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.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good. | instruction | 0 | 33,550 | 12 | 67,100 |
Tags: constructive algorithms, math
Correct Solution:
```
import os
import heapq
import sys
import math
import operator
from collections import defaultdict
from io import BytesIO, IOBase
"""def gcd(a,b):
if b==0:
return a
else:
return gcd(b,a%b)"""
"""def pw(a,b):
result=1
while(b>0):
if(b%2==1): result*=a
a*=a
b//=2
return result"""
def inpt():
return [int(k) for k in input().split()]
def main():
for _ in range(int(input())):
n=int(input())
for i in range(1,n+1):
print(i,end=' ')
print()
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")
if __name__ == "__main__":
main()
``` | output | 1 | 33,550 | 12 | 67,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good. | instruction | 0 | 33,551 | 12 | 67,102 |
Tags: constructive algorithms, math
Correct Solution:
```
for z in range(int(input())):
n=int(input())
arr=[x for x in range(1,n+1)]
print(*arr)
``` | output | 1 | 33,551 | 12 | 67,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good. | instruction | 0 | 33,552 | 12 | 67,104 |
Tags: constructive algorithms, math
Correct Solution:
```
for i in range(int(input())):
n = int(input())
lis = [0 for i in range(0,n+1)]
lis2 = 0
i = 0
x = 1
while (i < n):
lis[x] = 1
print(x,end=" ")
mx = 0
val = 0
for j in range(1,n+1):
if lis[j] != 1:
z = int(i|j)
if z == n:
val = j
mx = z
break
if mx <= z:
val = j
mx = z
x = val
i += 1
print()
``` | output | 1 | 33,552 | 12 | 67,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good. | instruction | 0 | 33,553 | 12 | 67,106 |
Tags: constructive algorithms, math
Correct Solution:
```
t=int(input())
for q in range(t):
n=int(input())
for x in range(1,n+1):
print(x,end=" ")
print()
``` | output | 1 | 33,553 | 12 | 67,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good. | instruction | 0 | 33,554 | 12 | 67,108 |
Tags: constructive algorithms, math
Correct Solution:
```
t=int(input())
for i in range(t):
#n,k=map(int,input().split())
#a=[int(v) for v in input().split()]
n=int(input())
res=[j+1 for j in range(n)]
print(' '.join(map(str,res)))
``` | output | 1 | 33,554 | 12 | 67,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good. | instruction | 0 | 33,555 | 12 | 67,110 |
Tags: constructive algorithms, math
Correct Solution:
```
for _ in " "*int(input()):
n=int(input())
for i in range(1,n+1):
print(i,end=" ")
``` | output | 1 | 33,555 | 12 | 67,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good. | instruction | 0 | 33,556 | 12 | 67,112 |
Tags: constructive algorithms, math
Correct Solution:
```
#Problem C Shreyansh
t=int(input())
while t:
t=t-1
n=int(input())
#n,m=map(int,input().split())
#a=list(map(int,input().split()))
for i in range(n,0,-1):
print(i,end=" ")
print()
``` | output | 1 | 33,556 | 12 | 67,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
Submitted Solution:
```
t=int(input())
while t!=0:
n=int(input())
for i in range(1,n+1):
print(i,end=" ")
print()
t=t-1
``` | instruction | 0 | 33,557 | 12 | 67,114 |
Yes | output | 1 | 33,557 | 12 | 67,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
ans=[]
j=n
while(j>=1):
ans.append(j)
j+=-1
print(*ans)
``` | instruction | 0 | 33,558 | 12 | 67,116 |
Yes | output | 1 | 33,558 | 12 | 67,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
Submitted Solution:
```
import sys
t = int(input().strip())
for line in sys.stdin:
n = int(line.strip())
print(' '.join(map(str, range(1, n + 1, 1))))
``` | instruction | 0 | 33,559 | 12 | 67,118 |
Yes | output | 1 | 33,559 | 12 | 67,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
Submitted Solution:
```
t = int(input())
for i in range(0, t):
n = int(input())
ns = [x for x in range(1, n+1)]
for element in ns:
print(element, end=" ")
``` | instruction | 0 | 33,560 | 12 | 67,120 |
Yes | output | 1 | 33,560 | 12 | 67,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
Submitted Solution:
```
import random
t = int(input())
for i in range(t):
n = int(input())
lis = []
for j in range(n):
lis.append(j+1)
flag = 1
while (flag == 1):
random.shuffle(lis)
permutation = pow(2, n) - 1
good_list = True
for p in range(1, n):
# noinspection PyUnreachableCode
if not good_list:
break
sublist = random.sample(lis, p)
res_OR = sublist[0]
if (len(sublist)) > 1:
for k in range(1, len(sublist)):
res_OR |= sublist[k]
if res_OR >= len(sublist):
continue
else:
good_list = False
if good_list:
print(lis)
break
``` | instruction | 0 | 33,561 | 12 | 67,122 |
No | output | 1 | 33,561 | 12 | 67,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = []
b = [i for i in range(1, n+1)]
c = [n-i for i in range(n)]
# print(c)
for i in range(n):
if i%2!=0:
a.append(b[i])
else:
a.append(c[i])
for i in a:
print(i, end=" ")
``` | instruction | 0 | 33,562 | 12 | 67,124 |
No | output | 1 | 33,562 | 12 | 67,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
Submitted Solution:
```
import sys
input = sys.stdin.readline
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1;
return x;
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k;
res = 1;
for i in range(k):
res = res * (n - i);
res = res / (i + 1);
return int(res);
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
if hi == None:
hi = len(a);
while lo < hi:
mid = (lo+hi)//2;
if a[mid] < x:
lo = mid+1;
else:
hi = mid;
return lo;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret;
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
for _ in range(int(input())):
n = int(input())
x = [x for x in range(1, n+1)]
for i in range(1, n):
print(x[i]|x[i-1])
print(*x)
``` | instruction | 0 | 33,563 | 12 | 67,126 |
No | output | 1 | 33,563 | 12 | 67,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) —
* (p_i OR p_{i+1} OR … OR p_{j-1} OR p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).
Output
For every test, output any good permutation of length n on a separate line.
