310)]*(310)
ansx=[]
ansy=[]
tmp=[[0]*(310)]*(310)
def chk():
val=[]
for i in range(0,n):
for j in range(0,m):
val.append(tmp[i][j])
Len=len(val)
for i in range(1,Len):
if val[i-1]>val[i]:
return False
return True
n,m=map(int,input().split())
for i in range(0,n):
sq[i]=list(map(int,input().split()))
for i in range(0,n):
tmp[i]=list(sq[i])
for i in range(0,m):
if tmp[0][i]==1:
ansy.append(1)
for j in range (0,n):
tmp[j][i]=1-tmp[j][i]
else:
ansy.append(0)
op=0
for i in range(0,n):
flag0=0
flag1=0
for j in range(0,m):
if tmp[i][j]==0:
flag0=1
else:
flag1=1
if flag0==1 and flag1==1:
op=1
if tmp[i][0]==1:
ansx.append(1)
for j in range (0,m):
tmp[i][j]=1-tmp[i][j]
else:
ansx.append(0)
elif flag0==1:
if op==0:
ansx.append(0)
elif op==1:
ansx.append(1)
for j in range(0,m):
tmp[i][j]=1-tmp[i][j]
elif flag1==1:
if op==0:
ansx.append(1)
for j in range(0,m):
tmp[i][j]=1-tmp[i][j]
else:
ansx.append(0)
if chk():
print("YES")
for i in range(0,n):
print(ansx[i],end="")
print()
for i in range(0,m):
print(ansy[i],end="")
sys.exit()
for i in range(0,n):
tmp[i]=list(sq[i])
ansx=[]
ansy=[]
for i in range(0,m):
if (tmp[n-1][i]==1):
ansy.append(0)
else:
ansy.append(1)
for j in range(0,n):
tmp[j][i]=1-tmp[j][i]
op=0
for i in range(0,n):
flag0=0
flag1=0
for j in range(0,m):
if tmp[i][j]==1:
flag1=1
else:
flag0=1
if flag1==1 and flag0==1:
op=1
if tmp[i][0]==1:
ansx.append(1)
for j in range(0,m):
tmp[i][j]=1-tmp[i][j]
else:
ansx.append(0)
elif flag0==1:
if op==0:
ansx.append(0)
else:
ansx.append(1)
for j in range(0,m):
tmp[i][j]=1-tmp[i][j]
elif flag1==1:
if op==1:
ansx.append(0)
else:
ansx.append(1)
for j in range(0,m):
tmp[i][j]=1-tmp[i][j]
if chk():
print("YES")
for i in range(0,n):
print(ansx[i],end="")
print()
for i in range(0,m):
print(ansy[i],end="")
sys.exit()
print("NO")
###
``` | output | 1 | 14,219 | 12 | 28,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary matrix a of size n Γ m. A binary matrix is a matrix where each element is either 0 or 1.
You may perform some (possibly zero) operations with this matrix. During each operation you can inverse the row of this matrix or a column of this matrix. Formally, inverting a row is changing all values in this row to the opposite (0 to 1, 1 to 0). Inverting a column is changing all values in this column to the opposite.
Your task is to sort the initial matrix by some sequence of such operations. The matrix is considered sorted if the array [a_{1, 1}, a_{1, 2}, ..., a_{1, m}, a_{2, 1}, a_{2, 2}, ..., a_{2, m}, ..., a_{n, m - 1}, a_{n, m}] is sorted in non-descending order.
Input
The first line of the input contains two integers n and m (1 β€ n, m β€ 200) β the number of rows and the number of columns in the matrix.
The next n lines contain m integers each. The j-th element in the i-th line is a_{i, j} (0 β€ a_{i, j} β€ 1) β the element of a at position (i, j).
Output
If it is impossible to obtain a sorted matrix, print "NO" in the first line.
Otherwise print "YES" in the first line. In the second line print a string r of length n. The i-th character r_i of this string should be '1' if the i-th row of the matrix is inverted and '0' otherwise. In the third line print a string c of length m. The j-th character c_j of this string should be '1' if the j-th column of the matrix is inverted and '0' otherwise. If there are multiple answers, you can print any.
Examples
Input
2 2
1 1
0 1
Output
YES
00
10
Input
3 4
0 0 0 1
0 0 0 0
1 1 1 1
Output
YES
010
0000
Input
3 3
0 0 0
1 0 1
1 1 0
Output
NO | instruction | 0 | 14,220 | 12 | 28,440 |
Tags: brute force, constructive algorithms
Correct Solution:
```
import sys
sys.setrecursionlimit(2000)
from collections import Counter
from functools import reduce
# sys.stdin.readline()
def get_info(row):
s = set(row)
if(len(s) == 1):
homogeneous = True
else:
homogeneous = False
is_sorted = True
for ind, val in enumerate(row[1:]):
if(val < row[ind]):
is_sorted = False
is_reverse_sorted = True
for ind in range(n-1, 1, -1):
if(row[ind-1] > row[ind]):
is_reversed_sorted = False
return (homogeneous, is_sorted, 0 in s, 1 in s, is_reverse_sorted)
def mc(A, col):
for row in A:
row[col] = 1 - row[col]
def mr(A, row):
for ind, val in enumerate(A[row]):
A[row][ind] = 1 - A[row][ind]
def is_pos(A, flip, start_row, end_row, rows, cols):
pos = True
for row in range(start_row, end_row+1):
info = get_info(A[row])
if(info[0]):
must_b = '0' if not flip else '1'
if(must_b == '0'):
if(info[2]):
pass
else:
mr(A, row)
rows[row] = '1'
else:
if(info[3]):
pass
else:
mr(A, row)
rows[row] = '1'
else:
if(flip):
pos = False
break
elif(info[1]):
flip = True
elif(info[4]):
mr(A, row)
rows[row] = '1'
flip = True
else:
pos = False
break
return pos
def force_row_to_val(A, row, val, rows, cols):
info = get_info(A[row])
check_ind = 2 if val == 1 else 3
if(info[0]):
if(info[check_ind]):
mr(A, row)
rows[row] = '1'
else:
pass
else:
for i in range(m):
if(A[row][i] != val):
mc(A, i)
cols[i] = '1'
if __name__ == "__main__":
# single variables
n, m = [int(val) for val in sys.stdin.readline().split()]
A = [[int(val) for val in sys.stdin.readline().split()] for row in range(n)]
# ans
rows1 = ['0'] * n
cols1 = ['0'] * m
rows2 = ['0'] * n
cols2 = ['0'] * m
from copy import deepcopy
A1 = deepcopy(A)
A2 = deepcopy(A)
force_row_to_val(A1, -1, 1, rows1, cols1)
pos1 = is_pos(A1, False, 0, n-2, rows1, cols1)
force_row_to_val(A2, 0, 0, rows2, cols2)
pos2 = is_pos(A2, False, 1, n-1, rows2, cols2)
if(pos1):
cols = cols1
rows = rows1
elif(pos2):
cols = cols2
rows = rows2
if(pos1 or pos2):
print("YES")
for val in rows:
print(val, end='')
print("")
for val in cols:
print(val, end='')
print("")
else:
print("NO")
``` | output | 1 | 14,220 | 12 | 28,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary matrix a of size n Γ m. A binary matrix is a matrix where each element is either 0 or 1.
You may perform some (possibly zero) operations with this matrix. During each operation you can inverse the row of this matrix or a column of this matrix. Formally, inverting a row is changing all values in this row to the opposite (0 to 1, 1 to 0). Inverting a column is changing all values in this column to the opposite.
Your task is to sort the initial matrix by some sequence of such operations. The matrix is considered sorted if the array [a_{1, 1}, a_{1, 2}, ..., a_{1, m}, a_{2, 1}, a_{2, 2}, ..., a_{2, m}, ..., a_{n, m - 1}, a_{n, m}] is sorted in non-descending order.
Input
The first line of the input contains two integers n and m (1 β€ n, m β€ 200) β the number of rows and the number of columns in the matrix.
The next n lines contain m integers each. The j-th element in the i-th line is a_{i, j} (0 β€ a_{i, j} β€ 1) β the element of a at position (i, j).
Output
If it is impossible to obtain a sorted matrix, print "NO" in the first line.
