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Count of all possible pairs of array elements with same parity

  • Difficulty Level : Easy
  • Last Updated : 05 Jul, 2021

Given an array A[] of integers, the task is to find the total number of pairs such that each pair contains either both even or both odd elements. A valid pair (A[ i ], A[ j ]) can only be formed if i != j

Examples: 

Input: A[ ] = {1, 2, 3, 1, 3} 
Output:
Explanation: 
Possible odd pairs = (1, 3), (1, 1), (1, 3), (3, 1), (3, 3), (1, 3) = 6 
Possible even pairs = 0 
Hence, total pairs = 6 + 0 = 6

Input: A[ ] = {8, 2, 3, 1, 4, 2} 
Output: 7 
Explanation: 
Possible odd pair = (3, 1) = 1 
Possible even pairs = (8, 2), (8, 4), (8, 2), (2, 4), (2, 2), (4, 2) = 6 
Hence, total pairs = 6 + 1 = 7 
 

Naive Approach: 
The simplest approach is to generate all possible pairs. For each pair, check if both elements are odd or both are even. If so, increment a counter. The final count will be the required answer. 



Time Complexity: O(N2)

Below is the implementation of the above approach:  

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the answer
int countPairs(int A[], int n)
{
    int count = 0, i, j;
     
    // Generate all possible pairs
    for(i = 0; i < n; i++)
    {
        for(j = i + 1; j < n; j++)
        {
             
            // Increment the count if
            // both even or both odd
            if ((A[i] % 2 == 0 &&
                 A[j] % 2 == 0) ||
                (A[i] % 2 != 0 &&
                 A[j] % 2 != 0))
                count++;
        }
    }
    return count;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 2, 3, 1, 3 };
    int n = sizeof(A) / sizeof(int);
     
    cout << countPairs(A, n);
}
 
// This code is contributed by jrishabh99

Java




// Java program for above approach
 
import java.util.*;
 
class GFG {
 
    static int countPairs(
        int[] A, int n)
    {
        int count = 0, i, j;
 
        // Generate all possible pairs
        for (i = 0; i < n; i++) {
            for (j = i + 1; j < n; j++) {
 
                // Increment the count if
                // both even or both odd
                if ((A[i] % 2 == 0
                     && A[j] % 2 == 0)
                    || (A[i] % 2 != 0
                        && A[j] % 2 != 0))
                    count++;
            }
        }
        return count;
    }
 
    // Driver Code
    public static void main(
        String[] args)
    {
        int[] A = { 1, 2, 3, 1, 3 };
        int n = A.length;
        System.out.println(
            countPairs(A, n));
    }
}

Python3




# Python3 program for
# the above approach
 
# Function to return the answer
def countPairs(A, n):
 
    count = 0
     
    # Generate all possible pairs
    for i in range (n):
        for j in range (i + 1, n):
       
            # Increment the count if
            # both even or both odd
            if ((A[i] % 2 == 0 and
                 A[j] % 2 == 0) or
                (A[i] % 2 != 0 and
                 A[j] % 2 != 0)):
                count += 1
     
    return count
 
# Driver Code
if __name__ == "__main__":
   
    A = [1, 2, 3, 1, 3]
    n = len(A)
    print(countPairs(A, n))
     
# This code is contributed by Chitranayal

C#




// C# program for above approach
using System;
class GFG{
 
static int countPairs(int[] A, int n)
{
    int count = 0, i, j;
 
    // Generate all possible pairs
    for (i = 0; i < n; i++)
    {
        for (j = i + 1; j < n; j++)
        {
 
            // Increment the count if
            // both even or both odd
            if ((A[i] % 2 == 0 && A[j] % 2 == 0) ||
                (A[i] % 2 != 0 && A[j] % 2 != 0))
                count++;
        }
    }
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] A = { 1, 2, 3, 1, 3 };
    int n = A.Length;
    Console.Write(countPairs(A, n));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript




<script>
 
// Javascript program for above approach
function countPairs(A, n)
{
    let count = 0, i, j;
 
    // Generate all possible pairs
    for(i = 0; i < n; i++)
    {
        for(j = i + 1; j < n; j++)
        {
             
