Check if all the set bits of the binary representation of N are at least K places away Last Updated : 21 Nov, 2022 Comments Improve Suggest changes Like Article Like Report Given numbers N and K, The task is to check if all the set bits of the binary representation of N are at least K places away. Examples: Input: N = 5, K = 1 Output: YES Explanation: Binary of 5 is 101. The 1's are 1 place far from each other. Input: N = 10, K = 2 Output: NO Explanation: Binary of 10 is 1010. The 1's are not at least 2 places far from each other. Approach: Iterate Over all the bits in the binary representation of N and maintain a variable 'count' initialize to 0.Whenever you find a set bit(1), check if count <= K. If not return false.If you find a unset bit(0), increases the value of count by 1. Below is the implementation of the above approach. C++ // C++ program to check if all the set // bits of the binary representation // of N are at least K places away. #include <bits/stdc++.h> using namespace std; bool CheckBits(int N, int K) { // Initialize check and count // with 0 int check = 0; int count = 0; for (int i = 31; i >= 0; i--) { // The i-th bit is a set bit if ((1 << i) & N) { // This is the first set bit so, // start calculating all the // distances between consecutive // bits from here if (check == 0) { check = 1; } else { // If count is less than K // return false if (count < K) { return false; } } count = 0; } else { // Adding the count as the // number of zeroes increase // between set bits count++; } } return true; } // Driver code int main() { int N = 5; int K = 1; if(CheckBits(N, K)) { cout << "YES"; } else { cout << "NO"; } return 0; } Java // Java program to check if all the set // bits of the binary representation // of N are at least K places away. import java.util.*; class GFG{ static boolean CheckBits(int N, int K) { // Initialize check and count // with 0 int check = 0; int count = 0; for(int i = 31; i >= 0; i--) { // The i-th bit is a set bit if (((1 << i) & N) > 0) { // This is the first set bit so, // start calculating all the // distances between consecutive // bits from here if (check == 0) { check = 1; } else { // If count is less than K // return false if (count < K) { return false; } } count = 0; } else { // Adding the count as the // number of zeroes increase // between set bits count++; } } return true; } // Driver code public static void main(String[] args) { int N = 5; int K = 1; if (CheckBits(N, K)) { System.out.print("YES"); } else { System.out.print("NO"); } } } // This code is contributed by shikhasingrajput Python3 # Python3 program to check if all the set # bits of the binary representation # of N are at least K places away. def CheckBits(N, K): # Initialize check and count # with 0 check = 0 count = 0 for i in range(31, -1, -1): # The i-th bit is a set bit if ((1 << i) & N): # This is the first set bit so, # start calculating all the # distances between consecutive # bits from here if (check == 0): check = 1 else: # If count is less than K # return false if (count < K): return False count = 0 else: # Adding the count as the # number of zeroes increase # between set bits count += 1 return True # Driver code if __name__ == "__main__": N = 5 K = 1 if (CheckBits(N, K)): print("YES") else: print("NO") # This code is contributed by chitranayal C# // C# program to check if all the set // bits of the binary representation // of N are at least K places away. using System; public class GFG{ static bool CheckBits(int N, int K) { // Initialize check and count // with 0 int check = 0; int count = 0; for(int i = 31; i >= 0; i--) { // The i-th bit is a set bit if (((1 << i) & N) > 0) { // This is the first set bit so, // start calculating all the // distances between consecutive // bits from here if (check == 0) { check = 1; } else { // If count is less than K // return false if (count < K) { return false; } } count = 0; } else { // Adding the count as the // number of zeroes increase // between set bits count++; } } return true; } // Driver code static public void Main () { int N = 5; int K = 1; if (CheckBits(N, K)) { Console.Write("YES"); } else { Console.Write("NO"); } } } // This code is contributed by avanitrachhadiya2155 JavaScript <script> // Javascript program to check if all the set // bits of the binary representation // of N are at least K places away. function CheckBits(N, K) { // Initialize check and count // with 0 var check = 0; var count = 0; for(var i = 31; i >= 0; i--) { // The i-th bit is a set bit if (((1 << i) & N) > 0) { // This is the first set bit so, // start calculating all the // distances between consecutive // bits from here if (check == 0) { check = 1; } else { // If count is less than K // return false if (count < K) { return false; } } count = 0; } else { // Adding the count as the // number of zeroes increase // between set bits count++; } } return true; } // Driver Code var N = 5; var K = 1; if (CheckBits(N, K)) { document.write("YES"); } else { document.write("NO"); } // This code is contributed by Ankita saini </script> Output: YES Time complexity: O(1)Auxiliary space: O(1) Comment More infoAdvertise with us V veedee Follow Improve Article Tags : Bit Magic DSA Practice Tags : Bit Magic 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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