Otherwise print "YES" in the first line. In the second line print a string r of length n. The i-th character r_i of this string should be '1' if the i-th row of the matrix is inverted and '0' otherwise. In the third line print a string c of length m. The j-th character c_j of this string should be '1' if the j-th column of the matrix is inverted and '0' otherwise. If there are multiple answers, you can print any.
Examples
Input
2 2
1 1
0 1
Output
YES
00
10
Input
3 4
0 0 0 1
0 0 0 0
1 1 1 1
Output
YES
010
0000
Input
3 3
0 0 0
1 0 1
1 1 0
Output
NO | instruction | 0 | 14,221 | 12 | 28,442 |
Tags: brute force, constructive algorithms
Correct Solution:
```
import sys
n=0
m=0
sq=[[0]*(310)]*(310)
ansx=[]
ansy=[]
tmp=[[0]*(310)]*(310)
def chk():
val=[]
for i in range(0,n):
for j in range(0,m):
val.append(tmp[i][j])
Len=len(val)
for i in range(1,Len):
if val[i-1]>val[i]:
return False
return True
n,m=map(int,input().split())
for i in range(0,n):
sq[i]=list(map(int,input().split()))
for i in range(0,n):
tmp[i]=list(sq[i])
for i in range(0,m):
if tmp[0][i]==1:
ansy.append(1)
for j in range (0,n):
tmp[j][i]=1-tmp[j][i]
else:
ansy.append(0)
op=0
for i in range(0,n):
flag0=0
flag1=0
for j in range(0,m):
if tmp[i][j]==0:
flag0=1
else:
flag1=1
if flag0==1 and flag1==1:
op=1
if tmp[i][0]==1:
ansx.append(1)
for j in range (0,m):
tmp[i][j]=1-tmp[i][j]
else:
ansx.append(0)
elif flag0==1:
if op==0:
ansx.append(0)
elif op==1:
ansx.append(1)
for j in range(0,m):
tmp[i][j]=1-tmp[i][j]
elif flag1==1:
if op==0:
ansx.append(1)
for j in range(0,m):
tmp[i][j]=1-tmp[i][j]
else:
ansx.append(0)
if chk():
print("YES")
for i in range(0,n):
print(ansx[i],end="")
print()
for i in range(0,m):
print(ansy[i],end="")
sys.exit()
for i in range(0,n):
tmp[i]=list(sq[i])
ansx=[]
ansy=[]
for i in range(0,m):
if (tmp[n-1][i]==1):
ansy.append(0)
else:
ansy.append(1)
for j in range(0,n):
tmp[j][i]=1-tmp[j][i]
op=0
for i in range(0,n):
flag0=0
flag1=0
for j in range(0,m):
if tmp[i][j]==1:
flag1=1
else:
flag0=1
if flag1==1 and flag0==1:
op=1
if tmp[i][0]==1:
ansx.append(1)
for j in range(0,m):
tmp[i][j]=1-tmp[i][j]
else:
ansx.append(0)
elif flag0==1:
if op==0:
ansx.append(0)
else:
ansx.append(1)
for j in range(0,m):
tmp[i][j]=1-tmp[i][j]
elif flag1==1:
if op==1:
ansx.append(0)
else:
ansx.append(1)
for j in range(0,m):
tmp[i][j]=1-tmp[i][j]
if chk():
print("YES")
for i in range(0,n):
print(ansx[i],end="")
print()
for i in range(0,m):
print(ansy[i],end="")
sys.exit()
print("NO")
``` | output | 1 | 14,221 | 12 | 28,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix a of size n Γ m. A binary matrix is a matrix where each element is either 0 or 1.
You may perform some (possibly zero) operations with this matrix. During each operation you can inverse the row of this matrix or a column of this matrix. Formally, inverting a row is changing all values in this row to the opposite (0 to 1, 1 to 0). Inverting a column is changing all values in this column to the opposite.
Your task is to sort the initial matrix by some sequence of such operations. The matrix is considered sorted if the array [a_{1, 1}, a_{1, 2}, ..., a_{1, m}, a_{2, 1}, a_{2, 2}, ..., a_{2, m}, ..., a_{n, m - 1}, a_{n, m}] is sorted in non-descending order.
Input
The first line of the input contains two integers n and m (1 β€ n, m β€ 200) β the number of rows and the number of columns in the matrix.
The next n lines contain m integers each. The j-th element in the i-th line is a_{i, j} (0 β€ a_{i, j} β€ 1) β the element of a at position (i, j).
Output
If it is impossible to obtain a sorted matrix, print "NO" in the first line.
Otherwise print "YES" in the first line. In the second line print a string r of length n. The i-th character r_i of this string should be '1' if the i-th row of the matrix is inverted and '0' otherwise. In the third line print a string c of length m. The j-th character c_j of this string should be '1' if the j-th column of the matrix is inverted and '0' otherwise. If there are multiple answers, you can print any.
Examples
Input
2 2
1 1
0 1
Output
YES
00
10
Input
3 4
0 0 0 1
0 0 0 0
1 1 1 1
Output
YES
010
0000
Input
3 3
0 0 0
1 0 1
1 1 0
Output
NO
Submitted Solution:
```
from collections import defaultdict
from collections.abc import Iterable
from copy import deepcopy
import sys
input = sys.stdin.readline
class BitSet:
BITS = 0
M = 0
def __init__(self, mask):
if isinstance(mask, Iterable):
mask = "".join(map(str, mask))
if isinstance(mask, str):
mask = int(mask, 2)
self.mask = mask
def comp(self):
return BitSet(~self.mask & (BitSet.M - 1))
def __eq__(self, other):
return self.mask == other.mask
def __hash__(self):
return hash(self.mask)
def __repr__(self):
return bin(self.mask)[2:].zfill(BitSet.BITS)
def __getitem__(self, bit):
if bit < 0:
raise TypeError("Bit position can't be negative")
return (self.mask >> bit) & 1
def valid_masks(rows):
d = defaultdict(lambda: 0)
for row in rows:
b = BitSet(row)
d[b] += 1
d[b.comp()] += 1
return { k for (k, v) in d.items() if v == len(rows) }
def isgoodrow(row, mask):
m = len(row)
c_row = [ row[i] ^ mask[m - i - 1] for i in range(m) ]
c = 0
for i in range(1, m):
if c_row[i - 1] > c_row[i]:
c += 1
for i in range(1, m):
if c_row[i - 1] < c_row[i]:
c += 1
if c <= 1:
change = 0
for i in range(1, m):
if c_row[i - 1] > c_row[i]:
change = 1
return (True, change)
return (False, False)
def solve(a, flip = False):
n = len(a)
m = len(a[0])
BitSet.BITS = m
BitSet.M = 1 << m
a.insert(n, [1] * m)
a.insert(0, [0] * m)
for i in range(1, n + 2):
up_rows = a[:i]
down_rows = a[i + 1:]
up = valid_masks(up_rows)
down = valid_masks(down_rows)
for mask in up:
comp = mask.comp()
goodrow, change = isgoodrow(a[i], mask)
if comp in down and goodrow:
r_ans = [0] * (n + 2)
c_ans = [ mask[m - c - 1] for c in range(m) ]
for r in range(len(up_rows)):
if BitSet(up_rows[r]) == comp:
r_ans[r] = 1
for r in range(len(down_rows)):
if BitSet(down_rows[r]) == mask:
r_ans[i + 1 + r] = 1
r_ans[i] = change
row_output = "".join(map(str, r_ans[1:n + 1]))
col_output = "".join(map(str, c_ans))
if flip:
row_output, col_output = col_output, row_output
print('YES')
print(row_output)
print(col_output)
exit(0)
def rotate(a):
n = len(a)
m = len(a[0])
b = [ [0] * n for i in range(m) ]
for i in range(n):
for j in range(m):
b[j][i] = a[i][j]
return b
def main():
n, m = map(int, input().split())
a = [ list(map(int, input().split())) for _ in range(n) ]
solve(deepcopy(a))
solve(rotate(a), True)
print('NO')
main()
``` | instruction | 0 | 14,222 | 12 | 28,444 |
No | output | 1 | 14,222 | 12 | 28,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix a of size n Γ m. A binary matrix is a matrix where each element is either 0 or 1.
You may perform some (possibly zero) operations with this matrix. During each operation you can inverse the row of this matrix or a column of this matrix. Formally, inverting a row is changing all values in this row to the opposite (0 to 1, 1 to 0). Inverting a column is changing all values in this column to the opposite.
