Sum of product of all pairs of a Binary Array

Given a binary array arr[] of size N, the task is to print the sum of product of all pairs of the given array elements.

Note: The binary array contains only 0 and 1.

Examples:

Input: arr[] = {0, 1, 1, 0, 1}
Output: 3
Explanation: Sum of product of all possible pairs are: {0 × 1 + 0 × 1 + 0 × 0 + 0 × 1 + 1 × 1 + 1 × 0 + 1 × 1 + 1 × 0 + 1 × 1 + 0 × 1}.
Therefore, the required output is 3.

Input: arr[] = {1, 1, 1, 1}
Output: 6

Naive Approach: The simplest approach to solve the problem is to use generate all possible pairs from the array and calculate the sum of their product. 

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, consider only those pairs in which both the elements are 1. Following are the observations:

If there is a pair (arr[i], arr[j]) where arr[i] × arr[j] = 1, then arr[i] and arr[j] must be 1.

Total number of pairs that satisfy (arr[i] × arr[j] = 1) are: 
=> 

\binom{cntOne}{2}      
 

=> cntOne × (cntOne - 1) / 2 
where, cntOne is the count of 1s in the given array

 

Follow the steps below to solve the problem: 

• Initialize the variable cntOne to store the count of 1s from the given array.
• Finally, return the value of cntOne * (cntOne - 1) / 2.

Below is the implementation of the above approach:

// C++ program to implement 
// the above approach 
#include <bits/stdc++.h> 
using namespace std;

// Function to print the sum of product
// of all pairs of the given array
int productSum(int arr[], int N)
{
    
    // Stores count of one in
    // the given array
    int cntOne = 0;
 
    for(int i = 0; i < N; i++)
    {
        
        // If current element is 1
        if (arr[i] == 1)
 
            // Increase count
            cntOne++;
    }
 
    // Return the sum of product
    // of all pairs
    return cntOne * (cntOne - 1) / 2;
}
 
// Driver Code
int main()
{
    int arr[] = { 0, 1, 1, 0, 1 };
    
    // Stores the size of
    // the given array
    int n = sizeof(arr) / sizeof(arr[0]);
    
    cout << productSum(arr, n) << endl;
}

// This code is contributed by code_hunt

// Java program to implement
// the above approach
import java.util.*;

class GFG {

    // Function to print the sum of product
    // of all pairs of the given array
    static int productSum(int[] arr)
    {

        // Stores count of one in
        // the given array
        int cntOne = 0;

        // Stores the size of
        // the given array
        int N = arr.length;

        for (int i = 0; i < N; i++) {

            // If current element is 1
            if (arr[i] == 1)

                // Increase count
                cntOne++;
        }

        // Return the sum of product
        // of all pairs
        return cntOne * (cntOne - 1) / 2;
    }

    // Driver Code
    public static void main(String[] args)
    {

        int[] arr = { 0, 1, 1, 0, 1 };

        System.out.println(productSum(arr));
    }
}

# Python3 program to implement 
# the above approach 

# Function to print the sum of product
# of all pairs of the given array
def productSum(arr):
 
    # Stores count of one in
    # the given array
    cntOne = 0
    
    # Stores the size of
    # the given array
    N = len(arr)
 
    for i in range(N):
 
        # If current element is 1
        if (arr[i] == 1):
 
            # Increase count
            cntOne += 1
    
    # Return the sum of product
    # of all pairs
    return cntOne * (cntOne - 1) // 2

# Driver Code
arr = [ 0, 1, 1, 0, 1 ]

print(productSum(arr))

# This code is contributed by code_hunt

// C# program to implement 
// the above approach 
using System;

class GFG{
 
// Function to print the sum of product
// of all pairs of the given array
static int productSum(int[] arr)
{
    
    // Stores count of one in
    // the given array
    int cntOne = 0;

    // Stores the size of
    // the given array
    int N = arr.Length;

    for(int i = 0; i < N; i++)
    {
        
        // If current element is 1
        if (arr[i] == 1)

            // Increase count
            cntOne++;
    }

    // Return the sum of product
    // of all pairs
    return cntOne * (cntOne - 1) / 2;
}

// Driver Code
public static void Main()
{
    int[] arr = { 0, 1, 1, 0, 1 };

    Console.Write(productSum(arr));
}
}

// This code is contributed by code_hunt

<script>

// Javascript program to implement
// the above approach

// Function to print the sum of product
// of all pairs of the given array
function productSum(arr, N)
{
    
    // Stores count of one in
    // the given array
    let cntOne = 0;

    for(let i = 0; i < N; i++)
    {
        
        // If current element is 1
        if (arr[i] == 1)

            // Increase count
            cntOne++;
    }

    // Return the sum of product
    // of all pairs
    return cntOne * (cntOne - 1) / 2;
}

// Driver Code

    let arr = [ 0, 1, 1, 0, 1 ];
    
    // Stores the size of
    // the given array
    let n = arr.length;
    
    document.write(productSum(arr, n) + "<br>");


// This code is contributed by Mayank Tyagi

</script>

3

Time Complexity: O(N) 
Auxiliary Space: O(1)