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:

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


Output:

Yes

Time complexity : 2(number of bits)



My Personal Notes arrow_drop_up