Next higher number with same number of set bits - GeeksforGeeks

29 Oct, 2021
Given a number x, find next number with same number of 1 bits in it’s binary representation.For example, consider x = 12, whose binary representation is 1100 (excluding leading zeros on 32 bit machine). It contains two logic 1 bits. The next higher number with two logic 1 bits is 17 (100012).Algorithm:When we observe the binary sequence from 0 to 2n – 1 (n is # of bits), right most bits (least significant) vary rapidly than left most bits. The idea is to find right most string of 1’s in x, and shift the pattern to right extreme, except the left most bit in the pattern. Shift the left most bit in the pattern (omitted bit) to left part of x by one position. An example makes it more clear,  
x = 156
10  
x = 10011100
(2)  
10011100
00011100 - right most string of 1's in x
00000011 - right shifted pattern except left most bit ------> [A]
00010000 - isolated left most bit of right most 1's pattern
00100000 - shiftleft-ed the isolated bit by one position ------> [B]
10000000 - left part of x, excluding right most 1's pattern ------> [C]
10100000 - add B and C (OR operation) ------> [D]
10100011 - add A and D which is required number 163
(10)After practicing with few examples, it easy to understand. Use the below given program for generating more sets.Program Design:We need to note few facts of binary numbers. The expression x & -x will isolate right most set bit in x (ensuring x will use 2’s complement form for negative numbers). If we add the result to x, right most string of 1’s in x will be reset, and the immediate ‘0’ left to this pattern of 1’s will be set, which is part [B] of above explanation. For example if x = 156, x & -x will result in 00000100, adding this result to x yields 10100000 (see part D). We left with the right shifting part of pattern of 1’s (part A of above explanation).There are different ways to achieve part A. Right shifting is essentially a division operation. What should be our divisor? Clearly, it should be multiple of 2 (avoids 0.5 error in right shifting), and it should shift the right most 1’s pattern to right extreme. The expression (x & -x) will serve the purpose of divisor. An EX-OR operation between the number X and expression which is used to reset right most bits, will isolate the rightmost 1’s pattern.A Correction Factor:Note that we are adding right most set bit to the bit pattern. The addition operation causes a shift in the bit positions. The weight of binary system is 2, one shift causes an increase by a factor of 2. Since the increased number (rightOnesPattern in the code) being used twice, the error propagates twice. The error needs to be corrected. A right shift by 2 positions will correct the result.The popular name for this program is same number of one bits.  
C++
Java
Python 3
C#
PHP
Javascript
#include<iostream> using namespace std; typedef unsigned int uint_t; // this function returns next higher number with same number of set bits as x.uint_t snoob(uint_t x){   uint_t rightOne;  uint_t nextHigherOneBit;  uint_t rightOnesPattern;   uint_t next = 0;   if(x)  {     // right most set bit    rightOne = x & -(signed)x;     // reset the pattern and set next higher bit    // left part of x will be here    nextHigherOneBit = x + rightOne;     // nextHigherOneBit is now part [D] of the above explanation.     // isolate the pattern    rightOnesPattern = x ^ nextHigherOneBit;     // right adjust pattern    rightOnesPattern = (rightOnesPattern)/rightOne;     // correction factor    rightOnesPattern >>= 2;     // rightOnesPattern is now part [A] of the above explanation.     // integrate new pattern (Add [D] and [A])    next = nextHigherOneBit | rightOnesPattern;  }   return next;} int main(){  int x = 156;  cout<<"Next higher number with same number of set bits is "<<snoob(x);   getchar();  return 0;}
// Java Implementation on above approachclass GFG{     // this function returns next higher// number with same number of set bits as x.static int snoob(int x){ int rightOne, nextHigherOneBit, rightOnesPattern, next = 0; if(x > 0){     // right most set bit    rightOne = x & -x;     // reset the pattern and set next higher bit    // left part of x will be here    nextHigherOneBit = x + rightOne;     // nextHigherOneBit is now part [D] of the above explanation.     // isolate the pattern    rightOnesPattern = x ^ nextHigherOneBit;     // right adjust pattern    rightOnesPattern = (rightOnesPattern)/rightOne;     // correction factor    rightOnesPattern >>= 2;     // rightOnesPattern is now part [A] of the above explanation.     // integrate new pattern (Add [D] and [A])    next = nextHigherOneBit | rightOnesPattern;} return next;} // Driver codepublic static void main (String[] args){    int x = 156;    System.out.println("Next higher number with same" +                    "number of set bits is "+snoob(x));}} // This code is contributed by mits