Example
Input
3
1
3
7
Output
1
3 1 2
4 3 5 2 7 1 6
Note
For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
* 3 OR 1 = 3 ≥ 2 (i = 1,j = 2)
* 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3)
* 1 OR 2 = 3 ≥ 2 (i = 2,j = 3)
* 1 ≥ 1 (i = 2,j = 2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
Submitted Solution:
```
def good_array(n):
arr = []
for i in range(n, n // 2, -1):
arr.append(i)
if n - i + 1> 0:
arr.append(n - i + 1)
return arr[::-1] if n % 2 == 0 else arr[::-1][1:]
for _ in range(int(input())):
n = int(input())
print(good_array(n))
``` | instruction | 0 | 33,564 | 12 | 67,128 |
No | output | 1 | 33,564 | 12 | 67,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3 | instruction | 0 | 33,581 | 12 | 67,162 |
Tags: dp, greedy, sortings
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
for _ in range(int(input())):
n,x = map(int,input().split())
a = list(map(int,input().split()))
if a == sorted(a):
print(0)
continue
i,op = 0,0
while i != n:
y = a[i:]
bb = sorted(y)!=y
if not bb:
break
if a[i] > x and bb:
op += 1
a[i],x = x,a[i]
i += 1
if a == sorted(a):
print(op)
else:
print(-1)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | output | 1 | 33,581 | 12 | 67,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3 | instruction | 0 | 33,582 | 12 | 67,164 |
Tags: dp, greedy, sortings
Correct Solution:
```
for _ in " " * int(input()):
n,x= map(int,input().split())
a=list(map(int,input().split()))
if a == sorted(a):
print(0)
else:
b = sorted(a)
cnt = 0
for i in range(n):
if a == sorted(a):
break
if a[i] > x:
x,a[i]=a[i],x
cnt+=1
if a == sorted(a):
print(cnt)
else:
print(-1)
``` | output | 1 | 33,582 | 12 | 67,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3 | instruction | 0 | 33,583 | 12 | 67,166 |
Tags: dp, greedy, sortings
Correct Solution:
```
from collections import defaultdict, deque, Counter, OrderedDict
# import threading
import sys
from bisect import bisect_right
def main():
t = int(input())
while t:
t -= 1
n, x = [int(i) for i in input().split()]
arr = [int(i) for i in input().split()]
dips, sdips = [], set()
for i in range(1, n):
if arr[i] < arr[i - 1]:
dips.append(i)
sdips.add(i)
if len(dips) == 0:
print(0)
continue
count, fl = 0, False
res = deque(dips)
while len(sdips) and len(res):
initial = bisect_right(arr, x, 0, res[0])
if initial == -1 or initial == n or initial in sdips:
print(-1)
fl = True
break
x, arr[initial] = arr[initial], x
count += 1
if initial + 1 < n and (initial + 1) in sdips:
if arr[initial + 1] < arr[initial]:
print(-1)
fl = True
break
sdips.remove(initial + 1)
res.popleft()
if not fl:
print(count)
if __name__ == "__main__":
"""sys.setrecursionlimit(400000)
threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()"""
main()
``` | output | 1 | 33,583 | 12 | 67,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3 | instruction | 0 | 33,584 | 12 | 67,168 |
Tags: dp, greedy, sortings
Correct Solution:
```
for _ in range(int(input())):
n, x = map(int, input().split())
A = list(map(int, input().split()))
if sorted(A) == A:
print(0)
else:
i = 1
ans = 0
flag = 0
while i < n:
if A[i] < A[i - 1]:
if x >= A[i - 1] and A[i] > x:
A[i], x = x, A[i]
ans += 1
else:
j = 0
for j in range(i):
if A[j] > x:
break
else:
flag = 1
break
for k in range(j, i):
if x >= A[i - 1]:
break
if A[k] > x:
A[k], x = x, A[k]
ans += 1
elif A[k] == x:
continue
if A[i] >= A[i - 1]: continue
if A[i] < A[i - 1] and A[i] > x:
A[i], x = x, A[i]
ans += 1
else:
flag = 1
break
i += 1
if flag:
print(-1)
else:
print(ans)
``` | output | 1 | 33,584 | 12 | 67,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3 | instruction | 0 | 33,585 | 12 | 67,170 |
Tags: dp, greedy, sortings
Correct Solution:
```
from math import *
from sys import *
from bisect import *
t=int(stdin.readline())
for _ in range(t):
n,k=map(int,stdin.readline().split())
a=list(map(int,stdin.readline().split()))
ans=0
for i in range(n-1):
if list(sorted(a))==a:
break
if a[i]>min(a[i:]) or a[i]>k:
if a[i]<=k:
break
a[i],k=k,a[i]
ans+=1
if list(sorted(a))==a:
print(ans)
else:
print(-1)
``` | output | 1 | 33,585 | 12 | 67,171 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3 | instruction | 0 | 33,586 | 12 | 67,172 |
Tags: dp, greedy, sortings
Correct Solution:
```
import math
for _ in range(int(input())):
n,x=map(int,input().split())
l=list(map(int,input().split()))
ll=sorted(l)
s=0
for i in range(n):
if sorted(l)==l:
break
else:
if l[i]>x:
l[i],x=x,l[i]
s+=1
if sorted(l)==l:
print(s)
else:
print(-1)
``` | output | 1 | 33,586 | 12 | 67,173 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3 | instruction | 0 | 33,587 | 12 | 67,174 |
Tags: dp, greedy, sortings
Correct Solution:
```
from collections import defaultdict
import sys
import bisect
input=sys.stdin.readline
t=int(input())
for ii in range(t):
n,x=map(int,input().split())
a=[int(i) for i in input().split() if i!='\n']
ans=0
if a==sorted(a):
print(0)
continue
for i in range(n):
if a[i]>x:
a[i],x=x,a[i]
ans+=1
if a==sorted(a):
break
else:
ans=-1
print(ans)
``` | output | 1 | 33,587 | 12 | 67,175 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3 | instruction | 0 | 33,588 | 12 | 67,176 |
Tags: dp, greedy, sortings
Correct Solution:
```
import sys
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
def solve():
n, x = mints()
a = list(mints())
b = a.copy()
b.append(x)
b.sort()
res = int(1e9)
#print(b)
for i in range(n+1):
ok = True
if b[i] >= x:
k = 0
xx = x
cnt = 0
for j in range(n):
if k == i:
k += 1
#print(k, j)
if b[k] != a[j]:
if xx != b[k] or a[j] < xx:
ok = False
break
xx = a[j]
cnt += 1
k += 1
if ok:
res = min(res, cnt)
if res >= int(1e9):
print(-1)
else:
print(res)
for t in range(mint()):
solve()
``` | output | 1 | 33,588 | 12 | 67,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3
Submitted Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n,x = list(map(int,input().split()))