Your task is to sort the initial matrix by some sequence of such operations. The matrix is considered sorted if the array [a_{1, 1}, a_{1, 2}, ..., a_{1, m}, a_{2, 1}, a_{2, 2}, ..., a_{2, m}, ..., a_{n, m - 1}, a_{n, m}] is sorted in non-descending order.
Input
The first line of the input contains two integers n and m (1 β€ n, m β€ 200) β the number of rows and the number of columns in the matrix.
The next n lines contain m integers each. The j-th element in the i-th line is a_{i, j} (0 β€ a_{i, j} β€ 1) β the element of a at position (i, j).
Output
If it is impossible to obtain a sorted matrix, print "NO" in the first line.
Otherwise print "YES" in the first line. In the second line print a string r of length n. The i-th character r_i of this string should be '1' if the i-th row of the matrix is inverted and '0' otherwise. In the third line print a string c of length m. The j-th character c_j of this string should be '1' if the j-th column of the matrix is inverted and '0' otherwise. If there are multiple answers, you can print any.
Examples
Input
2 2
1 1
0 1
Output
YES
00
10
Input
3 4
0 0 0 1
0 0 0 0
1 1 1 1
Output
YES
010
0000
Input
3 3
0 0 0
1 0 1
1 1 0
Output
NO
Submitted Solution:
```
def inverse_row(row):
inv_row[row]= not inv_row[row]
for i in range(m):
a[row][i]=not a[row][i]
def inverse_col(col):
inv_col[col]= not inv_col[col]
for i in range(n):
a[i][col]=not a[i][col]
def check_row(row):
if a[row][0]<a[row-1][m-1]:
return False
for i in range(1,m):
if a[row][i]<a[row][i-1]:
return False
return True
def check_all():
for i in range(1,n):
if check_row(i):
continue
inverse_row(i)
if check_row(i):
continue
return False
return True
def print_result():
print("YES")
for i in inv_row:
print(int(i),end="")
print("")
for i in inv_col:
print(int(i),end="")
print("")
n,m = [int(i) for i in input().split(" ")]
a=[]
inv_row=[False]*n
inv_col=[False]*m
had_result=False
for i in range(n):
a.append([bool(int(i)) for i in input().split(" ")])
for i in range(m):
if a[0][i]:
inverse_col(i)
if check_all():
print_result()
had_result=True
if not had_result:
for i in range(m-1,-1,-1):
inverse_col(i)
if check_all():
print_result()
had_result=True
break
if not had_result:
print("NO")
``` | instruction | 0 | 14,223 | 12 | 28,446 |
No | output | 1 | 14,223 | 12 | 28,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix a of size n Γ m. A binary matrix is a matrix where each element is either 0 or 1.
You may perform some (possibly zero) operations with this matrix. During each operation you can inverse the row of this matrix or a column of this matrix. Formally, inverting a row is changing all values in this row to the opposite (0 to 1, 1 to 0). Inverting a column is changing all values in this column to the opposite.
Your task is to sort the initial matrix by some sequence of such operations. The matrix is considered sorted if the array [a_{1, 1}, a_{1, 2}, ..., a_{1, m}, a_{2, 1}, a_{2, 2}, ..., a_{2, m}, ..., a_{n, m - 1}, a_{n, m}] is sorted in non-descending order.
Input
The first line of the input contains two integers n and m (1 β€ n, m β€ 200) β the number of rows and the number of columns in the matrix.
The next n lines contain m integers each. The j-th element in the i-th line is a_{i, j} (0 β€ a_{i, j} β€ 1) β the element of a at position (i, j).
Output
If it is impossible to obtain a sorted matrix, print "NO" in the first line.
Otherwise print "YES" in the first line. In the second line print a string r of length n. The i-th character r_i of this string should be '1' if the i-th row of the matrix is inverted and '0' otherwise. In the third line print a string c of length m. The j-th character c_j of this string should be '1' if the j-th column of the matrix is inverted and '0' otherwise. If there are multiple answers, you can print any.
Examples
Input
2 2
1 1
0 1
Output
YES
00
10
Input
3 4
0 0 0 1
0 0 0 0
1 1 1 1
Output
YES
010
0000
Input
3 3
0 0 0
1 0 1
1 1 0
Output
NO
Submitted Solution:
```
from collections import defaultdict
from collections.abc import Iterable
import sys
input = sys.stdin.readline
class BitSet:
M = 0
def __init__(self, mask):
if isinstance(mask, Iterable):
mask = "".join(mask)
if isinstance(mask, str):
mask = int(mask, 2)
self.mask = mask
def comp(self):
return BitSet(~self.mask & (BitSet.M - 1))
def __eq__(self, other):
return self.mask == other.mask
def __hash__(self):
return hash(self.mask)
def __repr__(self):
return bin(self.mask)[2:]
def __getitem__(self, bit):
if bit < 0:
raise TypeError("Bit position can't be negative")
return (self.mask >> bit) & 1
def valid_masks(rows):
d = defaultdict(lambda: 0)
for row in rows:
b = BitSet("".join(map(str, row)))
d[b] += 1
d[b.comp()] += 1
return { k for (k, v) in d.items() if v == len(rows) }
def isgoodrow(row, mask):
m = len(row)
c_row = [ row[i] ^ mask[m - i - 1] for i in range(m) ]
c = 0
for i in range(1, m):
if c_row[i - 1] > c_row[i]:
c += 1
for i in range(1, m):
if c_row[i - 1] < c_row[i]:
c += 1
return (c <= 1, c)
def main():
n, m = map(int, input().split())
a = [ list(map(int, input().split())) for _ in range(n) ]
BitSet.M = 1 << m
a.insert(n, [1] * m)
a.insert(0, [0] * m)
for i in range(1, n + 2):
up_rows = a[:i]
down_rows = a[i + 1:]
up = valid_masks(up_rows)
down = valid_masks(down_rows)
for mask in up:
comp = mask.comp()
goodrow, change = isgoodrow(a[i], mask)
if comp in down and goodrow:
r_ans = [0] * (n + 2)
c_ans = [ mask[m - c - 1] for c in range(m) ]
for r in range(len(up_rows)):
if BitSet("".join(map(str, up_rows[r]))) == comp:
r_ans[r] = 1
for r in range(len(down_rows)):
if BitSet("".join(map(str, down_rows[r]))) == mask:
r_ans[i + 1 + r] = 1
r_ans[r] = change
print('Yes')
print("".join(map(str, r_ans[1:n + 1])))
print("".join(map(str, c_ans)))
return
print('No')
main()
``` | instruction | 0 | 14,224 | 12 | 28,448 |
No | output | 1 | 14,224 | 12 | 28,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix a of size n Γ m. A binary matrix is a matrix where each element is either 0 or 1.
You may perform some (possibly zero) operations with this matrix. During each operation you can inverse the row of this matrix or a column of this matrix. Formally, inverting a row is changing all values in this row to the opposite (0 to 1, 1 to 0). Inverting a column is changing all values in this column to the opposite.
Your task is to sort the initial matrix by some sequence of such operations. The matrix is considered sorted if the array [a_{1, 1}, a_{1, 2}, ..., a_{1, m}, a_{2, 1}, a_{2, 2}, ..., a_{2, m}, ..., a_{n, m - 1}, a_{n, m}] is sorted in non-descending order.
Input
The first line of the input contains two integers n and m (1 β€ n, m β€ 200) β the number of rows and the number of columns in the matrix.
The next n lines contain m integers each. The j-th element in the i-th line is a_{i, j} (0 β€ a_{i, j} β€ 1) β the element of a at position (i, j).
Output
If it is impossible to obtain a sorted matrix, print "NO" in the first line.
Otherwise print "YES" in the first line. In the second line print a string r of length n. The i-th character r_i of this string should be '1' if the i-th row of the matrix is inverted and '0' otherwise. In the third line print a string c of length m. The j-th character c_j of this string should be '1' if the j-th column of the matrix is inverted and '0' otherwise. If there are multiple answers, you can print any.
