Given a number n and k (1 <= k <= 32), find the value of k-th bit in the binary representation of n. Bits are numbered from right (Least Significant Bit) to left (Most Significant Bit).

**Examples :**

Input :n = 13, k = 2Output :0Explanation:Binary representation of 13 is 1101. Second bit from right is 0.Input :n = 14, k = 3Output :1Explanation:Binary representation of 14 is 1110. Third bit from right is 1.

**Approach:**

1) Find a number with all 0s except k-th position. We get this number using ( 1 << (k-1) ). For example if k = 3, then ( 1 << 2) gives us (00..00100).

2) Do bitwise and of above-obtained number with n to find if k-th bit in n is set or not.

Below is the implementation of the above approach:

## C++

`// CPP program to find k-th bit from right` `#include <bits/stdc++.h>` `using` `namespace` `std;` `void` `printKthBit(unsigned ` `int` `n, unsigned ` `int` `k)` `{` ` ` `cout << ((n & (1 << (k - 1))) >> (k - 1));` `}` `// Driver Code` `int` `main()` `{` ` ` `unsigned ` `int` `n = 13, k = 2;` ` ` ` ` `// Function Call` ` ` `printKthBit(n, k);` ` ` `return` `0;` `}` |

## Java

`// Java program to find` `// k-th bit from right` `import` `java.io.*;` `class` `GFG {` ` ` `static` `void` `printKthBit(` `long` `n, ` `long` `k)` ` ` `{` ` ` `System.out.println(` ` ` `((n & (` `1` `<< (k - ` `1` `))) >> (k - ` `1` `)));` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `long` `n = ` `13` `, k = ` `2` `;` ` ` ` ` `// Function Call` ` ` `printKthBit(n, k);` ` ` `}` `}` `// This code is contributed by anuj_67.` |

## Python3

`# Python 3 program to find` `# k-th bit from right` `def` `printKthBit(n, k):` ` ` `print` `((n & (` `1` `<< (k ` `-` `1` `))) >> (k ` `-` `1` `))` `# Driver Code` `n ` `=` `13` `k ` `=` `2` `# Function Call` `printKthBit(n, k)` `# This code is contributed by Smitha` |

## C#

`// C# program to find k-th bit from right` `using` `System;` `class` `GFG {` ` ` `static` `void` `printKthBit(` `long` `n, ` `long` `k)` ` ` `{` ` ` `Console.WriteLine((n & (1 << (k - 1))) >> (k - 1));` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `long` `n = 13, k = 2;` ` ` ` ` `// Function Call` ` ` `printKthBit(n, k);` ` ` `}` `}` `// This code is contributed by anuj_67.` |

## PHP

`<?php` `// PHP program to find` `// k-th bit from right` `function` `printKthBit(` `$n` `, ` `$k` `)` `{` ` ` `echo` `(` `$n` `& (1 << (` `$k` `- 1)));` `}` `// Driver Code` `$n` `= 13; ` `$k` `= 2;` `printKthBit(` `$n` `, ` `$k` `);` `// This code is contributed` `// by anuj_67.` `?>` |

## Javascript

`<script>` `// JavaScript program to find k-th bit from right` `function` `printKthBit(n, k)` `{` ` ` `document.write((n & (1 << (k - 1))) >> (k - 1));` `}` `// Driver Code` `var` `n = 13, k = 2;` `// Function Call` `printKthBit(n, k);` `</script>` |

**Output :**

0

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