# 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++

 `// C++ program to implement` `// the above approach` `#include ` `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

 `// 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

 `# 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#

 `// 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

 `> 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

 ``

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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