arr = list(map(int,input().split()))
ans = 0
for i in range(1,len(arr)):
if arr[i]<arr[i-1]:
index = i-1
while index>=0 and arr[index]>x:
index-=1
ans+=1
index+=1
for j in range(index,i):
temp = arr[j]
if arr[j]==x:
ans-=1
arr[j]=x
x = temp
for i in range(1,len(arr)):
if arr[i]<arr[i-1]:
ans = -1
print(ans)
``` | instruction | 0 | 33,589 | 12 | 67,178 |
Yes | output | 1 | 33,589 | 12 | 67,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3
Submitted Solution:
```
import sys,math
from collections import deque,defaultdict
import operator as op
from functools import reduce
#sys.setrecursionlimit(10**6)
I=sys.stdin.readline
#s="abcdefghijklmnopqrstuvwxyz"
"""
x_move=[-1,0,1,0,-1,1,1,-1]
y_move=[0,1,0,-1,1,1,-1,-1]
"""
def ii():
return int(I().strip())
def li():
return list(map(int,I().strip().split()))
def mi():
return map(int,I().strip().split())
def ncr(n, r):
r = min(r, n-r)
numer = reduce(op.mul, range(n, n-r, -1), 1)
denom = reduce(op.mul, range(1, r+1), 1)
return numer // denom
def gcd(x, y):
while y:
x, y = y, x % y
return x
def valid(row,col,rows,cols,rcross,lcross):
return rows[row]==0 and cols[col]==0 and rcross[col+row]==0 and lcross[col-row]==0
def div(n):
tmp=[]
for i in range(2,int(n**.5)+1):
if n%i==0:
cnt=0
while(n%i==0):
n=n//i
cnt+=1
tmp.append((i,cnt))
return tmp
def main():
ans=""
for _ in range(ii()):
n,x=mi()
arr=li()
cnt=0
flag=0
while (sorted(arr)!=arr):
cnt+=1
for i in range(n):
if arr[i]>x:
arr[i],x=x,arr[i]
break
else:
flag=1
break
if flag:
ans+="-1\n"
else:
ans+=str(cnt)+"\n"
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 33,590 | 12 | 67,180 |
Yes | output | 1 | 33,590 | 12 | 67,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3
Submitted Solution:
```
from sys import stdin,stdout
from math import gcd,sqrt,factorial,inf
from collections import deque,defaultdict
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\n')
L=lambda:list(R())
P=lambda x:stdout.write(x)
lcm=lambda x,y:(x*y)//gcd(x,y)
hg=lambda x,y:((y+x-1)//x)*x
pw=lambda x:0 if x==1 else 1+pw(x//2)
chk=lambda x:chk(x//2) if not x%2 else True if x==1 else False
sm=lambda x:(x**2+x)//2
N=10**9+7
def is_sort(a):
p=a[0]
for i in a[1:]:
if i<p:return False
p=i
return True
for _ in range(I()):
n,k=R()
*a,=R()
ans=0
for i in range(n):
if is_sort(a):break
if a[i]>k:
a[i],k=k,a[i]
ans+=1
if is_sort(a):print(ans)
else:print(-1)
``` | instruction | 0 | 33,591 | 12 | 67,182 |
Yes | output | 1 | 33,591 | 12 | 67,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3
Submitted Solution:
```
import math
t = int(input())
for _ in range(t):
n, x = map(int, input().split())
a = list(map(int, input().split()))
cnt = 0
for i in range(1, n):
if a[i] >= a[i - 1]:
cnt += 1
if cnt == (n - 1):
print(0)
continue
ans = -1
for i in range(n):
b = []
for j in range(n):
if j != i:
b.append(a[j])
else:
b.append(x)
b.sort()
c = []
for j in range(n):
if a[j] != b[j]:
c.append((a[j], j))
c.sort(key=lambda k: k[0])
curr = 0
xx = x
for aj, j in c:
if b[j] != xx or aj <= xx:
curr = -1
break
else:
xx = aj
if xx != a[i]:
continue
if curr != -1:
curr = len(c)
if ans == -1:
ans = curr
else:
ans = min(curr, ans)
print(ans)
``` | instruction | 0 | 33,592 | 12 | 67,184 |
Yes | output | 1 | 33,592 | 12 | 67,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3
Submitted Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
import time
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
start_time = time.time()
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
def getMat(n):
return [getInts() for _ in range(n)]
MOD = 998244353
"""
If we have no inversions, we're done immediately
If we have an inversion, swap X
0 1 4 5 3, x = 2
0 1 2 5 3, x = 4
0 1 2 4 3, x = 5
0 1 3 5 4, x = 2
0 1 2 5 4, x = 3
0 1 2 3 4, x = 5
81 324 218 413 324
18 81 218 413 324 - 324 - two steps
18 81 218 324 324
work left to right
identify first inversion
11 9 - 10
10 9 - 11
it only cures the inversion if X < problem number
3 8 6 - 4
3 4 6
3 8 6 - 7
3 7 6 - 8
to cure inversion, number needs to go in in the right order
"""
def solve():
N, X = getInts()
A = getInts()
#if A == sorted(A): return 0
ans = 0
i = 0
while i < N-1:
#find an inversion
if A[i+1] < A[i]:
if X > A[i]:
return -1
#print(A[i], A[i+1])
rel_ind = bisect_right(A,X)
#print(i+1,rel_ind)
ans += i+1-rel_ind
X = A[i+1]
A[i+1] = X
i += 1
return ans
for _ in range(getInt()):
print(solve())
#solve()
#print(time.time()-start_time)á
``` | instruction | 0 | 33,593 | 12 | 67,186 |
No | output | 1 | 33,593 | 12 | 67,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**5)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
def solve():
cnt=0
pa=-1
first=True
for a,b in zip(aa,bb):
if a==b:continue
if first:
if b!=x:return -1
first=False
if a<b:return -1
cnt+=1
pa=a
return cnt
for _ in range(II()):
n,x=MI()
aa=LI()
if aa==sorted(aa):
print(0)
continue
bb=aa+[x]
bb.sort()
bb=bb[:-1]
print(solve())
``` | instruction | 0 | 33,594 | 12 | 67,188 |
No | output | 1 | 33,594 | 12 | 67,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3
Submitted Solution:
```
from sys import stdin ,stdout
from os import path
rd = lambda:stdin.readline().strip()
wr = stdout.write
if(path.exists('input.txt')):
stdin = open("input.txt","r")
#------------------------------------
for _ in range(int(rd())):
x,y = map(int,rd().split())
mylist = list(map(int,rd().split()))
ar = mylist.copy()
ar.sort()
flag = False
counter = 0
for i in range(x):
if mylist[i] != ar[i]:
flag = True
counter +=1
if flag == False:
print(0)
elif counter %2 !=0 :
print(-1)
else:
counter = 0
for i in range(x):
if mylist[i] > y:
temp = mylist[i]
mylist[i] = y
y = temp
counter+=1
flag = True
for i in range(1,x):
if mylist[i-1] > mylist[i]:
flag = False
break
if flag ==False:
print(-1)
else:
print(counter)
``` | instruction | 0 | 33,595 | 12 | 67,190 |
No | output | 1 | 33,595 | 12 | 67,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).
