Iterating over all possible combinations in an Array using Bits Last Updated : 03 Jun, 2021 Comments Improve Suggest changes Like Article Like Report 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) Comment More infoAdvertise with us Next Article Analysis of Algorithms N noob_coders_ka_baap Follow Improve Article Tags : DSA Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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