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Iterating over all possible combinations in an Array using Bits

  • Difficulty Level : Medium
  • Last Updated : 03 Jun, 2021

There arise several situations while solving a problem where we need to iterate over all possible combinations of an array. In this article, we will discuss the method of using bits to do so.
For the purpose of explaining, consider the following question: 

Given an array b[] = {2, 1, 4}. The task is to check if there exists any combination of elements of this array whose sum of elements is equal to k = 6.  

Solution using Bit operations
As there are 3 elements in this array, hence we need 3 bits to represent each of the numbers. A bit set as 1 corresponding to the element means it is included while calculating the sum, and not if it is 0. 
The possible combinations are: 

000 : No element is selected.
001 : 4 is selected.
010 : 1 is selected.
011 : 1 and 4 are selected.
100 : 2 is selected.
101 : 2 and 4 are selected.
110 : 2 and 1 are selected.
111 : All elements are selected.

Hence, the range required to access all these bits is 0 – 7. We iterate over each bit of each of the possible combinations, and check for each combination if the sum of chosen elements is equal to the required sum or not.
 

Examples: 



Input : A = {3, 4, 1, 2} and k = 6 
Output : YES
Here, the combination of using 3, 1 and 2 yields 
the required sum.

Input : A = {3, 4, 1, 2} and k = 11
Output : NO

Below is the implementation of the above approach: 

C++




// C++ program to iterate over all possible
// combinations of array elements
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if any combination of
// elements of the array sums to k
bool checkSum(int a[], int n, int k)
{
    // Flag variable to check if
    // sum exists
    int flag = 0;
 
    // Calculate number of bits
    int range = (1 << n) - 1;
 
    // Generate combinations using bits
    for (int i = 0; i <= range; i++) {
 
        int x = 0, y = i, sum = 0;
 
        while (y > 0) {
 
            if (y & 1 == 1) {
 
                // Calculate sum
                sum = sum + a[x];
            }
            x++;
            y = y >> 1;
        }
 
        // If sum is found, set flag to 1
        // and terminate the loop
        if (sum == k)
           return true;
    }
 
    return false;
}
 
// Driver Code
int main()
{
    int k = 6;
    int a[] = { 3, 4, 1, 2 };
    int n = sizeof(a)/sizeof(a[0]);
    if (checkSum(a, n, k))
       cout << "Yes";
    else
       cout << "No";
 
    return 0;
}

Java




// Java program to iterate over all possible
// combinations of array elements
class GFG
{
     
// Function to check if any combination
// of elements of the array sums to k
static boolean checkSum(int a[], int n, int k)
{
    // Flag variable to check if
    // sum exists
    int flag = 0;
 
    // Calculate number of bits
    int range = (1 << n) - 1;
 
    // Generate combinations using bits
    for (int i = 0; i <= range; i++)
    {
        int x = 0, y = i, sum = 0;
 
        while (y > 0)
        {
            if ((y & 1) == 1)
            {
 
                // Calculate sum
                sum = sum + a[x];
            }
            x++;
            y = y >> 1;
        }
 
        // If sum is found, set flag to 1
        // and terminate the loop
        if (sum == k)
        return true;
    }
 
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    int k = 6;
    int a[] = { 3, 4, 1, 2 };
    int n = a.length;
    if (checkSum(a, n, k))
    System.out.println("Yes");
    else
    System.out.println("No");
 
}
}
 
// This code is contributed
// by Code_Mech

Python3




# Python 3 program to iterate over all
# possible combinations of array elements
 
# Function to check if any combination of
# elements of the array sums to k
def checkSum(a, n, k):
     
    # Flag variable to check if
    # sum exists
    flag = 0
 
    # Calculate number of bits
    range__ = (1 << n) - 1
 
    # Generate combinations using bits
    for i in range(range__ + 1):
        x = 0
        y = i
        sum = 0
 
        while (y > 0):
            if (y & 1 == 1):
                 
