Program to find higher number with same number of set bits as n in Python?

Suppose we have a number n; we have to find the smallest next higher number with the same number of 1s as n in binary form.

So, if the input is like n = 7, then the output will be 11, as 7 in binary is 0111 and next higher from 7 with three ones would be 11 which is 1011 in binary.

To solve this, we will follow these steps:

• copy := n, zeros := 0, ones := 0

• while copy is not 0 and copy is even, do

  • zeros := zeros + 1

  • copy = copy / 2

• while copy is odd, do

  • ones := ones + 1

  • copy = copy / 2

• right := ones + zeros

• n := n OR (2^right)

• n := n AND invert of ((2^right) - 1)

• n := n OR((2 ^ (ones - 1)) - 1

• return n

Let us see the following implementation to get better understanding:

Example

 Live Demo

class Solution:
   def solve(self, n):
      copy = n
      zeros = 0
      ones = 0
      while copy and not copy & 1:
         zeros += 1
         copy >>= 1
      while copy & 1:
         ones += 1
         copy >>= 1
      right = ones + zeros
      n |= 1 << right
      n &= ~((1 << right) - 1)
      n |= (1 << ones - 1) - 1
      return n

ob = Solution()
n = 7
print(ob.solve(n))

Input

7

Output

11

Arnab Chakraborty

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Updated on: 10-Nov-2020

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