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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:

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))

7

11

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