Check whether K-th bit is set or not Last Updated : 12 Mar, 2025 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice Given a number n and a bit position k, check if the kth bit of n is set or not. A bit is called set if it is 1. Note: Indexing starts with 0 from LSB (least significant bit) side in the binary representation of the number.Examples: Input: n = 7, k = 2Output: YesExplanation: 7 is represented as 111 in binary and bit at position 2 is set.Input: n = 5, k = 1Output: NoExplanation: 5 is represented as 101 in binary and bit at position 1 is not set.Using Left Shift Operator - O(k) time and O(1) spaceThe idea is to leverage bitwise operations to check if a specific bit is set. Create a number that has only the k-th bit set (by left-shifting 1 by k positions). Then we perform a bitwise AND operation between this number and the given number n. If the result is non-zero, it means the k-th bit in n is set to 1, otherwise it's 0. C++ // CPP program to check if k-th bit // of a given number is set or not #include <iostream> using namespace std; bool checkKthBit(int n, int k) { // Value whose only kth bit // is set. int val = (1<<k); // If AND operation of n and // value is non-zero, it means // k'th bit is set if ((n&val) != 0) { return true; } return false; } int main() { int n = 7, k = 2; if (checkKthBit(n, k)) { cout << "Yes" << endl; } else { cout << "No" << endl; } return 0; } Java // Java program to check if k-th bit // of a given number is set or not class GfG { static boolean checkKthBit(int n, int k) { // Value whose only kth bit // is set. int val = (1 << k); // If AND operation of n and // value is non-zero, it means // k'th bit is set if ((n & val) != 0) { return true; } return false; } public static void main(String[] args) { int n = 7, k = 2; if (checkKthBit(n, k)) { System.out.println("Yes"); } else { System.out.println("No"); } } } Python # Python program to check if k-th bit # of a given number is set or not def checkKthBit(n, k): # Value whose only kth bit # is set. val = (1 << k) # If AND operation of n and # value is non-zero, it means # k'th bit is set if (n & val) != 0: return True return False if __name__ == "__main__": n, k = 7, 2 if checkKthBit(n, k): print("Yes") else: print("No") C# // C# program to check if k-th bit // of a given number is set or not using System; class GfG { static bool checkKthBit(int n, int k) { // Value whose only kth bit // is set. int val = (1 << k); // If AND operation of n and // value is non-zero, it means // k'th bit is set if ((n & val) != 0) { return true; } return false; } static void Main() { int n = 7, k = 2; if (checkKthBit(n, k)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } } } JavaScript // JavaScript program to check if k-th bit // of a given number is set or not function checkKthBit(n, k) { // Value whose only kth bit // is set. let val = (1 << k); // If AND operation of n and // value is non-zero, it means // k'th bit is set if ((n & val) !== 0) { return true; } return false; } let n = 7, k = 2; if (checkKthBit(n, k)) { console.log("Yes"); } else { console.log("No"); } OutputYes Using Right Shift Operator - O(k) time and O(1) spaceThe idea is to shift the bits of the given number to the right by k positions, which brings the k-th bit to the rightmost position (0th position). Then we perform a bitwise AND operation with 1. If this result is non-zero (meaning the bit is 1), then the k-th bit in the original number was set, otherwise it wasn't. C++ // CPP program to check if k-th bit // of a given number is set or not #include <iostream> using namespace std; bool checkKthBit(int n, int k) { // Right shift n by k n = n >> k; // If 0th bit is set if ((n & 1) != 0) { return true; } return false; } int main() { int n = 7, k = 2; if (checkKthBit(n, k)) { cout << "Yes" << endl; } else { cout << "No" << endl; } return 0; } Java // Java program to check if k-th bit // of a given number is set or not class GfG { static boolean checkKthBit(int n, int k) { // Right shift n by k n = n >> k; // If 0th bit is set if ((n & 1) != 0) { return true; } return false; } public static void main(String[] args) { int n = 7, k = 2; if (checkKthBit(n, k)) { System.out.println("Yes"); } else { System.out.println("No"); } } } Python # Python program to check if k-th bit # of a given number is set or not def checkKthBit(n, k): # Right shift n by k n = n >> k # If 0th bit is set if (n & 1) != 0: return True return False if __name__ == "__main__": n, k = 7, 2 if checkKthBit(n, k): print("Yes") else: print("No") C# // C# program to check if k-th bit // of a given number is set or not using System; class GfG { static bool checkKthBit(int n, int k) { // Right shift n by k n = n >> k; // If 0th bit is set if ((n & 1) != 0) { return true; } return false; } static void Main() { int n = 7, k = 2; if (checkKthBit(n, k)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } } } JavaScript // JavaScript program to check if k-th bit // of a given number is set or not function checkKthBit(n, k) { // Right shift n by k n = n >> k; // If 0th bit is set if ((n & 1) !== 0) { return true; } return false; } let n = 7, k = 2; if (checkKthBit(n, k)) { console.log("Yes"); } else { console.log("No"); } OutputYes Check whether K-th bit is set or not | DSA Problem Comment More infoAdvertise with us Next Article Types of Asymptotic Notations in Complexity Analysis of Algorithms S Sakshi_Tiwari Improve Article Tags : Misc Strings Bit Magic DSA Microsoft +1 More Practice Tags : MicrosoftBit MagicMiscStrings Similar Reads Basics & PrerequisitesTime Complexity and Space ComplexityMany times there are more than one ways to solve a problem with different algorithms and we need a way to compare multiple ways. 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