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:
=>
=> 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> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1)