Examples
Input
2 2
1 1
0 1
Output
YES
00
10
Input
3 4
0 0 0 1
0 0 0 0
1 1 1 1
Output
YES
010
0000
Input
3 3
0 0 0
1 0 1
1 1 0
Output
NO
Submitted Solution:
```
import sys
def change(x, y):
aim_matrix = []
for i in range(n):
t_list = []
for j in range(m):
t_list.append(0 if i * m + j < x * m + y else 1)
aim_matrix.append(t_list)
change_r[0] = aim_matrix[0][0] ^ matrix[0][0]
for i in range(m):
change_c[i] = aim_matrix[0][i] ^ matrix[0][i] ^ change_r[0]
for i in range(1, n):
change_r[i] = aim_matrix[i][0] ^ matrix[i][0] ^ change_c[0]
for j in range(m):
if aim_matrix[i][j] != (matrix[i][j] ^ change_r[i] ^ change_c[j]):
return False
return True
n, m = input().split()
n, m = int(n), int(m)
change_r = [0] * n
change_c = [0] * m
matrix = []
mark = False
for i in range(n):
t = input().split()
t_list = []
for val in t:
t_list.append(int(val))
matrix.append(t_list)
for i in range(n):
for j in range(m):
res = change(i, j)
if res:
print("Yes")
mark = True
for t in range(n):
print(change_r[t], end='')
if t != n - 1:
print(" ", end='')
else:
print()
for t in range(m):
print(change_c[t], end='')
if t != m - 1:
print(" ", end='')
else:
print()
if mark:
break
if not mark:
print("No")
``` | instruction | 0 | 14,225 | 12 | 28,450 |
No | output | 1 | 14,225 | 12 | 28,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a be an array consisting of n numbers. The array's elements are numbered from 1 to n, even is an array consisting of the numerals whose numbers are even in a (eveni = a2i, 1 β€ 2i β€ n), odd is an array consisting of the numberals whose numbers are odd in Π° (oddi = a2i - 1, 1 β€ 2i - 1 β€ n). Then let's define the transformation of array F(a) in the following manner:
* if n > 1, F(a) = F(odd) + F(even), where operation " + " stands for the arrays' concatenation (joining together)
* if n = 1, F(a) = a
Let a be an array consisting of n numbers 1, 2, 3, ..., n. Then b is the result of applying the transformation to the array a (so b = F(a)). You are given m queries (l, r, u, v). Your task is to find for each query the sum of numbers bi, such that l β€ i β€ r and u β€ bi β€ v. You should print the query results modulo mod.
Input
The first line contains three integers n, m, mod (1 β€ n β€ 1018, 1 β€ m β€ 105, 1 β€ mod β€ 109). Next m lines describe the queries. Each query is defined by four integers l, r, u, v (1 β€ l β€ r β€ n, 1 β€ u β€ v β€ 1018).
Please do not use the %lld specificator to read or write 64-bit integers in C++. Use %I64d specificator.
Output
Print m lines each containing an integer β remainder modulo mod of the query result.
Examples
Input
4 5 10000
2 3 4 5
2 4 1 3
1 2 2 4
2 3 3 5
1 3 3 4
Output
0
5
3
3
3
Input
2 5 10000
1 2 2 2
1 1 4 5
1 1 2 5
1 1 1 3
1 2 5 5
Output
2
0
0
1
0
Note
Let's consider the first example. First let's construct an array b = F(a) = F([1, 2, 3, 4]).
* Step 1. F([1, 2, 3, 4]) = F([1, 3]) + F([2, 4])
* Step 2. F([1, 3]) = F([1]) + F([3]) = [1] + [3] = [1, 3]
* Step 3. F([2, 4]) = F([2]) + F([4]) = [2] + [4] = [2, 4]
* Step 4. b = F([1, 2, 3, 4]) = F([1, 3]) + F([2, 4]) = [1, 3] + [2, 4] = [1, 3, 2, 4]
Thus b = [1, 3, 2, 4]. Let's consider the first query l = 2, r = 3, u = 4, v = 5. The second and third positions in the array b do not have numbers in the range [4, 5], so the sum obviously equals zero. Let's consider the second query l = 2, r = 4, u = 1, v = 3. The second and third positions have two numbers that belong to the range [1, 3], their sum equals 5.
Submitted Solution:
```
import sys
line = sys.stdin.readline().split(" ")
N = int(line[0])
queryc = int(line[1])
MOD = int(line[2])
def find (length, idx):
if length == 1:
return 0
if idx < (length + 1) // 2:
return 2 * find((length + 1) // 2, idx)
else:
return 2 * find(length // 2, idx - (length + 1) // 2) + 1
MAX_LG = 60
def n_suff (suff, lens, v):
add = 0
if suff <= v % (1 << lens):
add = 1
return (v >> lens) + add
def sum_n (n):
return n * (n - 1) // 2
def sum_le (x, v):
cnt = 0
sums = 0
if x <= v:
sums += x
cnt += 1
for i in range(MAX_LG - 1, -1, -1):
if x & 1 << i:
suff = x % (1 << i)
ns = n_suff(suff, i + 1, v)
cnt += ns
sums += ns * suff
sums += sum_n(ns) << (i + 1)
return sums + cnt
def query_up (r, v):
if r < 0:
return 0
if v < 0:
return 0
return sum_le(find(N, r), min(v, N - 1))
def query (l, r, u, v):
return query_up(r, v) - query_up(l - 1, v) - query_up(r, u - 1) + query_up(l - 1, u - 1)
for i in range(queryc):
q = sys.stdin.readline().split(" ")
sys.stdout.write(str(query(int(q[0]) - 1, int(q[1]) - 1, int(q[2]) - 1, int(q[3]) - 1)))
sys.stdout.write('\n')
``` | instruction | 0 | 14,226 | 12 | 28,452 |
No | output | 1 | 14,226 | 12 | 28,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4. | instruction | 0 | 14,281 | 12 | 28,562 |
Tags: implementation, math
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
c=0
for i in range(n):
if a[i]==0:
a[i]+=1
c+=1
s=sum(a)
if s==0:
c+=1
print(c)
``` | output | 1 | 14,281 | 12 | 28,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4. | instruction | 0 | 14,282 | 12 | 28,564 |
Tags: implementation, math
Correct Solution:
```
for t in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
if 0 not in l:
if sum(l)!=0:
print(0)
else:
print(1)
else:
k=l.count(0)
p=sum(l)+k
if p!=0:
print(k)
else:
print(k+1)
``` | output | 1 | 14,282 | 12 | 28,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4. | instruction | 0 | 14,283 | 12 | 28,566 |
Tags: implementation, math
Correct Solution:
```
"""
Author : co_devil Chirag Garg
Institute : JIIT
"""
from math import *
from sys import stdin, stdout
import itertools
import os
import sys
import threading
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import *
# from math import ceil, floor, log, sqrt, factorial, pow, pi, gcd
# from bisect import bisect_left,bisect_right
# from decimal import *,threading
from fractions import Fraction
mod = int(pow(10, 9)+7)
def ii(): return int(input())
def si(): return str(input())
def mi(): return map(int, input().split())
def li(): return list(mi())
def fii(): return int(stdin.readline())
def fsi(): return str(stdin.readline())
def fmi(): return map(int, stdin.readline().split())
def fli(): return list(fmi())
abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
'z': 25}
def getKey(item): return item[0]
def sort2(l): return sorted(l, key=getKey)
def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo(x): return (x and (not (x & (x - 1))))
def decimalToBinary(n): return bin(n).replace("0b", "")
def ntl(n): return [int(i) for i in str(n)]
def powerMod(x, y, p):
res = 1
x %= p
while y > 0:
if y & 1:
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
graph = defaultdict(list)
visited = [0] * 1000000
col = [-1] * 1000000
def bfs(d, v):
q = []
q.append(v)
visited[v] = 1
while len(q) != 0:
x = q[0]
q.pop(0)
for i in d[x]:
if visited[i] != 1:
visited[i] = 1
q.append(i)
print(x)
def make_graph(e):
d = {}
for i in range(e):
x, y = mi()
if x not in d:
d[x] = [y]
else:
d[x].append(y)
if y not in d:
d[y] = [x]
else:
d[y].append(x)
return d
def gr2(n):
d = defaultdict(list)
for i in range(n):
x, y = mi()
d[x].append(y)
return d
def connected_components(graph):
seen = set()
def dfs(v):
vs = set([v])
component = []
while vs:
v = vs.pop()
seen.add(v)
vs |= set(graph[v]) - seen
component.append(v)
return component
ans = []
for v in graph:
if v not in seen:
d = dfs(v)
ans.append(d)
return ans
def primeFactors(n):
s = set()
while n % 2 == 0:
s.add(2)
n = n // 2
