Sum of product of all pairs of a Binary Array

Last Updated : 01 Jun, 2021

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++

// 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

// 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

# 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

JavaScript

<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>

Output:

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

Sum of Bitwise OR of all pairs in a given array

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Sum of product of all pairs of a Binary Array

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