BEST PYTHON SOLUTION WITH 59MS OF RUNTIME !
\n# Code\npython3 []\nclass Solution:\n    def minBitwiseArray(self, nums: List[int]) -> List[int]:\n        ans = [0] * len(nums)  # Initialize ans array\n    
\n# Code\n```python3 []\nclass Solution:\n    def minBitwiseArray(self, nums: List[int]) -> List[int]:\n        ans = [0] * len(nums)  # Initialize ans array\n        for i, num in enumerate(nums):\n            # Convert the number to binary (without the \'0b\' prefix)\n            b = bin(num)[2:]\n            \n            # If the number ends with \'0\', it is impossible to find such x\n            if b[-1] == \'0\':\n                ans[i] = -1\n                continue\n            \n            # Count consecutive trailing \'1\'s in the binary representation\n            count = 0\n            for k in range(-1, -len(b) - 1, -1):\n                if b[k] == \'1\':\n                    count += 1\n                else:\n                    break\n            \n            # The index of the first bit that is not part of the trailing \'1\'s\n            front = len(b) - count\n            \n            # Form the result by turning the last \'1\' into a \'0\' and keeping other bits the same\n            ans[i] = int(b[:front] + \'0\' + (\'1\' * (count - 1)), 2)\n        \n        return ans\n\n```
['Python3']