To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.
For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:
1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2;
2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3;
3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).
The sum of values of n over all test cases in the input does not exceed 500.
Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.
Example
Input
6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324
Output
3
0
0
-1
1
3
Submitted Solution:
```
t = int(input())
for z in range(t):
n, x = [int(num) for num in input().split()]
vals = [int(val) for val in input().split()]
vals_sorted = [val for val in vals]
vals_sorted.append(x)
vals_sorted.sort()
ans = n
found = False
for i in range(1+n):
cur = 0
valid = True
turns = 0
for j in range(n):
if cur == i:
if (vals_sorted[cur] == x):
turns += 1
cur += 1
val = vals_sorted[cur]
if val != vals[j]:
turns += 1
if (vals[j] <= x):
valid = False
break
cur += 1
if (valid):
ans = min(ans, turns)
found = True
if found:
print(ans)
else:
print(-1)
``` | instruction | 0 | 33,596 | 12 | 67,192 |
No | output | 1 | 33,596 | 12 | 67,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0 | instruction | 0 | 33,597 | 12 | 67,194 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
import collections
for i in range(int(input())):
n = int(input())
a = [int(j) for j in input().split()]
a.sort()
counter = collections.Counter(a)
b=list(counter.values())
b.sort(reverse=True)
if b[0]<=n/2:
print(n%2)
elif b[0]>n/2:
print(b[0]-(n-b[0]))
"""543630ccb2c185694d103df7ff024c49ed9f83d6ccaaeecd2327eb3c3a8a135aExample in C/C++/Java:// 543630ccb2c185694d103df7ff024c49ed9f83d6ccaaeecd2327eb3c3a8a135aExample in Python:# 543630ccb2c185694d103df7ff024c49ed9f83d6ccaaeecd2327eb3c3a8a135a"""
``` | output | 1 | 33,597 | 12 | 67,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0 | instruction | 0 | 33,598 | 12 | 67,196 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
import math,random,time
from collections import *
from itertools import *
def ass(a, b): print(f"Assertion error: {a} != {b}" if a != b else 'OK')
def nr(): return int(input())
def nrs(): return [int(i) for i in input().split()]
def nrss(n): return [nrs() for _ in range(n)]
def word(): return input()
def words(): return [w for w in input().split()]
def dump(lst):
for i, item in enumerate(lst): print(i, item)
def binsearch(arr, a, b):
m = (a + b) // 2
if a >= b: return a - 1 if arr[a] == 0 else a
return binsearch(arr, a, m - 1) if arr[m] == 0 else binsearch(arr, m + 1, b)
def gcd(x,y): return x if y == 0 else gcd(y, x % y)
def lcm(x,y): return x*y//gcd(x,y)
def sign(a): return 0 if a==0 else int(a/abs(a))
def f(arr):
arr.sort()
antal = 0
n = len(arr)
orig = n
if n%2==1:
arr.pop(n//2)
n -= 1
n2 = n//2
for i in range(n2):
if arr[i] != arr[i+n2]: antal+=2
return orig-antal
# print(f([4,5,4,5,4]))
for i in range(nr()):
n = nrs()
print(f(nrs()))
``` | output | 1 | 33,598 | 12 | 67,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0 | instruction | 0 | 33,599 | 12 | 67,198 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
"""
Author - Satwik Tiwari .
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt,log2
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def testcase(t):
for pp in range(t):
solve(pp)
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def modInverse(b):
g = gcd(b, mod)
if (g != 1):
# print("Inverse doesn't exist")
return -1
else:
# If b and m are relatively prime,
# then modulo inverse is b^(m-2) mode m
return pow(b, mod - 2, mod)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))
def solve(case):
n = int(inp())
a = lis()
cnt = {}
for i in range(n):
if(a[i] not in cnt):
cnt[a[i]]= 1
else:
cnt[a[i]] += 1
h = []
for i in cnt:
heappush(h,-cnt[i])
cnt = 0
while(len(h)>1):
f = heappop(h)
s = heappop(h)
if(f == 0):
heappush(h,s)
continue
if(s == 0):
heappush(h,f)
continue
cnt += 2
heappush(h,f+1)
heappush(h,s+1)
print(n - cnt)
# testcase(1)
testcase(int(inp()))
``` | output | 1 | 33,599 | 12 | 67,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0 | instruction | 0 | 33,600 | 12 | 67,200 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
from sys import stdin
import heapq
input=stdin.readline
class MaxHeap:
def __init__(self):
self.data = []
def top(self):
return -self.data[0]
def push(self, val):
heapq.heappush(self.data, -val)
def pop(self):
return -heapq.heappop(self.data)
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
dic={}
for d in arr:
if d not in dic:
dic[d]=0
dic[d]+=1
new=[]
for d in dic:
new.append(dic[d])
# new.sort(reverse=True)
k=len(new)
maxheap=MaxHeap()
for d in dic:
maxheap.push(dic[d])
while True:
if k>=2:
k-=2
x=maxheap.pop()
y=maxheap.pop()
x-=1
y-=1
if x>0:
maxheap.push(x)
k+=1
if y>0:
maxheap.push(y)
k+=1
elif k==1:
print(maxheap.pop())
break
else:
print(0)
break
``` | output | 1 | 33,600 | 12 | 67,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0 | instruction | 0 | 33,601 | 12 | 67,202 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
#input template
from sys import stdin, stdout
from collections import Counter, OrderedDict, defaultdict
import math
import re
cin = stdin.readline
cout = stdout.write
mp = lambda: list(map(int, cin().split()))
def chars(): #function for taking string input as character array since string in python is immutable
s = cin()
return(list(s[:len(s) - 1]))
#print list
def pl(a):
for val in a:
cout(str(val) + ' ')
cout('\n')