for i in range(3, int(sqrt(n)) + 1, 2):
while n % i == 0:
s.add(i)
n = n // i
if n > 2:
s.add(n)
return s
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1:
return
yield start
start += len(sub)
def SieveOfEratosthenes(n, isPrime):
isPrime[0] = isPrime[1] = False
for i in range(2, n):
isPrime[i] = True
p = 2
while (p * p <= n):
if (isPrime[p] == True):
i = p * p
while (i <= n):
isPrime[i] = False
i += p
p += 1
return isPrime
def dijkstra(edges, f, t):
g = defaultdict(list)
for l, r, c in edges:
g[l].append((c, r))
q, seen, mins = [(0, f, ())], set(), {f: 0}
while q:
(cost, v1, path) = heappop(q)
if v1 not in seen:
seen.add(v1)
path = (v1, path)
if v1 == t:
return (cost, path)
for c, v2 in g.get(v1, ()):
if v2 in seen:
continue
prev = mins.get(v2, None)
next = cost + c
if prev is None or next < prev:
mins[v2] = next
heappush(q, (next, v2, path))
return float("inf")
def binsearch(a, l, r, x):
while l <= r:
mid = l + (r-1)//2
if a[mid]:
return mid
elif a[mid] > x:
l = mid-1
else:
r = mid+1
return -1
t = ii()
for _ in range(t):
z = 0
n = ii()
l = li()
for i in range(n):
if l[i] == 0:
l[i] += 1
z += 1
s = sum(l)
if s != 0:
print(z)
else:
print(z+1)
``` | output | 1 | 14,283 | 12 | 28,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4. | instruction | 0 | 14,284 | 12 | 28,568 |
Tags: implementation, math
Correct Solution:
```
for _ in range(int(input())):
input()
a = list(map(int, input().split()))
x = a.count(0)
print(x+1 if x + sum(a) == 0 else x)
``` | output | 1 | 14,284 | 12 | 28,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4. | instruction | 0 | 14,285 | 12 | 28,570 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
for i in range(n):
t = int(input())
l = list(map(int,input().split()))
count = l.count(0)
a = [1 if x == 0 else x for x in l]
if sum(a) == 0:
count+=1
print(count)
``` | output | 1 | 14,285 | 12 | 28,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4. | instruction | 0 | 14,286 | 12 | 28,572 |
Tags: implementation, math
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
l=list(map(int,input().split()))
l.sort()
s=0
m=1
c=0
for i in l:
s+=i
m*=i
#print(s,m)
if(s!=0 and m!=0):
#print('fj')
c=0
if(s==0 and m!=0):
#print("dk")
c+=1
if(m==0 and s!=0):
#("sh")
k=l.count(0)
c+=k
if(s+c==0):
c+=1
if(s==0 and m==0):
#print("sdjh")
k=l.count(0)
c+=k
if(s+c==0):
c+=1
print(c)
``` | output | 1 | 14,286 | 12 | 28,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4. | instruction | 0 | 14,287 | 12 | 28,574 |
Tags: implementation, math
Correct Solution:
```
# @author
import sys
class ANonZero:
def solve(self):
for _ in range(int(input())):
n = int(input())
a = [int(_) for _ in input().split()]
s = sum(a)
v = a.count(0)
ans = v + (1 if (s + v == 0) else 0)
print(ans)
solver = ANonZero()
input = sys.stdin.readline
solver.solve()
``` | output | 1 | 14,287 | 12 | 28,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4. | instruction | 0 | 14,288 | 12 | 28,576 |
Tags: implementation, math
Correct Solution:
```
from collections import Counter
def min_steps(n, arr):
aux = Counter(arr)
counter = 0
counter += aux.get(0) or 0
if aux.get(1) is None:
aux[1] = 0
aux[1] += counter
if sum([k*v for k, v in aux.items()]) == 0:
counter += 1
print(counter)
return
T = int(input().strip())
for t in range(T):
n = int(input().strip())
s = input().strip().split(' ')
arr = list(map(int, s))
min_steps(n, arr)
``` | output | 1 | 14,288 | 12 | 28,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4.
Submitted Solution:
```
a=int(input( ))
for i in range(a):
n=int(input( ))
c=list(map(int,input().split()))
prod=1
for i in c:
prod=prod*i
total=sum(c)
if total==0:
p=0
if prod!=0:
for i in range(n):
if c[i]>0:
c[i]=c[i]+1
p+=1
print(p)
break
else:
for i in range(n):
if c[i]==0:
c[i]=c[i]+1
p+=1
total=sum(c)
if total==0:
for i in range(n):
if c[i]>0:
c[i]=c[i]+1
p+=1
print(p)
break
else:
print(p)
else:
p=0
if prod==0:
for i in range(n):
if c[i]==0:
c[i]=c[i]+1
p+=1
total=sum(c)
if total==0:
for i in range(n):
if c[i]>0:
c[i]=c[i]+1
p+=1
print(p)
break
else:
print(p)
else:
print(p)
``` | instruction | 0 | 14,289 | 12 | 28,578 |
Yes | output | 1 | 14,289 | 12 | 28,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4.
Submitted Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
def sumproductzero():
for _ in range(t):
n = int(input())
*a, = [int(x) for x in input().split()]
n=a.count(0)
sumall=sum(a)+n
if sumall==0:
yield n+1
else:
yield n
if __name__ == '__main__':
t= int(input())
ans = sumproductzero()
print(*ans,sep='\n')
``` | instruction | 0 | 14,290 | 12 | 28,580 |
Yes | output | 1 | 14,290 | 12 | 28,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4.
Submitted Solution:
```
cases = int(input())
for i in range(cases):
_ = input()
L = map(int, input().split())
s = 0
z = 0
for e in L:
s += e
if e == 0: z += 1
if s + z == 0:
z += 1
print(z)
``` | instruction | 0 | 14,291 | 12 | 28,582 |
Yes | output | 1 | 14,291 | 12 | 28,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4.
Submitted Solution:
```
for z in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
k=a.count(0)
c=0
if k==0:
if sum(a)==0:
c=1
else:
for i in range(len(a)):
if a[i]==0:
a[i]+=1
c+=1
if sum(a)==0:
c+=1
print(c)
``` | instruction | 0 | 14,292 | 12 | 28,584 |
Yes | output | 1 | 14,292 | 12 | 28,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = [int(i) for i in input().split()]
s = sum(a)
z = a.count(0)
if z == 0 and s != 0:
print(0)
elif z == 0 and s == 0:
print(1)
elif z>0:
print(max(z, -s+1))
``` | instruction | 0 | 14,293 | 12 | 28,586 |
No | output | 1 | 14,293 | 12 | 28,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4.
Submitted Solution:
```
def multiply(l):
m = 1
for x in l:
m *= x
return m
t=int(input())
for i in range(t):
n=int(input())
l=list(map(int,input().strip().split()))
s=0
M=multiply(l)
for j in range(n):
if l[j]==0:
s=s+1
l[j]=1
elif sum(l)==0:
s=s+1
l.append(1)
print(s)
``` | instruction | 0 | 14,294 | 12 | 28,588 |
No | output | 1 | 14,294 | 12 | 28,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4.
Submitted Solution:
```
n=int(input())
for i in range(n):
s1=int(input())
s2=list(map(int,input().split()))
p=1
s=0
count=0
for i in range(s1):
s=s+s2[i]
p=p*s2[i]
if p==0:
if s2[i]==0:
s2[i]=s2[i]+1
count+=1
s+=count
if s == 0:
# s2[0]=s2[0]+1
print(count+1)
else:
print(count)
# print(s2)
``` | instruction | 0 | 14,295 | 12 | 28,590 |
No | output | 1 | 14,295 | 12 | 28,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas have an array a of n integers [a_1, a_2, ..., a_n]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1 β€ i β€ n) and do a_i := a_i + 1.
If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.
What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a_1 + a_2 + ... + a_n β 0 and a_1 β
a_2 β
... β
a_n β 0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10^3). The description of the test cases follows.
The first line of each test case contains an integer n (1 β€ n β€ 100) β the size of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-100 β€ a_i β€ 100) β elements of the array .
Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.
Example
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,-1,-1], the sum will be equal to 1 and the product will be equal to 3.
In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [-1,1,1,1], the sum will be equal to 2 and the product will be equal to -1. It can be shown that fewer steps can't be enough.
