Position of the K-th set bit in a number

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++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns the Kth set bit
int 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 Code
int 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 program to implement
// the above approach
import java.util.*;

class GFG
{

// Function that returns the Kth set bit
static 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 Code
public 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 implementation of the approach 

# Function that returns the Kth set bit
def 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 Code
if __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# program to implement
// the above approach
using System;

class GFG
{

// Function that returns the Kth set bit
static 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 Code
public 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 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
?>

<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>

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.