def binSearch(l, val, start, end):
if end < start:
return False
mid = (start + end) // 2
if l[mid] == val:
return True
elif l[mid] < val:
return binSearch(l, val, mid+1, end)
elif l[mid] > val:
return binSearch(l, val, start, mid-1)
def myLog(x, y):
z = int(math.log(x, y))
if y**(z+1) > x:
return z
elif y**(z+1) == x:
return z+1
#main
t, = mp()
for _ in range(t):
n, = mp()
a = mp()
d = {}
maxi = 0
for item in a:
if item in d:
d[item] += 1
else:
d[item] = 1
maxi = max(maxi, d[item])
if maxi <= n // 2:
if n % 2:
cout('1\n')
else:
cout('0\n')
else:
cout(str(2*maxi-n) + '\n')
``` | output | 1 | 33,601 | 12 | 67,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0 | instruction | 0 | 33,602 | 12 | 67,204 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
import collections,heapq
def sign(val):
return -val
def solve(a, n):
hm = collections.defaultdict(int)
for val in a:
hm[val] += 1
h = []
for k,v in hm.items():
heapq.heappush(h, sign(v))
ans = n
while len(h) > 1:
c1 = sign(heapq.heappop(h))
c2 = sign(heapq.heappop(h))
c1 -= 1
c2 -= 1
ans -= 2
if c1 > 0 :
heapq.heappush(h, sign(c1))
if c2 > 0:
heapq.heappush(h, sign(c2))
return ans
local_ans = []
local_mode = False
def getArr():
return list(map(int, input().split()))
def getNums():
return map(int, input().split())
# local_mode = True
t = int(input())
for _ in range(t):
n = int(input())
a = getArr()
if local_mode:
local_ans.append(solve(a, n))
else:
print(solve(a, n))
if local_mode:
def printAll(sol):
for val in sol:
print(val)
printAll(local_ans)
``` | output | 1 | 33,602 | 12 | 67,205 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0 | instruction | 0 | 33,603 | 12 | 67,206 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
from collections import Counter
for _ in range(int(input())):
n= int(input())
arr= list(map(int, input().split()))
count = Counter(arr)
to = (list(count.values()))
to.sort()
f = to[-1]
res = 0
if f>n-f:
res = f-(n-f)
if res==0 and n%2==1:
res = 1
print(res)
``` | output | 1 | 33,603 | 12 | 67,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0 | instruction | 0 | 33,604 | 12 | 67,208 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from bisect import bisect_right
from math import gcd,log
from collections import Counter, defaultdict, deque
from pprint import pprint
from itertools import permutations
from bisect import bisect_left
import heapq
def main(tnum):
n=int(input())
arr=list(map(int,input().split()))
arr=list(Counter(arr).values())
arr=[-x for x in arr]
heapq.heapify(arr)
while len(arr)>1:
n1=-heapq.heappop(arr)
n2=-heapq.heappop(arr)
n1-=1
n2-=1
if n1>0:
heapq.heappush(arr,-n1)
if n2>0:
heapq.heappush(arr,-n2)
print(-sum(arr))
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__":
for _ in range(int(input())):
main(_+1)
``` | output | 1 | 33,604 | 12 | 67,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0
Submitted Solution:
```
import io
import os
from collections import defaultdict
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def solve():
n = int(input())
a = list(map(int, input().split()))
counts = defaultdict(int)
for x in a:
counts[x] += 1
counts = counts.values()
m = max(counts)
if m >= n/2:
print(m-(n-m))
return
print(n%2)
t = int(input())
for _ in range(t):
solve()
``` | instruction | 0 | 33,605 | 12 | 67,210 |
Yes | output | 1 | 33,605 | 12 | 67,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0
Submitted Solution:
```
import sys
input=sys.stdin.readline
t = int(input())
for i in range(t):
n=int(input())
a=[int(x) for x in input().split()]
a.sort()
temp=1
if n==1:
print(1)
continue
count=[]
for i in range(1,n):
if a[i]==a[i-1]:
temp+=1
if i==n-1:
count.append(temp)
else:
count.append(temp)
temp=1
if i==n-1:
count.append(temp)
count.sort()
if len(count)==1:
print(count[0])
continue
if count[-1]<=sum(count[:len(count)-1]):
print(n%2)
else:
print(count[-1]-sum(count[:len(count)-1]))
``` | instruction | 0 | 33,606 | 12 | 67,212 |
Yes | output | 1 | 33,606 | 12 | 67,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0
Submitted Solution:
```
def divisors(M):
d=[]
i=1
while M>=i**2:
if M%i==0:
d.append(i)
if i**2!=M:
d.append(M//i)
i=i+1
return d
def popcount(x):
x = x - ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
return x & 0x0000007f
def eratosthenes(n):
res=[0 for i in range(n+1)]
prime=set([])
for i in range(2,n+1):
if not res[i]:
prime.add(i)
for j in range(1,n//i+1):
res[i*j]=1
return prime
def factorization(n):
res=[]
for p in prime:
if n%p==0:
while n%p==0:
n//=p
res.append(p)
if n!=1:
res.append(n)
return res
def euler_phi(n):
res = n
for x in range(2,n+1):
if x ** 2 > n:
break
if n%x==0:
res = res//x * (x-1)
while n%x==0:
n //= x
if n!=1:
res = res//n * (n-1)
return res
def ind(b,n):
res=0
while n%b==0:
res+=1
n//=b
return res
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(n):
res = [1]
prime = primeFactor(n)
for p in prime:
newres = []
for d in res:
for j in range(prime[p]+1):
newres.append(d*p**j)
res = newres
res.sort()
return res
def xorfactorial(num):#排他的論理和の階乗
if num==0:
return 0
elif num==1:
return 1
elif num==2:
return 3
elif num==3:
return 0
else:
x=baseorder(num)
return (2**x)*((num-2**x+1)%2)+function(num-2**x)
def xorconv(n,X,Y):
if n==0:
res=[(X[0]*Y[0])%mod]
return res
x=[X[i]+X[i+2**(n-1)] for i in range(2**(n-1))]
y=[Y[i]+Y[i+2**(n-1)] for i in range(2**(n-1))]
z=[X[i]-X[i+2**(n-1)] for i in range(2**(n-1))]
w=[Y[i]-Y[i+2**(n-1)] for i in range(2**(n-1))]
res1=xorconv(n-1,x,y)
res2=xorconv(n-1,z,w)