In the third test case, both sum and product are non-zero, we don't need to do anything.
In the fourth test case, after adding 1 twice to the first element the array will be [2,-2,1], the sum will be 1 and the product will be -4.
Submitted Solution:
```
t= int(input())
for _ in range(t):
n = int(input())
a = list(map(int,input().split()))
if a.count(0) >0:
print(1)
continue
if sum(a)==0:
print(1)
continue
print(0)
``` | instruction | 0 | 14,296 | 12 | 28,592 |
No | output | 1 | 14,296 | 12 | 28,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes. | instruction | 0 | 14,297 | 12 | 28,594 |
Tags: brute force, strings
Correct Solution:
```
def binary_search(arr, low, high, x):
# Check base case
if high >= low:
mid = (high + low) // 2
# If element is present at the middle itself
if arr[mid] == x:
return mid
# If element is smaller than mid, then it can only
# be present in left subarray
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
# Else the element can only be present in right subarray
else:
return binary_search(arr, mid + 1, high, x)
else:
# Element is not present in the array
return -1
from collections import defaultdict
t=int(input())
for _ in range(t):
n=int(input())
l=list(map(int,input().split()))
x=set(l)
d=defaultdict(list)
s=defaultdict(list)
for i in range(n):
d[l[i]].append(i)
for j in x:
if j==l[i]:
if len(s[j])>0:
s[j].append(s[j][-1]+1)
else:
s[j].append(1)
else:
if len(s[j])>0:
s[j].append(s[j][-1])
else:
s[j].append(0)
ans=1
for i in range(n):
j=binary_search(d[l[i]],0,len(d[l[i]])-1,i)+1
k=d[l[i]][-j]
if i<k:
for z in x:
ans=max(ans,2*j+s[z][k]-s[z][i]) if z!=l[i] else max(ans,2*j+s[z][k]-s[z][i]-1)
print(ans)
``` | output | 1 | 14,297 | 12 | 28,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes. | instruction | 0 | 14,298 | 12 | 28,596 |
Tags: brute force, strings
Correct Solution:
```
import sys
def main():
t = int(sys.stdin.readline())
for _ in range(t):
n = int(sys.stdin.readline())
a = [int(x) for x in sys.stdin.readline().split(" ")]
ans = "NO"
for i, e in enumerate(a):
if e in a[i+2:]:
ans = "YES"
break
print(ans)
main()
``` | output | 1 | 14,298 | 12 | 28,597 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes. | instruction | 0 | 14,299 | 12 | 28,598 |
Tags: brute force, strings
Correct Solution:
```
import collections
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
table = collections.defaultdict(list)
table2 = collections.defaultdict(list)
for x in a:
table[x].append([len(table[j]) for j in range(1, 27)])
for x in a[::-1]:
table2[x].append([len(table2[j]) for j in range(1, 27)])
ans = 0
for x in table:
ans = max(ans, len(table[x]))
for x in table: # a:center
for y in range(26):
if x == y+1:
continue
l, r = 0, 0
L = len(table[x])
while(l+r <= L-1):
if table[x][l][y] == table2[x][r][y]:
ans = max(ans, table[x][l][y]*2 + L - l - r)
l += 1
elif table[x][l][y] > table2[x][r][y]:
ans = max(ans, table2[x][r][y]*2 + L - l - r)
r += 1
else:
ans = max(ans, table[x][l][y]*2 + L - l - r)
l += 1
# print(t)
print(ans)
``` | output | 1 | 14,299 | 12 | 28,599 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes. | instruction | 0 | 14,300 | 12 | 28,600 |
Tags: brute force, strings
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
li = list(map(int, input().split()))
arr = [0] * (max(li))
for i in range(len(li) - 1):
if li[i] != li[i + 1]:
arr[li[i] - 1] += 1
arr[li[-1] - 1] += 1
count = 0
abc = False
xyz = False
for i in range(len(li) - 2):
if li[i] == li[i + 2]:
abc = True
xyz = True
break
for i in range(len(arr)):
if arr[i] > 1:
abc = True
xyz = True
break
if not xyz:
if len(set(li)) == 1:
print("YES")
abc = True
continue
elif not abc:
print("NO")
else:
print("YES")
``` | output | 1 | 14,300 | 12 | 28,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes. | instruction | 0 | 14,301 | 12 | 28,602 |
Tags: brute force, strings
Correct Solution:
```
import sys
from array import array
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
s = [int(i) for i in input().split()]
prev = [array('l', (0 for i in range(n))) for i in range(200)]
pos = [[] for i in range(200)]
for i in range(n):
for j in range(200):
if i == 0: continue
prev[j][i] = prev[j][i-1]
prev[s[i]][i] += 1
pos[s[i]].append(i)
ans = 0
for i in range(200):
# if x == 0
for k in range(200):
if len(pos[i]):
ans = max(ans, prev[k][pos[i][-1]] - prev[k][pos[i][0]] + 1)
# if x != 0
for j in range(1, (len(pos[i]) // 2) + 1):
m = 0
for k in range(200):
m = max(m, prev[k][pos[i][-j]-1] - prev[k][pos[i][j-1]])
ans = max(ans, m + 2 * j)
print(ans)
``` | output | 1 | 14,301 | 12 | 28,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes. | instruction | 0 | 14,302 | 12 | 28,604 |
Tags: brute force, strings
Correct Solution:
```
import sys
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
ORDA = 97 # a
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return [int(i) for i in input().split()]
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r
def divs(n, start=2):
r = []
for i in range(start, int(math.sqrt(n) + 1)):
if (n % i == 0):
if (n / i == i):
r.append(i)
else:
r.extend([i, n // i])
return r
def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
def prime(n):
if n == 2: return True
if n % 2 == 0 or n <= 1: return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0: return False
return True
def convn(number, base):
new_number = 0
while number > 0:
new_number += number % base
number //= base
return new_number
def cdiv(n, k): return n // k + (n % k != 0)
def ispal(s): # Palindrome
for i in range(len(s) // 2 + 1):
if s[i] != s[-i - 1]:
return False
return True
# a = [1,2,3,4,5] -----> print(*a) ----> list print krne ka new way
#rr = sieve(int(1000**0.5))
def main():
for i in range(ii()):
d = collections.defaultdict(int)
n = ii()
arr = li()
for ele in arr:
d[ele] += 1
for key in list(d.keys()):
if d[key] == 2:
j = arr.index(key)
if j+1 <= n-1 and arr[j] != arr[j+1]:
break
elif d[key] > 2:
break
else:
print('NO')
continue
print('YES')
main()
``` | output | 1 | 14,302 | 12 | 28,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes. | instruction | 0 | 14,303 | 12 | 28,606 |
Tags: brute force, strings
Correct Solution:
```
from collections import Counter
from collections import defaultdict
import math
import random
import heapq as hq
from math import sqrt
import sys
from functools import reduce
def input():
return sys.stdin.readline().strip()
def iinput():
return int(input())
def tinput():
return input().split()
def rinput():
return map(int, tinput())
def rlinput():
return list(rinput())
mod = int(1e9)+7
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
# ----------------------------------------------------
if __name__ == "__main__":
for _ in range(iinput()):
n=iinput()
a=rlinput()
d=defaultdict(list)
j=0
for i in a:
d[i].append(j)
j+=1
flag=False
for i in d:
if len(d[i])>=2 and d[i][-1]-d[i][0]>1:
flag=True
break
print('YES' if flag else 'NO')
``` | output | 1 | 14,303 | 12 | 28,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes. | instruction | 0 | 14,304 | 12 | 28,608 |
Tags: brute force, strings
Correct Solution:
```
for u in range(int(input())):
n = int(input())
x = list(map(int, input().split()))
ind = [[] for i in range(n+1)]
flag = 1
for i in range(n):
ind[x[i]] += [i]
if len(ind[x[i]]) == 2 and ind[x[i]][-1] - ind[x[i]][-2] > 1 or len(ind[x[i]]) >= 3:
print('YES')
flag = 0
break
if flag:
print('NO')
``` | output | 1 | 14,304 | 12 | 28,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
Submitted Solution:
```
n = int(input())
for _ in range(n):
m = int(input())
s = set()
mat = list(map(int, input().split()))
for i in range(1,m):
if mat[i] in s:
print("YES")
break
s.add(mat[i-1])
else:
print("NO")
``` | instruction | 0 | 14,305 | 12 | 28,610 |
Yes | output | 1 | 14,305 | 12 | 28,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
pass
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
def test():
n = int(input())
a = list(map(int, input().split()))
for i in range(n):
for j in range(i, n - 2):
if a[j + 2] == a[i]:
return 'YES'
return 'NO'
t = int(input())
for _ in range(t):
print(test())
``` | instruction | 0 | 14,306 | 12 | 28,612 |
Yes | output | 1 | 14,306 | 12 | 28,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
Submitted Solution:
```
# cook your dish here
t = int(input())
out = []
for i in range(t):
n = int(input())
# NOTE: If same number gte 3, pass
# NOTE: same number gte 2 with space pass
nums = {}
flag = False
for i, num in enumerate(input().split()):
if not num in nums:
nums[num] = [i]
else:
if len(nums[num]) >= 2:
