# M-th smallest number having k number of set bits.

Given two non-negative integers **m** and **k**. The problem is to find the **m-th** smallest number having **k** number of set bits.**Constraints:** 1 <= m, k.**Examples:**

Input : m = 4, k = 2 Output : 9 (9)_{10}= (1001)_{2}, it is the4thsmallest number having2set bits. Input : m = 6, k = 4 Output : 39

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**Approach:** Following are the steps:

- Find the smallest number having
**k**number of set bits. Let it be**num**, where**num**= (1 << k) – 1. - Loop for
**m-1**times and each time replace**num**with the next higher number than ‘num’ having same number of bits as in ‘num’. Refer this post to find the required next higher number. - Finally return
**num**.

## C++

`// C++ implementation to find the mth smallest` `// number having k number of set bits` `#include <bits/stdc++.h>` `using` `namespace` `std;` `typedef` `unsigned ` `int` `uint_t;` `// function to find the next higher number` `// with same number of set bits as in 'x'` `uint_t nxtHighWithNumOfSetBits(uint_t x)` `{` ` ` `uint_t rightOne;` ` ` `uint_t nextHigherOneBit;` ` ` `uint_t rightOnesPattern;` ` ` `uint_t next = 0;` ` ` `/* the approach is same as discussed in` ` ` `*/` ` ` `if` `(x) {` ` ` `rightOne = x & -(` `signed` `)x;` ` ` `nextHigherOneBit = x + rightOne;` ` ` `rightOnesPattern = x ^ nextHigherOneBit;` ` ` `rightOnesPattern = (rightOnesPattern) / rightOne;` ` ` `rightOnesPattern >>= 2;` ` ` `next = nextHigherOneBit | rightOnesPattern;` ` ` `}` ` ` `return` `next;` `}` `// function to find the mth smallest number` `// having k number of set bits` `int` `mthSmallestWithKSetBits(uint_t m, uint_t k)` `{` ` ` `// smallest number having 'k'` ` ` `// number of set bits` ` ` `uint_t num = (1 << k) - 1;` ` ` `// finding the mth smallest number` ` ` `// having k set bits` ` ` `for` `(` `int` `i = 1; i < m; i++)` ` ` `num = nxtHighWithNumOfSetBits(num);` ` ` `// required number` ` ` `return` `num;` `}` `// Driver program to test above` `int` `main()` `{` ` ` `uint_t m = 6, k = 4;` ` ` `cout << mthSmallestWithKSetBits(m, k);` ` ` `return` `0;` `}` |

## Python3

`# Python3 implementation to find the mth` `# smallest number having k number of set bits` `# function to find the next higher number` `# with same number of set bits as in 'x'` `def` `nxtHighWithNumOfSetBits(x):` ` ` `rightOne ` `=` `0` ` ` `nextHigherOneBit ` `=` `0` ` ` `rightOnesPattern ` `=` `0` ` ` `next` `=` `0` ` ` `""" the approach is same as discussed in` ` ` `http:#www.geeksforgeeks.org/next-higher-number-with-same-number-of-set-bits/` ` ` `"""` ` ` `if` `(x):` ` ` `rightOne ` `=` `x & (` `-` `x)` ` ` `nextHigherOneBit ` `=` `x ` `+` `rightOne` ` ` `rightOnesPattern ` `=` `x ^ nextHigherOneBit` ` ` `rightOnesPattern ` `=` `(rightOnesPattern) ` `/` `/` `rightOne` ` ` `rightOnesPattern >>` `=` `2` ` ` `next` `=` `nextHigherOneBit | rightOnesPattern` ` ` `return` `next` `# function to find the mth smallest` `# number having k number of set bits` `def` `mthSmallestWithKSetBits(m, k):` ` ` `# smallest number having 'k'` ` ` `# number of set bits` ` ` `num ` `=` `(` `1` `<< k) ` `-` `1` ` ` `# finding the mth smallest number` ` ` `# having k set bits` ` ` `for` `i ` `in` `range` `(` `1` `, m):` ` ` `num ` `=` `nxtHighWithNumOfSetBits(num)` ` ` `# required number` ` ` `return` `num` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `m ` `=` `6` ` ` `k ` `=` `4` ` ` `print` `(mthSmallestWithKSetBits(m, k))` `# This code is contributed by` `# Shubham Singh(SHUBHAMSINGH10)` |

**Output:**

39

**Time Complexity:** O(m)

This article is contributed by **Ayush Jauhari**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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