former=[(res1[i]+res2[i])*inv for i in range(2**(n-1))]
latter=[(res1[i]-res2[i])*inv for i in range(2**(n-1))]
former=list(map(lambda x:x%mod,former))
latter=list(map(lambda x:x%mod,latter))
return former+latter
def merge_sort(A,B):
pos_A,pos_B = 0,0
n,m = len(A),len(B)
res = []
while pos_A < n and pos_B < m:
a,b = A[pos_A],B[pos_B]
if a < b:
res.append(a)
pos_A += 1
else:
res.append(b)
pos_B += 1
res += A[pos_A:]
res += B[pos_B:]
return res
class UnionFindVerSize():
def __init__(self, N):
self._parent = [n for n in range(0, N)]
self._size = [1] * N
self.group = N
def find_root(self, x):
if self._parent[x] == x: return x
self._parent[x] = self.find_root(self._parent[x])
stack = [x]
while self._parent[stack[-1]]!=stack[-1]:
stack.append(self._parent[stack[-1]])
for v in stack:
self._parent[v] = stack[-1]
return self._parent[x]
def unite(self, x, y):
gx = self.find_root(x)
gy = self.find_root(y)
if gx == gy: return
self.group -= 1
if self._size[gx] < self._size[gy]:
self._parent[gx] = gy
self._size[gy] += self._size[gx]
else:
self._parent[gy] = gx
self._size[gx] += self._size[gy]
def get_size(self, x):
return self._size[self.find_root(x)]
def is_same_group(self, x, y):
return self.find_root(x) == self.find_root(y)
class WeightedUnionFind():
def __init__(self,N):
self.parent = [i for i in range(N)]
self.size = [1 for i in range(N)]
self.val = [0 for i in range(N)]
self.flag = True
self.edge = [[] for i in range(N)]
def dfs(self,v,pv):
stack = [(v,pv)]
new_parent = self.parent[pv]
while stack:
v,pv = stack.pop()
self.parent[v] = new_parent
for nv,w in self.edge[v]:
if nv!=pv:
self.val[nv] = self.val[v] + w
stack.append((nv,v))
def unite(self,x,y,w):
if not self.flag:
return
if self.parent[x]==self.parent[y]:
self.flag = (self.val[x] - self.val[y] == w)
return
if self.size[self.parent[x]]>self.size[self.parent[y]]:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[x] += self.size[y]
self.val[y] = self.val[x] - w
self.dfs(y,x)
else:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[y] += self.size[x]
self.val[x] = self.val[y] + w
self.dfs(x,y)
class Dijkstra():
class Edge():
def __init__(self, _to, _cost):
self.to = _to
self.cost = _cost
def __init__(self, V):
self.G = [[] for i in range(V)]
self._E = 0
self._V = V
@property
def E(self):
return self._E
@property
def V(self):
return self._V
def add_edge(self, _from, _to, _cost):
self.G[_from].append(self.Edge(_to, _cost))
self._E += 1
def shortest_path(self, s):
import heapq
que = []
d = [10**15] * self.V
d[s] = 0
heapq.heappush(que, (0, s))
while len(que) != 0:
cost, v = heapq.heappop(que)
if d[v] < cost: continue
for i in range(len(self.G[v])):
e = self.G[v][i]
if d[e.to] > d[v] + e.cost:
d[e.to] = d[v] + e.cost
heapq.heappush(que, (d[e.to], e.to))
return d
#Z[i]:length of the longest list starting from S[i] which is also a prefix of S
#O(|S|)
def Z_algorithm(s):
N = len(s)
Z_alg = [0]*N
Z_alg[0] = N
i = 1
j = 0
while i < N:
while i+j < N and s[j] == s[i+j]:
j += 1
Z_alg[i] = j
if j == 0:
i += 1
continue
k = 1
while i+k < N and k + Z_alg[k]<j:
Z_alg[i+k] = Z_alg[k]
k += 1
i += k
j -= k
return Z_alg
class BIT():
def __init__(self,n,mod=0):
self.BIT = [0]*(n+1)
self.num = n
self.mod = mod
def query(self,idx):
res_sum = 0
mod = self.mod
while idx > 0:
res_sum += self.BIT[idx]
if mod:
res_sum %= mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod = self.mod
while idx <= self.num:
self.BIT[idx] += x
if mod:
self.BIT[idx] %= mod
idx += idx&(-idx)
return
class dancinglink():
def __init__(self,n,debug=False):
self.n = n
self.debug = debug
self._left = [i-1 for i in range(n)]
self._right = [i+1 for i in range(n)]
self.exist = [True for i in range(n)]
def pop(self,k):
if self.debug:
assert self.exist[k]
L = self._left[k]
R = self._right[k]
if L!=-1:
if R!=self.n:
self._right[L],self._left[R] = R,L
else:
self._right[L] = self.n
elif R!=self.n:
self._left[R] = -1
self.exist[k] = False
def left(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._left[res]
if res==-1:
break
k -= 1
return res
def right(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._right[res]
if res==self.n:
break
k -= 1
return res
class SparseTable():
def __init__(self,A,merge_func,ide_ele):
N=len(A)
n=N.bit_length()
self.table=[[ide_ele for i in range(n)] for i in range(N)]
self.merge_func=merge_func
for i in range(N):
self.table[i][0]=A[i]
for j in range(1,n):
for i in range(0,N-2**j+1):
f=self.table[i][j-1]
s=self.table[i+2**(j-1)][j-1]
self.table[i][j]=self.merge_func(f,s)
def query(self,s,t):
b=t-s+1
m=b.bit_length()-1
return self.merge_func(self.table[s][m],self.table[t-2**m+1][m])
class BinaryTrie:
class node:
def __init__(self,val):
self.left = None
self.right = None
self.max = val
def __init__(self):
self.root = self.node(-10**15)
def append(self,key,val):
pos = self.root
for i in range(29,-1,-1):
pos.max = max(pos.max,val)
if key>>i & 1:
if pos.right is None:
pos.right = self.node(val)
pos = pos.right
else:
pos = pos.right
else:
if pos.left is None:
pos.left = self.node(val)
pos = pos.left
else:
pos = pos.left
pos.max = max(pos.max,val)
def search(self,M,xor):
res = -10**15
pos = self.root
for i in range(29,-1,-1):
if pos is None:
break
if M>>i & 1:
if xor>>i & 1:
if pos.right:
res = max(res,pos.right.max)
pos = pos.left
else:
if pos.left:
res = max(res,pos.left.max)
pos = pos.right
else:
if xor>>i & 1:
pos = pos.right
else:
pos = pos.left
if pos:
res = max(res,pos.max)
return res
def solveequation(edge,ans,n,m):
#edge=[[to,dire,id]...]