flag = True
break
elif nums[num][-1] + 1 == i:
# consecutive
nums[num].append(i)
else:
flag = True
break
if flag:
out.append("YES")
else:
out.append("NO")
print("\n".join(out))
``` | instruction | 0 | 14,307 | 12 | 28,614 |
Yes | output | 1 | 14,307 | 12 | 28,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
Submitted Solution:
```
t = int(input())
for k in range(t):
n = int(input())
arr = list(input().split())
check = False
for i in range(n):
for j in range(n):
if arr[i] == arr[j] and abs(i - j) > 1:
check = True
break
if check:
break
print('YES' if check else 'NO')
``` | instruction | 0 | 14,308 | 12 | 28,616 |
Yes | output | 1 | 14,308 | 12 | 28,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
Submitted Solution:
```
n=int(input())
c=[]
for i in range(n):
b=int(input())
a=list(map(int,input().split()))
a.reverse()
c=a
if a==c:
print("YES")
else:
print("NO")
``` | instruction | 0 | 14,309 | 12 | 28,618 |
No | output | 1 | 14,309 | 12 | 28,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
Submitted Solution:
```
t=int(input())
l1=[]
for i in range(t):
l=[]
n=int(input())
s=list(map(int,input().split()))
for j in range(n):
l.append([s[j],j])
l1.append(l)
for k in l1:
k.sort()
for m in range(1,len(k)):
if k[m-1][0]==k[m][0] and k[m][1]-k[m-1][1]>1:
print("YES")
break
else:
if m==len(k)-1:
print("NO")
``` | instruction | 0 | 14,310 | 12 | 28,620 |
No | output | 1 | 14,310 | 12 | 28,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
Submitted Solution:
```
from itertools import combinations as comb
def check():
for n in range(N):
for m in range(n+2, N):
if number[n] == number[m]:
l = number[n:m+1]
rl = l[::-1]
if l == rl: return 'YES'
return 'NO'
for case in range(int(input())):
N = int(input())
number = list(map(int, input().split()))
print(check())
``` | instruction | 0 | 14,311 | 12 | 28,622 |
No | output | 1 | 14,311 | 12 | 28,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
Your task is to determine if a has some subsequence of length at least 3 that is a palindrome.
Recall that an array b is called a subsequence of the array a if b can be obtained by removing some (possibly, zero) elements from a (not necessarily consecutive) without changing the order of remaining elements. For example, [2], [1, 2, 1, 3] and [2, 3] are subsequences of [1, 2, 1, 3], but [1, 1, 2] and [4] are not.
Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array a of length n is the palindrome if a_i = a_{n - i - 1} for all i from 1 to n. For example, arrays [1234], [1, 2, 1], [1, 3, 2, 2, 3, 1] and [10, 100, 10] are palindromes, but arrays [1, 2] and [1, 2, 3, 1] are not.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β the number of test cases.
Next 2t lines describe test cases. The first line of the test case contains one integer n (3 β€ n β€ 5000) β the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 5000 (β n β€ 5000).
Output
For each test case, print the answer β "YES" (without quotes) if a has some subsequence of length at least 3 that is a palindrome and "NO" otherwise.
Example
Input
5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
Output
YES
YES
NO
YES
NO
Note
In the first test case of the example, the array a has a subsequence [1, 2, 1] which is a palindrome.
In the second test case of the example, the array a has two subsequences of length 3 which are palindromes: [2, 3, 2] and [2, 2, 2].
In the third test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
In the fourth test case of the example, the array a has one subsequence of length 4 which is a palindrome: [1, 2, 2, 1] (and has two subsequences of length 3 which are palindromes: both are [1, 2, 1]).
In the fifth test case of the example, the array a has no subsequences of length at least 3 which are palindromes.
Submitted Solution:
```
a=int(input())
letters=['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
for i in range (a) :
leng=int(input())
arr=list(input().split())
flag=0
for k in range (int(int(leng+1)/2)) :
if arr.count(arr[k])>2 :
flag=1
elif arr.count(arr[k])>1 :
flag2=0
for j in range (leng-1) :
if arr[j]==arr[j+1] and arr[j]==arr[k] :
flag2=1
if flag2==0 :
flag=1
if flag==1 :
print ("YES")
else :
print ("NO")
``` | instruction | 0 | 14,312 | 12 | 28,624 |
No | output | 1 | 14,312 | 12 | 28,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashish has n elements arranged in a line.
These elements are represented by two integers a_i β the value of the element and b_i β the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.
He can perform the following operation any number of times:
* Select any two elements i and j such that b_i β b_j and swap them. That is, he can only swap two elements of different types in one move.
Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 500) β the size of the arrays.
The second line contains n integers a_i (1 β€ a_i β€ 10^5) β the value of the i-th element.
The third line containts n integers b_i (b_i β \{0, 1\}) β the type of the i-th element.
Output
For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.
You may print each letter in any case (upper or lower).
Example
Input
5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1
Output
Yes
Yes
Yes
No
Yes
Note
For the first case: The elements are already in sorted order.
For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.
For the third case: The elements are already in sorted order.
For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i β b_j. The elements cannot be sorted.
For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2. | instruction | 0 | 14,329 | 12 | 28,658 |
Tags: constructive algorithms, implementation
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key,lru_cache
from itertools import *
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# import sys
# input = sys.stdin.readline
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip().split()]
def st():return str(input().rstrip())[2:-1]
def val():return int(input().rstrip())
def li2():return [str(i)[2:-1] for i in input().rstrip().split()]
def li3():return [int(i) for i in st()]
for _ in range(val()):
n = val()
l1 = li()
l2 = li()
l3 = sorted(l1)
ans = 'NO'
if l3 == l1 or max(l2) != min(l2):
ans = 'YES'
print(ans)
``` | output | 1 | 14,329 | 12 | 28,659 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashish has n elements arranged in a line.
These elements are represented by two integers a_i β the value of the element and b_i β the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.
He can perform the following operation any number of times:
* Select any two elements i and j such that b_i β b_j and swap them. That is, he can only swap two elements of different types in one move.
Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 500) β the size of the arrays.
The second line contains n integers a_i (1 β€ a_i β€ 10^5) β the value of the i-th element.
The third line containts n integers b_i (b_i β \{0, 1\}) β the type of the i-th element.
Output
For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.
You may print each letter in any case (upper or lower).
Example
Input
5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1
Output
Yes
Yes
Yes
No
Yes
Note
For the first case: The elements are already in sorted order.
For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.
For the third case: The elements are already in sorted order.
For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i β b_j. The elements cannot be sorted.
For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2. | instruction | 0 | 14,330 | 12 | 28,660 |
Tags: constructive algorithms, implementation
Correct Solution:
```
def main():
for t in range(int(input())):
n = int(input())
a = list(zip(list(map(int, input().split())), list(map(int, input().split()))))
cnt = [0, 0]
for i in range(n):
cnt[a[i][1]] += 1
is_sorted = True
for i in range(1, n):
if a[i - 1][0] > a[i][0]:
is_sorted = False
break
ans = 'Yes' if is_sorted or (not is_sorted and cnt[0] != 0 and cnt[1] != 0) else 'No'
print(ans)
if __name__ == "__main__":
main()
``` | output | 1 | 14,330 | 12 | 28,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashish has n elements arranged in a line.
These elements are represented by two integers a_i β the value of the element and b_i β the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.
He can perform the following operation any number of times:
* Select any two elements i and j such that b_i β b_j and swap them. That is, he can only swap two elements of different types in one move.
Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 500) β the size of the arrays.
The second line contains n integers a_i (1 β€ a_i β€ 10^5) β the value of the i-th element.
The third line containts n integers b_i (b_i β \{0, 1\}) β the type of the i-th element.
Output
For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.
You may print each letter in any case (upper or lower).
Example
Input
5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1
Output
Yes
Yes
Yes
No
Yes
Note
For the first case: The elements are already in sorted order.
For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.
For the third case: The elements are already in sorted order.
For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i β b_j. The elements cannot be sorted.
For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2. | instruction | 0 | 14,331 | 12 | 28,662 |
Tags: constructive algorithms, implementation
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
l1=list(map(int,input().split()))
zeros,ones=0,0
for i in l1:
if i==0:
zeros+=1
else:
ones+=1
if zeros==0 or ones==0:
l2=[]
for i in l:
l2.append(i)
#print(l2)
l2.sort()
#print(l2)
#print(l)
if l==l2:
print("Yes")
else:
print("No")
else:
print("Yes")
``` | output | 1 | 14,331 | 12 | 28,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashish has n elements arranged in a line.
These elements are represented by two integers a_i β the value of the element and b_i β the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.
He can perform the following operation any number of times:
* Select any two elements i and j such that b_i β b_j and swap them. That is, he can only swap two elements of different types in one move.
Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 500) β the size of the arrays.
The second line contains n integers a_i (1 β€ a_i β€ 10^5) β the value of the i-th element.
The third line containts n integers b_i (b_i β \{0, 1\}) β the type of the i-th element.
Output
For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.
You may print each letter in any case (upper or lower).
Example
Input
5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1
Output
Yes
Yes
Yes
No
Yes
Note
For the first case: The elements are already in sorted order.
For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.
For the third case: The elements are already in sorted order.
For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i β b_j. The elements cannot be sorted.
For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2. | instruction | 0 | 14,332 | 12 | 28,664 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
from sys import stdin
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
def neo(): return map(int,input().split())
def Neo(): return list(map(int,input().split()))
for _ in range(int(input())):
n=int(input());A=Neo();B=Neo();mark={};C=A.copy();C.sort()
if C==A:print('Yes')
else:
if B.count(0)==0 or B.count(1)==0:print('No')
else:print('Yes')
``` | output | 1 | 14,332 | 12 | 28,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashish has n elements arranged in a line.
These elements are represented by two integers a_i β the value of the element and b_i β the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.
He can perform the following operation any number of times:
* Select any two elements i and j such that b_i β b_j and swap them. That is, he can only swap two elements of different types in one move.
Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 500) β the size of the arrays.
The second line contains n integers a_i (1 β€ a_i β€ 10^5) β the value of the i-th element.
The third line containts n integers b_i (b_i β \{0, 1\}) β the type of the i-th element.
Output
For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.
You may print each letter in any case (upper or lower).
Example
Input
5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1
Output
Yes
Yes
Yes
No
Yes
Note
For the first case: The elements are already in sorted order.
For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.
For the third case: The elements are already in sorted order.
For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i β b_j. The elements cannot be sorted.
For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2. | instruction | 0 | 14,333 | 12 | 28,666 |
Tags: constructive algorithms, implementation
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
if a == sorted(a):
print("Yes")
elif b.count(0)==len(b) or b.count(1)==len(b):
print("No")
else:
l = []
for i in range(len(a)):
l.append([a[i],b[i]])
l.sort()
temp = []
for j in range(len(l)):
temp.append(l[i][1])
if temp!=b:
print("Yes")
else:
print("No")
``` | output | 1 | 14,333 | 12 | 28,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashish has n elements arranged in a line.
These elements are represented by two integers a_i β the value of the element and b_i β the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.
He can perform the following operation any number of times:
* Select any two elements i and j such that b_i β b_j and swap them. That is, he can only swap two elements of different types in one move.
Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 500) β the size of the arrays.
The second line contains n integers a_i (1 β€ a_i β€ 10^5) β the value of the i-th element.
The third line containts n integers b_i (b_i β \{0, 1\}) β the type of the i-th element.
Output
For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.
You may print each letter in any case (upper or lower).
Example
Input
5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1
Output
Yes
Yes
Yes
No
Yes
Note
For the first case: The elements are already in sorted order.
For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.
For the third case: The elements are already in sorted order.
For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i β b_j. The elements cannot be sorted.
For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2. | instruction | 0 | 14,334 | 12 | 28,668 |
Tags: constructive algorithms, implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = [int(i) for i in input().split(' ')]
b = [int(i) for i in input().split(' ')]
prev = a[0]
sorted = True
for cur in a:
if prev > cur:
sorted = False
prev = b[0]
one_type = True
for cur in b:
if prev != cur:
one_type = False
if sorted or not one_type:
print('Yes')
else:
print('No')
``` | output | 1 | 14,334 | 12 | 28,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashish has n elements arranged in a line.
These elements are represented by two integers a_i β the value of the element and b_i β the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.
He can perform the following operation any number of times:
* Select any two elements i and j such that b_i β b_j and swap them. That is, he can only swap two elements of different types in one move.
Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 500) β the size of the arrays.
The second line contains n integers a_i (1 β€ a_i β€ 10^5) β the value of the i-th element.
The third line containts n integers b_i (b_i β \{0, 1\}) β the type of the i-th element.
Output
For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.
You may print each letter in any case (upper or lower).
Example
Input
5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1
Output
Yes
Yes
Yes
No
Yes
Note
For the first case: The elements are already in sorted order.
For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.
For the third case: The elements are already in sorted order.
For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i β b_j. The elements cannot be sorted.
For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2. | instruction | 0 | 14,335 | 12 | 28,670 |
Tags: constructive algorithms, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
if a == sorted(a):
print('Yes')
else:
ch0 = 0
ch1 = 0
for i in b:
if i == 0:
ch0 += 1
else:
ch1 += 1
if ch1 == 0 or ch0 == 0:
print('No')
else:
print('Yes')
``` | output | 1 | 14,335 | 12 | 28,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashish has n elements arranged in a line.
These elements are represented by two integers a_i β the value of the element and b_i β the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.
He can perform the following operation any number of times:
* Select any two elements i and j such that b_i β b_j and swap them. That is, he can only swap two elements of different types in one move.
Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 500) β the size of the arrays.
The second line contains n integers a_i (1 β€ a_i β€ 10^5) β the value of the i-th element.
The third line containts n integers b_i (b_i β \{0, 1\}) β the type of the i-th element.
Output
For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.
You may print each letter in any case (upper or lower).
Example
Input
5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1
Output
Yes
Yes
Yes
No
Yes
Note
For the first case: The elements are already in sorted order.
For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.
For the third case: The elements are already in sorted order.
For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i β b_j. The elements cannot be sorted.
For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2. | instruction | 0 | 14,336 | 12 | 28,672 |
Tags: constructive algorithms, implementation
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
o = b.count(1)
z = b.count(0)
ans = "Yes"
for i in range(n-1):
if (o==0 or z==0) and a[i] > a[i+1]:
ans="No"
break
print(ans)
``` | output | 1 | 14,336 | 12 | 28,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ashish has n elements arranged in a line.
These elements are represented by two integers a_i β the value of the element and b_i β the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.
He can perform the following operation any number of times:
* Select any two elements i and j such that b_i β b_j and swap them. That is, he can only swap two elements of different types in one move.
Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 500) β the size of the arrays.
The second line contains n integers a_i (1 β€ a_i β€ 10^5) β the value of the i-th element.
The third line containts n integers b_i (b_i β \{0, 1\}) β the type of the i-th element.
Output
For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.
You may print each letter in any case (upper or lower).
Example
Input
5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1
Output
Yes
Yes
Yes
No
Yes
Note
For the first case: The elements are already in sorted order.
For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.
For the third case: The elements are already in sorted order.
For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i β b_j. The elements cannot be sorted.
For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2.
Submitted Solution:
```
t = int(input())
results = []
for case in range(t):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
ans = 'Yes'
if (1 in b and 0 not in b):
ans = 'No'
elif (0 in b and 1 not in b):
ans = 'No'
if (a == sorted(a)):
ans = 'Yes'
results.append(ans)
for result in results:
print (result)
``` | instruction | 0 | 14,337 | 12 | 28,674 |
Yes | output | 1 | 14,337 | 12 | 28,675 |
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