Given two numbers N and K, The task is to find the index of the K-th set bit in the number from the right.
Note: Indexing in the binary representation starts from 0 from the right. For example in the binary number “000011”, the first set bit is at index 0 from the right, and the second set bit is at index 1 from the right.
Examples:
Input: N = 15, K = 3 Output: 2 15 is "1111", hence the third bit is at index 2 from right. Input: N = 19, K = 2 Output: 1 19 is "10011", hence the second set bit is at index 1 from right.
Approach: Initialize a counter 0, and increase it if the last bit is set in the number. For accessing the next bit, right-shift the number by 1. When the counter’s value is equal to K, then we return the index of the number which was being incremented on every right shift.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function that returns the Kth set bitint FindIndexKthBit(int n, int k){ int cnt = 0; int ind = 0; // Traverse in the binary while (n) { // Check if the last // bit is set or not if (n & 1) cnt++; // Check if count is equal to k // then return the index if (cnt == k) return ind; // Increase the index // as we move right ind++; // Right shift the number by 1 n = n >> 1; } return -1;}// Driver Codeint main(){ int n = 15, k = 3; int ans = FindIndexKthBit(n, k); if (ans != -1) cout << ans; else cout << "No k-th set bit"; return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function that returns the Kth set bitstatic int FindIndexKthBit(int n, int k){ int cnt = 0; int ind = 0; // Traverse in the binary while (n > 0) { // Check if the last // bit is set or not if ((n & 1 ) != 0) cnt++; // Check if count is equal to k // then return the index if (cnt == k) return ind; // Increase the index // as we move right ind++; // Right shift the number by 1 n = n >> 1; } return -1;}// Driver Codepublic static void main(String args[]){ int n = 15, k = 3; int ans = FindIndexKthBit(n, k); if (ans != -1) System.out.println(ans); else System.out.println("No k-th set bit");}}// This code is contributed by // Surendra_Gangwar |
Python3
# Python3 implementation of the approach # Function that returns the Kth set bitdef FindIndexKthBit(n, k): cnt, ind = 0, 0 # Traverse in the binary while n > 0: # Check if the last # bit is set or not if n & 1: cnt += 1 # Check if count is equal to k # then return the index if cnt == k: return ind # Increase the index # as we move right ind += 1 # Right shift the number by 1 n = n >> 1 return -1# Driver Codeif __name__ == "__main__": n, k = 15, 3 ans = FindIndexKthBit(n, k) if ans != -1: print(ans) else: print("No k-th set bit") # This code is contributed by # Rituraj Jain |
C#
// C# program to implement// the above approachusing System;class GFG{// Function that returns the Kth set bitstatic int FindIndexKthBit(int n, int k){ int cnt = 0; int ind = 0; // Traverse in the binary while (n > 0) { // Check if the last // bit is set or not if ((n & 1 ) != 0) cnt++; // Check if count is equal to k // then return the index if (cnt == k) return ind; // Increase the index // as we move right ind++; // Right shift the number by 1 n = n >> 1; } return -1;}// Driver Codepublic static void Main(){ int n = 15, k = 3; int ans = FindIndexKthBit(n, k); if (ans != -1) Console.WriteLine(ans); else Console.WriteLine("No k-th set bit");}}// This code is contributed by // Code_Mech. |
PHP
<?php// PHP program to implement // the above approach // Function that returns the Kth set bit function FindIndexKthBit($n, $k) { $cnt = 0; $ind = 0; // Traverse in the binary while ($n) { // Check if the last // bit is set or not if ($n & 1) $cnt++; // Check if count is equal to k // then return the index if ($cnt == $k) return $ind; // Increase the index // as we move right $ind++; // Right shift the number by 1 $n = $n >> 1; } return -1; } // Driver Code $n = 15;$k = 3; $ans = FindIndexKthBit($n, $k); if ($ans != -1) echo $ans; else echo "No k-th set bit";// This code is contributed by Ryuga?> |
Javascript
<script> // Javascript program to implement // the above approach // Function that returns the Kth set bit function FindIndexKthBit(n, k) { let cnt = 0; let ind = 0; // Traverse in the binary while (n > 0) { // Check if the last // bit is set or not if (n & 1 > 0) cnt++; // Check if count is equal to k // then return the index if (cnt == k) return ind; // Increase the index // as we move right ind++; // Right shift the number by 1 n = n >> 1; } return -1; } let n = 15, k = 3; let ans = FindIndexKthBit(n, k); if (ans != -1) document.write(ans); else document.write("No k-th set bit");</script> |
Output:
2
Time Complexity: O(logN), as we are using a while loop to traverse and in each traversal we are right shifting N by 1 place which is equivalent to floor division of N by 2. So, the effective time will be 1+1/2+1/4+…..+1/2^N which is equivalent to logN.
Auxiliary Space: O(1), as we are not using any extra space.
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