Next greater integer having one more number of set bits

Given a positive integer 'n' having 'x' number of set bits in its binary representation. The problem is to find the next greater integer(smallest integer greater than n), having (x+1) number of set bits in its binary representation.
Examples : 
 

Input : 10 Output : 11 (10)10 = (1010)2 is having 2 set bits. (11)10 = (1011)2 is having 3 set bits and is the next greater. Input : 39 Output : 47

Approach: Following are the steps: 

1. Find the position of the rightmost unset bit(considering last bit at position 0, second last bit at position 1 and so on) in the binary representation of n.
2. Let the position be represented by pos.
3. Set the bit at position pos. Refer this post.
4. If there are no unset bits in the binary representation, then perform bitwise left shift by 1 on the given number and then add 1 to it.

How to get the position of rightmost unset bit? 

1. Perform bitwise not on the given number(operation equivalent to 1's complement).Let it be num = ~n.
2. Get the position of rightmost set bit of num.

// C++ implementation to find the next greater integer // with one more number of set bits #include <bits/stdc++.h> using namespace std; // function to find the position of rightmost // set bit. Returns -1 if there are no set bits int getFirstSetBitPos(int n) {  return (log2(n&-n)+1) - 1; } // function to find the next greater integer int nextGreaterWithOneMoreSetBit(int n) {  // position of rightmost unset bit of n  // by passing ~n as argument  int pos = getFirstSetBitPos(~n);    // if n consists of unset bits, then  // set the rightmost unset bit  if (pos > -1)  return (1 << pos) | n;    //n does not consists of unset bits   return ((n << 1) + 1);  } // Driver program to test above int main() {  int n = 10;  cout << "Next greater integer = "  << nextGreaterWithOneMoreSetBit(n);  return 0; }  

// Java implementation to find the next greater integer // with one more number of set bits class GFG {    // function to find the position of rightmost  // set bit. Returns -1 if there are no set bits  static int getFirstSetBitPos(int n)  {  return ((int)(Math.log(n & -n) / Math.log(2)) + 1) - 1;  }  // function to find the next greater integer  static int nextGreaterWithOneMoreSetBit(int n)  {    // position of rightmost unset bit of n  // by passing ~n as argument  int pos = getFirstSetBitPos(~n);  // if n consists of unset bits, then  // set the rightmost unset bit  if (pos > -1)  return (1 << pos) | n;  // n does not consists of unset bits  return ((n << 1) + 1);  }    // Driver code  public static void main(String[] args)  {  int n = 10;  System.out.print("Next greater integer = "  + nextGreaterWithOneMoreSetBit(n));  } } // This code is contributed by Anant Agarwal. 

# Python3 implementation to find # the next greater integer with # one more number of set bits import math # Function to find the position # of rightmost set bit. Returns -1 # if there are no set bits def getFirstSetBitPos(n): return ((int)(math.log(n & -n) / math.log(2)) + 1) - 1 # Function to find the next greater integer def nextGreaterWithOneMoreSetBit(n): # position of rightmost unset bit of  # n by passing ~n as argument pos = getFirstSetBitPos(~n) # if n consists of unset bits, then # set the rightmost unset bit if (pos > -1): return (1 << pos) | n # n does not consists of unset bits  return ((n << 1) + 1) # Driver code n = 10 print("Next greater integer = ", nextGreaterWithOneMoreSetBit(n)) # This code is contributed by Anant Agarwal. 

// C# implementation to find the next greater // integer with one more number of set bits using System; class GFG {    // function to find the position of rightmost  // set bit. Returns -1 if there are no set bits  static int getFirstSetBitPos(int n)  {  return ((int)(Math.Log(n & -n) / Math.Log(2))  + 1) - 1;  }    // function to find the next greater integer  static int nextGreaterWithOneMoreSetBit(int n)  {    // position of rightmost unset bit of n  // by passing ~n as argument  int pos = getFirstSetBitPos(~n);    // if n consists of unset bits, then  // set the rightmost unset bit  if (pos > -1)  return (1 << pos) | n;    // n does not consists of unset bits   return ((n << 1) + 1);   }    // Driver code  public static void Main()  {  int n = 10;    Console.Write("Next greater integer = "  + nextGreaterWithOneMoreSetBit(n));  } } // This code is contributed by Anant Agarwal. 

<?php // PHP implementation to find the // next greater integer with // one more number of set bits // function to find the position // of rightmost set bit. Returns  // -1 if there are no set bits function getFirstSetBitPos($n) { return (log($n & -$n + 1)) - 1; } // function to find the // next greater integer function nextGreaterWithOneMoreSetBit($n) { // position of rightmost unset bit of n // by passing ~n as argument $pos = getFirstSetBitPos(~$n); // if n consists of unset bits, then // set the rightmost unset bit if ($pos > -1) return (1 << $pos) | $n; //n does not consists of unset bits  return (($n << 1) + 1); } // Driver Code $n = 10; echo "Next greater integer = ", nextGreaterWithOneMoreSetBit($n); // This code is contributed by Ajit ?> 

<script> // Javascript implementation to find the next greater integer  // with one more number of set bits  // function to find the position of rightmost  // set bit. Returns -1 if there are no set bits  function getFirstSetBitPos(n)  {   return (parseInt(Math.log(n&-n)/Math.log(2))+1) - 1;  }  // function to find the next greater integer  function nextGreaterWithOneMoreSetBit(n)  {   // position of rightmost unset bit of n   // by passing ~n as argument   var pos = getFirstSetBitPos(~n);     // if n consists of unset bits, then   // set the rightmost unset bit   if (pos > -1)   return (1 << pos) | n;     //n does not consists of unset bits   return ((n << 1) + 1);  }  // Driver program to test above  var n = 10;  document.write("Next greater integer = " + nextGreaterWithOneMoreSetBit(n));    </script> 

Output : 

Next greater integer = 11

Time Complexity: O(log(log N))
Auxiliary Space: O(1)