Iterating over all possible combinations in an Array using Bits

There arises 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 tasks 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 possible 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right


Output:

Yes

Time complexity : 2(number of bits)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.