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 the 4th smallest
number having 2 set bits.


Input : m = 6, k = 4
Output : 39

Approach: Following are the steps:



  1. Find the smallest number having k number of set bits. Let it be num, where num = (1 << k) – 1.
  2. 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.
  3. Finally return num.

C++

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// 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 dicussed in
       http://www.geeksforgeeks.org/next-higher-number-with-same-number-of-set-bits/  
    */
    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;
}

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Python3














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# 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 dicussed 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) 



















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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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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