x=[0]*m
used=[False]*n
for v in range(n):
if used[v]:
continue
y = dfs(v)
if y!=0:
return False
return x
def dfs(v):
used[v]=True
r=ans[v]
for to,dire,id in edge[v]:
if used[to]:
continue
y=dfs(to)
if dire==-1:
x[id]=y
else:
x[id]=-y
r+=y
return r
class Matrix():
mod=10**9+7
def set_mod(m):
Matrix.mod=m
def __init__(self,L):
self.row=len(L)
self.column=len(L[0])
self._matrix=L
for i in range(self.row):
for j in range(self.column):
self._matrix[i][j]%=Matrix.mod
def __getitem__(self,item):
if type(item)==int:
raise IndexError("you must specific row and column")
elif len(item)!=2:
raise IndexError("you must specific row and column")
i,j=item
return self._matrix[i][j]
def __setitem__(self,item,val):
if type(item)==int:
raise IndexError("you must specific row and column")
elif len(item)!=2:
raise IndexError("you must specific row and column")
i,j=item
self._matrix[i][j]=val
def __add__(self,other):
if (self.row,self.column)!=(other.row,other.column):
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(self.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(self.column):
res[i][j]=self._matrix[i][j]+other._matrix[i][j]
res[i][j]%=Matrix.mod
return Matrix(res)
def __sub__(self,other):
if (self.row,self.column)!=(other.row,other.column):
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(self.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(self.column):
res[i][j]=self._matrix[i][j]-other._matrix[i][j]
res[i][j]%=Matrix.mod
return Matrix(res)
def __mul__(self,other):
if type(other)!=int:
if self.column!=other.row:
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(other.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(other.column):
temp=0
for k in range(self.column):
temp+=self._matrix[i][k]*other._matrix[k][j]
res[i][j]=temp%Matrix.mod
return Matrix(res)
else:
n=other
res=[[(n*self._matrix[i][j])%Matrix.mod for j in range(self.column)] for i in range(self.row)]
return Matrix(res)
def __pow__(self,m):
if self.column!=self.row:
raise MatrixPowError("the size of row must be the same as that of column")
n=self.row
res=Matrix([[int(i==j) for i in range(n)] for j in range(n)])
while m:
if m%2==1:
res=res*self
self=self*self
m//=2
return res
def __str__(self):
res=[]
for i in range(self.row):
for j in range(self.column):
res.append(str(self._matrix[i][j]))
res.append(" ")
res.append("\n")
res=res[:len(res)-1]
return "".join(res)
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import log,gcd
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
ans = []
for _ in range(int(input())):
n = int(input())
A = li()
dic = {a:0 for a in A}
for a in A:
dic[a] += 1
m = max(dic[a] for a in A)
if 2*m > n:
ans.append(2*m-n)
else:
ans.append(n%2)
print(*ans,sep="\n")
``` | instruction | 0 | 33,607 | 12 | 67,214 |
Yes | output | 1 | 33,607 | 12 | 67,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0
Submitted Solution:
```
from collections import defaultdict
import math
for cases in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
d=defaultdict(int)
for i in range(0,len(l)):
d[l[i]]+=1
max1=0
for i in d.keys():
max1=max(max1,d[i])
if(max1<=math.ceil(n/2)):
if(n%2==0):
print(0)
else:
print(1)
else:
cnt=max1
for i in d.keys():
if(d[i]!=max1):
cnt=cnt-d[i]
print(cnt)
``` | instruction | 0 | 33,608 | 12 | 67,216 |
Yes | output | 1 | 33,608 | 12 | 67,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0
Submitted Solution:
```
def hash(a):
stack = []
for i in a:
stack.append(i)
if len(stack) > 1 and stack[len(stack)-1]!= stack[len(stack)-2]:
stack.pop()
stack.pop()
return len(stack)
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
print(hash(a))
``` | instruction | 0 | 33,609 | 12 | 67,218 |
No | output | 1 | 33,609 | 12 | 67,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0
Submitted Solution:
```
from collections import Counter
def solve(N,A):
d=Counter(A)
mc=d.most_common()
x=mc[0][1]
return max(0,x-(N-x))
T=int(input())
for i in range(T):
N=int(input())
A=list(map(int,input().split()))
print(solve(N,A))
``` | instruction | 0 | 33,610 | 12 | 67,220 |
No | output | 1 | 33,610 | 12 | 67,221 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0
Submitted Solution:
```
import heapq
t=int(input())
for q in range(t):
n=int(input())
arr=list(map(int,input().split()))
di={}
temp=0
for i in range(n):
if di.__contains__(arr[i])==False:
di[arr[i]]=1
else:
temp=di[arr[i]]
di[arr[i]]=temp+1
val=di.values()
lis=list(val)
#print(lis)
for i in range(len(lis)):
lis[i]*=-1
if len(lis)>1:
while (True):
x = heapq.heappop(lis)
y = heapq.heappop(lis)
if x == 0 or y == 0:
heapq.heappush(lis, x)
heapq.heappush(lis, y)
break
x = min(x+1,0)
y= min(y+1,0)
heapq.heappush(lis, x)
heapq.heappush(lis, y)
print(sum(lis)*-1)
else:
print(lis[0]*(-1))
``` | instruction | 0 | 33,611 | 12 | 67,222 |
No | output | 1 | 33,611 | 12 | 67,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:
* you select two different numbers in the array a_i and a_j;
* you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations:
* select i=1, j=5. The array a becomes equal to [6, 1, 1, 4];
* select i=1, j=2. The array a becomes equal to [1, 4].
What can be the minimum size of the array after applying some sequence of operations to it?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.
Example
Input
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
Output
0
0
2
1
0
Submitted Solution:
```
for t in range(int(input())):
n = int(input())
s = list(map(int, input().split()))
d = dict()
for i in range(n):
d[s[i]] = d.get(s[i], 0) + 1
m = max(d.values())
if m <= n // 2:
print(0)
else:
print(2 * m - n)
``` | instruction | 0 | 33,612 | 12 | 67,224 |
No | output | 1 | 33,612 | 12 | 67,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has the sequence consisting of n integers. Vasya consider the pair of integers x and y k-interesting, if their binary representation differs from each other exactly in k bits. For example, if k = 2, the pair of integers x = 5 and y = 3 is k-interesting, because their binary representation x=101 and y=011 differs exactly in two bits.
Vasya wants to know how many pairs of indexes (i, j) are in his sequence so that i < j and the pair of integers ai and aj is k-interesting. Your task is to help Vasya and determine this number.