            // Increment the count if
            // both even or both odd
            if ((A[i] % 2 == 0 && A[j] % 2 == 0) ||
                (A[i] % 2 != 0 && A[j] % 2 != 0))
                count++;
        }
    }
    return count;
}
 
// Driver code
let A = [ 1, 2, 3, 1, 3 ];
let n = A.length;
 
document.write(countPairs(A, n));
 
// This code is contributed by rameshtravel07
 
</script>
Output: 
6

 

Efficient Approach: 
Traverse the array and count and store even and odd numbers in the array and calculate the possible pairs from respective counts and display their sum.

Let the count of even and odd elements in the array be EC and OC respectively. 
Count of even pairs = ( EC * ( EC – 1 ) ) / 2 
Count of odd pairs = ( OC * ( OC – 1 ) ) / 2 
Hence, total number of possible pairs = (( EC * ( EC – 1 ) ) + ( OC * ( OC – 1 ) ))/ 2 
 

Below is the implementation of the above approach:  

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
int countPairs(int A[], int n)
{
     
    // Store count of
    // even and odd elements
    int even = 0, odd = 0;
     
    for(int i = 0; i < n; i++)
    {
        if (A[i] % 2 == 0)
            even++;
        else
            odd++;
    }
 
    return (even * (even - 1)) / 2 +
             (odd * (odd - 1)) / 2;
}
 
// Driver code
int main()
{
    int A[] = { 1, 2, 3, 1, 3 };
    int n = sizeof(A) / sizeof(int);
     
    cout << countPairs(A, n);
}
 
// This code is contributed by jrishabh99

Java




// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
    static int countPairs(
        int[] A, int n)
    {
        // Store count of
        // even and odd elements
        int even = 0, odd = 0;
 
        for (int i = 0; i < n; i++) {
 
            if (A[i] % 2 == 0)
                even++;
            else
                odd++;
        }
 
        return (even * (even - 1)) / 2
            + (odd * (odd - 1)) / 2;
    }
 
    // Driver Program
    public static void main(
        String[] args)
    {
 
        int[] A = { 1, 2, 3, 1, 3 };
        int n = A.length;
        System.out.println(
            countPairs(A, n));
    }
}

Python3




# Python3 program for the above approach
 
# Function to return count of pairs
def countPairs(A, n):
     
    # Store count of
    # even and odd elements
    even, odd = 0, 0
     
    for i in range(0, n):
        if A[i] % 2 == 0:
            even = even + 1
        else:
            odd = odd + 1
 
    return ((even * (even - 1)) // 2 +
              (odd * (odd - 1)) // 2)
 
# Driver code
A = [ 1, 2, 3, 1, 3 ]
n = len(A)
 
print(countPairs(A, n))
 
# This code is contributed by jrishabh99

C#




// C# program for the above approach
using System;
 
class GFG{
 
static int countPairs(int[] A, int n)
{
     
    // Store count of
    // even and odd elements
    int even = 0, odd = 0;
 
    for(int i = 0; i < n; i++)
    {
    if (A[i] % 2 == 0)
        even++;
    else
        odd++;
    }
 
    return (even * (even - 1)) / 2 +
            (odd * (odd - 1)) / 2;
}
 
// Driver code
public static void Main()
{
    int[] A = { 1, 2, 3, 1, 3 };
    int n = A.Length;
     
    Console.Write(countPairs(A, n));
}
}
 
// This code is contributed by nidhi_biet

Javascript




<script>
 
    // Javascript program for the above approach
     
    function countPairs(A, n)
    {
 
        // Store count of
        // even and odd elements
        let even = 0, odd = 0;
 
        for(let i = 0; i < n; i++)
        {
            if (A[i] % 2 == 0)
                even++;
            else
                odd++;
        }
 
        return (even * (even - 1)) / 2 + (odd * (odd - 1)) / 2;
    }
     
    let A = [ 1, 2, 3, 1, 3 ];
    let n = A.length;
      
    document.write(countPairs(A, n));
     
</script>
Output: 
6

 

Time complexity: O(N)
 

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