Count number of ordered pairs with Even and Odd Sums in C++

We are given an array of n positive numbers.The goal is to count the ordered pairs (arr[x], arr[y]) with the sum of arr[x] and arr[y] is even or odd. Pair ( arr[i],arr[j] ) and ( arr[j],arr[i] are counted as separate.

We will traverse the array using two for loops for each number of pairs. Now calculate sum, if it is even increment count by 2 for even sums else increment count by 2 for odd sums.

Let’s understand with examples.

Input− Arr[]= { 1,1,2,3 } N=4

Output− Count of even product sums − 6 Count of odd sum pairs − 6

Explanation − Valid odd sum pairs are −

Arr[0] & Arr[1] → (1,1) Arr[1] & Arr[0] → (1,1) count=2
Arr[0] & Arr[3] → (1,3) Arr[3] & Arr[0] → (3,1) count=2
Arr[1] & Arr[3] → (1,3) Arr[3] & Arr[1] → (3,1) count=2 Total=6
Valid even sum pairs are:
Arr[0] & Arr[2] → (1,2) Arr[2] & Arr[0] → (2,1) count=2
Arr[1] & Arr[2] → (1,2) Arr[2] & Arr[1] → (2,1) count=2
Arr[2] & Arr[3] → (2,3) Arr[3] & Arr[2] → (3,2) count=2 Total=6

Input− Arr[]= { 2,2,2 } N=3

Output − Count of even sum pairs − 6 Count of odd sum pairs − 0

Explanation − Valid even product pairs are −

Arr[0] & Arr[1] → (2,2) Arr[1] & Arr[0] → (2,2) count=2
Arr[1] & Arr[2] → (2,2) Arr[2] & Arr[1] → (2,2) count=2
Arr[2] & Arr[3] → (2,2) Arr[3] & Arr[2] → (2,2) count=2 Total=6
No odd sum as all elements are even.

Approach used in the below program is as follows

We take an integer array arr[] initialized with random numbers.
Take a variable n which stores the length of Arr[].
Function countPairs(int arr[], int n) takes an array, its length as input and prints the count of pairs with even and odd sums.
Traverse array using two for loops for each element of the pair.
Outer Loop from 0<=i<n-1, inner loop i<j<n
Check if arr[i]+arr[j]%2==0. Increment count1 for count of even sum pairs by 2 as arr[i],arr[j] and arr[j],arr[i] will be two pairs.
If the above condition is false increment count2 for odd sum pairs by 2.
At the end of all loops count1 will have a total number of pairs that have even sum and count2 will have a total number of pairs that have odd sum
Print the count1 and count2 as result.

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
void countPairs(int arr[], int n){
   int count1=0; //even sum pairs
   int count2=0; //odd sum pairs
   int sum=0;
   for(int i=0;i<n-1;i++){
      for(int j=i+1;j<n;j++){
         sum=arr[i]+arr[j];
         if(sum%2==0) //sum is even
            { count1+=2; } //(a,b) and (b,a) as two pairs
         else
            { count2+=2; }
      }
   }
   cout<<"Even Sum pairs: "<<count1;
   cout<<endl<<"Odd Sum pairs: "<<count2;
}
int main(){
   int arr[] = { 1,2,3,2 };
   int n = sizeof(arr) / sizeof(int);
   countPairs(arr, n);
   return 0;
}