# This function returns next# higher number with same# number of set bits as x.def snoob(x):         next = 0    if(x):                 # right most set bit        rightOne = x & -(x)                 # reset the pattern and        # set next higher bit        # left part of x will        # be here        nextHigherOneBit = x + int(rightOne)                 # nextHigherOneBit is        # now part [D] of the        # above explanation.        # isolate the pattern        rightOnesPattern = x ^ int(nextHigherOneBit)                 # right adjust pattern        rightOnesPattern = (int(rightOnesPattern) /                            int(rightOne))                 # correction factor        rightOnesPattern = int(rightOnesPattern) >> 2                 # rightOnesPattern is now part        # [A] of the above explanation.                 # integrate new pattern        # (Add [D] and [A])        next = nextHigherOneBit | rightOnesPattern    return next # Driver Codex = 156print("Next higher number with " +      "same number of set bits is",                          snoob(x)) # This code is contributed by Smita
// C# Implementation on above approachusing System;class GFG{     // this function returns next higher// number with same number of set bits as x.static int snoob(int x){     int rightOne, nextHigherOneBit,        rightOnesPattern, next = 0;         if(x > 0)    {             // right most set bit        rightOne = x & -x;             // reset the pattern and set next higher bit        // left part of x will be here        nextHigherOneBit = x + rightOne;             // nextHigherOneBit is now part [D]        // of the above explanation.             // isolate the pattern        rightOnesPattern = x ^ nextHigherOneBit;             // right adjust pattern        rightOnesPattern = (rightOnesPattern) / rightOne;             // correction factor        rightOnesPattern >>= 2;             // rightOnesPattern is now part [A]        // of the above explanation.             // integrate new pattern (Add [D] and [A])        next = nextHigherOneBit | rightOnesPattern;    }    return next;} // Driver codestatic void Main(){    int x = 156;    Console.WriteLine("Next higher number with same" +                      "number of set bits is " + snoob(x));}} // This code is contributed by mits
<?php // This function returns next higher number// with same number of set bits as x.function snoob($x){    $next = 0;         if($x)    {             // right most set bit        $rightOne = $x & - $x;             // reset the pattern and set next higher        // bit left part of x will be here        $nextHigherOneBit = $x + $rightOne;             // nextHigherOneBit is now part [D] of        // the above explanation.             // isolate the pattern        $rightOnesPattern = $x ^ $nextHigherOneBit;             // right adjust pattern        $rightOnesPattern = intval(($rightOnesPattern) /                                            $rightOne);             // correction factor        $rightOnesPattern >>= 2;             // rightOnesPattern is now part [A]        // of the above explanation.             // integrate new pattern (Add [D] and [A])        $next = $nextHigherOneBit | $rightOnesPattern;    }     return $next;} // Driver Code$x = 156;echo "Next higher number with same " .     "number of set bits is " . snoob($x); // This code is contributed by ita_c?>
<script>     // this function returns next higher     // number with same number of set bits as x.    function snoob(x)    {        let rightOne, nextHigherOneBit, rightOnesPattern, next = 0;        if(x > 0)        {            // right most set bit            rightOne = x & -x;                       // reset the pattern and set next higher bit            // left part of x will be here            nextHigherOneBit = x + rightOne;                       // nextHigherOneBit is now part [D] of the above explanation.                       // isolate the pattern            rightOnesPattern = x ^ nextHigherOneBit;                       // right adjust pattern            rightOnesPattern = (rightOnesPattern)/rightOne;                       // correction factor            rightOnesPattern >>= 2;                       // rightOnesPattern is now part [A] of the above explanation.                       // integrate new pattern (Add [D] and [A])            next = nextHigherOneBit | rightOnesPattern;        }        return next;    }         // Driver code    let x = 156;    document.write("Next higher number with same " +                     "number of set bits is "+snoob(x));         // This code is contributed by avanitrachhadiya2155     </script>
Output: 
Next higher number with same number of set bits is 163
Time Complexity: O(1)
Auxiliary Space: O(1)
Usage: Finding/Generating subsets.Variations:  
Write a program to find a number immediately smaller than given, with same number of logic 1 bits? (Pretty simple)How to count or generate the subsets available in the given set?
Write a program to find a number immediately smaller than given, with same number of logic 1 bits? (Pretty simple)
How to count or generate the subsets available in the given set?
References:  
A nice presentation here.Hackers Delight by Warren (An excellent and short book on various bit magic algorithms, a must for enthusiasts)C A Reference Manual by Harbison and Steele (A good book on standard C, you can access code part of this post here).
A nice presentation here.
Hackers Delight by Warren (An excellent and short book on various bit magic algorithms, a must for enthusiasts)
C A Reference Manual by Harbison and Steele (A good book on standard C, you can access code part of this post here).
– Venki. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Smitha Dinesh Semwal
ukasp
Mithun Kumar
shubham_singh
avanitrachhadiya2155
souravmahato348
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