Input
The first line contains two integers n and k (2 ≤ n ≤ 105, 0 ≤ k ≤ 14) — the number of integers in Vasya's sequence and the number of bits in which integers in k-interesting pair should differ.
The second line contains the sequence a1, a2, ..., an (0 ≤ ai ≤ 104), which Vasya has.
Output
Print the number of pairs (i, j) so that i < j and the pair of integers ai and aj is k-interesting.
Examples
Input
4 1
0 3 2 1
Output
4
Input
6 0
200 100 100 100 200 200
Output
6
Note
In the first test there are 4 k-interesting pairs:
* (1, 3),
* (1, 4),
* (2, 3),
* (2, 4).
In the second test k = 0. Consequently, integers in any k-interesting pair should be equal to themselves. Thus, for the second test there are 6 k-interesting pairs:
* (1, 5),
* (1, 6),
* (2, 3),
* (2, 4),
* (3, 4),
* (5, 6). | instruction | 0 | 33,907 | 12 | 67,814 |
Tags: *special, bitmasks, brute force, meet-in-the-middle
Correct Solution:
```
def binary(n):
curr=0
while(n>0):
if(n%2):
curr+=1
n=n//2
return curr
l=input().split()
n=int(l[0])
k=int(l[1])
l=input().split()
li=[int(i) for i in l]
arr=[]
for i in range(2**15):
if(binary(i)==k):
arr.append(i)
hashi=dict()
for i in li:
if i in hashi:
hashi[i]+=1
else:
hashi[i]=1
count=0
for i in hashi:
for j in arr:
if((i^j) in hashi):
if((i^j)==i):
count=count+(hashi[i]*(hashi[i]-1))
else:
count=count+(hashi[i]*hashi[i^j])
print(count//2)
``` | output | 1 | 33,907 | 12 | 67,815 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has the sequence consisting of n integers. Vasya consider the pair of integers x and y k-interesting, if their binary representation differs from each other exactly in k bits. For example, if k = 2, the pair of integers x = 5 and y = 3 is k-interesting, because their binary representation x=101 and y=011 differs exactly in two bits.
Vasya wants to know how many pairs of indexes (i, j) are in his sequence so that i < j and the pair of integers ai and aj is k-interesting. Your task is to help Vasya and determine this number.
Input
The first line contains two integers n and k (2 ≤ n ≤ 105, 0 ≤ k ≤ 14) — the number of integers in Vasya's sequence and the number of bits in which integers in k-interesting pair should differ.
The second line contains the sequence a1, a2, ..., an (0 ≤ ai ≤ 104), which Vasya has.
Output
Print the number of pairs (i, j) so that i < j and the pair of integers ai and aj is k-interesting.
Examples
Input
4 1
0 3 2 1
Output
4
Input
6 0
200 100 100 100 200 200
Output
6
Note
In the first test there are 4 k-interesting pairs:
* (1, 3),
* (1, 4),
* (2, 3),
* (2, 4).
In the second test k = 0. Consequently, integers in any k-interesting pair should be equal to themselves. Thus, for the second test there are 6 k-interesting pairs:
* (1, 5),
* (1, 6),
* (2, 3),
* (2, 4),
* (3, 4),
* (5, 6). | instruction | 0 | 33,908 | 12 | 67,816 |
Tags: *special, bitmasks, brute force, meet-in-the-middle
Correct Solution:
```
from collections import defaultdict
n, k = [int(i) for i in input().split()]
A = [int(i) for i in input().split()]
A_dict = defaultdict(int)
for i in A:
A_dict[i] += 1
def bitCount(x):
cur = 0
while x > 0:
if x % 2:
cur += 1
x //= 2
return cur
mask = []
for i in range(2**15):
if bitCount(i) == k:
mask.append(i)
ans = 0
for i in A_dict:
for j in mask:
if i^j in A_dict:
if i^j == i:
ans += A_dict[i] * (A_dict[i]-1)
else:
ans += A_dict[i] * A_dict[i^j]
print(ans//2)
``` | output | 1 | 33,908 | 12 | 67,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has the sequence consisting of n integers. Vasya consider the pair of integers x and y k-interesting, if their binary representation differs from each other exactly in k bits. For example, if k = 2, the pair of integers x = 5 and y = 3 is k-interesting, because their binary representation x=101 and y=011 differs exactly in two bits.
Vasya wants to know how many pairs of indexes (i, j) are in his sequence so that i < j and the pair of integers ai and aj is k-interesting. Your task is to help Vasya and determine this number.
Input
The first line contains two integers n and k (2 ≤ n ≤ 105, 0 ≤ k ≤ 14) — the number of integers in Vasya's sequence and the number of bits in which integers in k-interesting pair should differ.
The second line contains the sequence a1, a2, ..., an (0 ≤ ai ≤ 104), which Vasya has.
Output
Print the number of pairs (i, j) so that i < j and the pair of integers ai and aj is k-interesting.
Examples
Input
4 1
0 3 2 1
Output
4
Input
6 0
200 100 100 100 200 200
Output
6
Note
In the first test there are 4 k-interesting pairs:
* (1, 3),
* (1, 4),
* (2, 3),
* (2, 4).
In the second test k = 0. Consequently, integers in any k-interesting pair should be equal to themselves. Thus, for the second test there are 6 k-interesting pairs:
* (1, 5),
* (1, 6),
* (2, 3),
* (2, 4),
* (3, 4),
* (5, 6).
Submitted Solution:
```
from itertools import combinations
from collections import defaultdict
from math import log2
def flip_bit(x, position):
mask = 1 << position
return x ^ mask
def flip(inds, n):
for i in inds:
n = flip_bit(n, i)
return n
def make_diffs(n, K, m):
#length = 14
length = round(log2(m) + 1)
#length = len("{0:b}".format(n)) + K if len("{0:b}".format(n)) + K < 15 else 14
arr = []
difs = combinations(range(length), K)
for el in difs:
arr.append(flip(el, n))
return arr
def main():
N, K = [int(i) for i in input().strip().split()]
nums = [int(i) for i in input().strip().split()]
m = max(nums)
d = defaultdict(list)
for i, e in enumerate(nums):
d[e].append(i)
n = 0
for el in range(N):
vars = make_diffs(nums[el], K, m)
n += sum([1 for i in vars if any(True for ind in d[i] if ind >= el)])
print(n)
main()
``` | instruction | 0 | 33,909 | 12 | 67,818 |
No | output | 1 | 33,909 | 12 | 67,819 |
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