Solution 1

  • TimeO(n)
  • SpaceO(1)

where, n is length of nums
DID NOT GO THROUGH THE SOLUTION YET!

Python · 2026-01-21
class Solution:
    '''
    Time Complexity: O(n)
    Space Complexity: O(1)
    where, n is length of nums
    DID NOT GO THROUGH THE SOLUTION YET!
    '''
    def minBitwiseArray(self, nums: List[int]) -> List[int]:
        ans = []
        for n in nums:
            if n % 2 == 0:
                ans.append(-1)
                continue

            t = 0
            temp = n
            while temp & 1:
                t += 1
                temp >>= 1

            ans.append(n - (1 << (t - 1)))
        return ans
Leet Code/python.py · L4435–4457