To determine if there exists an array ( B ) such that the array ( C ) constructed as ( C_i = A_i + B_i ) is a permutation of the numbers from 1 to ( n ), we need to check certain conditions for each test case.
Here’s how we can approach this problem:Identify the Constraints:Array ( C ) must be a permutation of the numbers from 1 to ( n ), meaning it must contain each number from 1 to ( n ) exactly once.
Analyze Array ( A ): We need to check if it’s possible to add integers ( B_i ) such that each element of ( C ) falls within the range [1, n] and all values are unique.
Steps to Solve:For each element ( A_i ), determine the range of possible values of ( B_i ) such that ( C_i = A_i + B_i ) lies between 1 and ( n ).
Ensure that all values from 1 to ( n ) can be covered without overlap, meaning each possible ( C_i ) must be unique.
Sorting Array ( A ):
Sort ( A ) to easily check if ( B_i ) can be chosen to make ( C ) a permutation.
Checking Range:For each sorted value ( A_i ), check if adding ( i ) (from 0 to ( n-1 )) keeps ( A_i + i ) within the range [1, n].If any value fails to stay within this range, output “NO”.
Input and Output Handling:The function reads multiple test cases and processes each one.This approach ensures that we check for the possibility of forming the permutation correctly and efficiently
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
b=sorted(a)
f=1
for i in range(0,n-1):
if b[i]>i+1:
f=0
else:
continue
if not f:
print("No")
else:
print("Yes")