                # Calculate sum
                sum = sum + a[x]
 
            x += 1
            y = y >> 1
 
        # If sum is found, set flag to 1
        # and terminate the loop
        if (sum == k):
            return True
 
    return False
 
# Driver Code
if __name__ == '__main__':
    k = 6
    a = [3, 4, 1, 2]
    n = len(a)
    if (checkSum(a, n, k)):
        print("Yes")
    else:
        print("No")
         
# This code is contributed by
# Surendra_Gangwar

C#




// C# program to iterate over all possible
// combinations of array elements
using System;
class GFG
{
// Function to check if any combination
// of elements of the array sums to k
static bool checkSum(int[] a, int n, int k)
{
    // Flag variable to check if
    // sum exists
    int // C# program to iterate over all possible
// combinations of array elements
using System;
 
class GFG
{
     
// Function to check if any combination
// of elements of the array sums to k
static bool checkSum(int[] a, int n, int k)
{
    // Flag variable to check if
    // sum exists
    int flag = 0;
 
    // Calculate number of bits
    int range = (1 << n) - 1;
 
    // Generate combinations using bits
    for (int i = 0; i <= range; i++)
    {
        int x = 0, y = i, sum = 0;
 
        while (y > 0)
        {
            if ((y & 1) == 1)
            {
 
                // Calculate sum
                sum = sum + a[x];
            }
            x++;
            y = y >> 1;
        }
 
        // If sum is found, set flag to 1
        // and terminate the loop
        if (sum == k)
        return true;
    }
 
    return false;
}
 
// Driver Code
public static void Main()
{
    int k = 6;
    int[] a = { 3, 4, 1, 2 };
    int n = a.Length;
    if (checkSum(a, n, k))
    Console.WriteLine("Yes");
    else
    Console.WriteLine("No");
}
}
 
// This code is contributed
// by Code_Mech

PHP




<?php
// PHP program to iterate over all possible
// combinations of array elements
 
// Function to check if any combination of
// elements of the array sums to k
function checkSum($a, $n, $k)
{
    // Flag variable to check if
    // sum exists
    $flag = 0;
 
    // Calculate number of bits
    $range = (1 << $n) - 1;
 
    // Generate combinations using bits
    for ($i = 0; $i <= $range; $i++)
    {
 
        $x = 0;
        $y = $i;
        $sum = 0;
 
        while ($y > 0)
        {
 
            if ($y & 1 == 1)
            {
 
                // Calculate sum
                $sum = $sum + $a[$x];
            }
            $x++;
            $y = $y >> 1;
        }
 
        // If sum is found, set flag to 1
        // and terminate the loop
        if ($sum == $k)
        return true;
    }
 
    return false;
}
 
    // Driver Code
    $k = 6;
    $a = array( 3, 4, 1, 2 );
    $n = sizeof($a);
    if (checkSum($a, $n, $k))
        echo "Yes";
    else
        echo "No";
 
    // This code is contributed by Ryuga
?>

Javascript




<script>
    // Javascript program to iterate over all possible
    // combinations of array elements
     
    // Function to check if any combination
    // of elements of the array sums to k
    function checkSum(a, n, k)
    {
        // Flag variable to check if
        // sum exists
        let flag = 0;
 
        // Calculate number of bits
        let range = (1 << n) - 1;
 
        // Generate combinations using bits
        for (let i = 0; i <= range; i++)
        {
            let x = 0, y = i, sum = 0;
 
            while (y > 0)
            {
                if ((y & 1) == 1)
                {
 
                    // Calculate sum
                    sum = sum + a[x];
                }
                x++;
                y = y >> 1;
            }
 
            // If sum is found, set flag to 1
            // and terminate the loop
            if (sum == k)
            return true;
        }
 
        return false;
    }
     
    let k = 6;
    let a = [ 3, 4, 1, 2 ];
    let n = a.length;
    if (checkSum(a, n, k))
        document.write("Yes");
    else
        document.write("No");
 
     
</script>
Output: 
Yes

 

Time complexity : 2(number of bits)
 

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