Rishab and Pairs || RSBPRS

Difficulty:

easy
Hint:
The array must contain an even number of odd and even integers.
Explanation:
Since (even + even) and (odd + odd) give an even integer, the number of even and odd integers must be divisible by 2.
Basically, for every odd integer, there must be another odd integer to make a pair with an even sum, symmetrically for every even integer there must be another even integer to make a pair with an even sum.

Setter’s Solution:

#include <bits/stdc++.h> 
using namespace std; 
int main(){ 
int T; 
cin>>T; 
while(T--){ 
int N; 
cin>>N; 
int eve = 0, odd = 0; 
for(int i = 0; i< 2*N ; i++){ int k; 
cin>>k; 
if(k%2==0) 
eve++; 
else 
odd++; 
} 
if(eve%2==0 && odd%2==0) 
cout<<"TRUE"<<"\n"; 
else 
cout<<"FALSE"<<"\n"; 
} 
return 0; 
}

Time Complexity:
O(2N)
Space Complexity:
O(2
N) for storing the given array.

Python Implementation:

                   
for t in range(int(input())):
    
    n = int(input())
    
    arr = list(map(int, input().split()))
    
    odd = 0
    even = 0
    
    for a in arr:
        
        if a%2==0:
            even+=1
        
        else:
            odd+=1
            
    if odd%2 == 0 and even%2 == 0:
        print("TRUE")
    
    else:
        print("FALSE")
Author: Siddharth Bharmoria - papa_pengu1n

Tester: Ramandeep - ramandeep8421
Editorialist: Siddharth Bharmoria - papa_pengu1n