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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
    */
    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;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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)

chevron_right



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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : SHUBHAMSINGH10



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.