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:
- 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++ 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
https://www.geeksforgeeks.org/dsa/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;
}
// Java implementation to find the mth
// smallest number having k number of set bits
import java.util.*;
class GFG
{
// function to find the next higher number
// with same number of set bits as in 'x'
static int nxtHighWithNumOfSetBits(int x)
{
int rightOne = 0;
int nextHigherOneBit = 0;
int rightOnesPattern = 0;
int next = 0;
if (x > 0) {
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
static int mthSmallestWithKSetBits(int m, int k)
{
// smallest number having 'k'
// number of set bits
int 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 Code
public static void main(String[] args)
{
int m = 6;
int k = 4;
// Function call
System.out.println(mthSmallestWithKSetBits(m, k));
}
}
// This code is contributed by phasing17
# 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)
// C# implementation to find the mth
// smallest number having k number of set bits
using System;
class GFG {
// function to find the next higher number
// with same number of set bits as in 'x'
static int nxtHighWithNumOfSetBits(int x)
{
int rightOne = 0;
int nextHigherOneBit = 0;
int rightOnesPattern = 0;
int next = 0;
if (x > 0) {
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
static int mthSmallestWithKSetBits(int m, int k)
{
// smallest number having 'k'
// number of set bits
int 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 Code
public static void Main(string[] args)
{
int m = 6;
int k = 4;
// Function call
Console.Write(mthSmallestWithKSetBits(m, k));
}
}
// This code is contributed by phasing17
//JS 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'
function nxtHighWithNumOfSetBits(x)
{
var rightOne = 0;
var nextHigherOneBit = 0;
var rightOnesPattern = 0;
var next = 0;
if (x > 0)
{
rightOne = x & (-x);
nextHigherOneBit = x + rightOne;
rightOnesPattern = x ^ nextHigherOneBit;
rightOnesPattern = Math.floor((rightOnesPattern) / rightOne);
rightOnesPattern >>= 2;
next = nextHigherOneBit | rightOnesPattern;
}
return next;
}
// function to find the mth smallest
// number having k number of set bits
function mthSmallestWithKSetBits(m, k)
{
// smallest number having 'k'
// number of set bits
var num = (1 << k) - 1;
// finding the mth smallest number
// having k set bits
for (var i = 1; i < m; i++)
num = nxtHighWithNumOfSetBits(num);
// required number
return num;
}
// Driver Code
var m = 6;
var k = 4;
//Function call
console.log(mthSmallestWithKSetBits(m, k));
// This code is contributed by phasing17
Output:
39
Time Complexity: O(m)
Space Complexity: O(1)