Easy Approach for Beginners
Intuition  \n Describe your first thoughts on how to solve this problem.   \nGiven a list of numbers, the goal is to find the smallest integer \( j \) such that
# Intuition  \n<!-- Describe your first thoughts on how to solve this problem. -->  \nGiven a list of numbers, the goal is to find the smallest integer \\( j \\) such that \\( j | (j + 1) = i \\) for each \\( i \\) in the input list. If no such \\( j \\) exists, we return -1 for that element. The bitwise OR operation between \\( j \\) and \\( j + 1 \\) is always greater than or equal to both \\( j \\) and \\( j + 1 \\).  \n\n# Approach  \n<!-- Describe your approach to solving the problem. -->  \nFor each element \\( i \\) in the input list, we are looking for the smallest integer \\( j \\) such that \\( j | ( j + 1 ) = i \\). We do this because \\( j  | ( j + 1 ) \\) is always greater than \\( j \\) and \\( j + 1 \\). This means that as we iterate from \\( j = 0 \\) up to \\( i - 1 \\), we will find a \\( j \\) that satisfies the condition or determine that no such \\( j \\) exists. If we find a valid \\( j \\), we append it to the answer list; otherwise, we append -1 if no valid \\( j \\) exists after checking all possibilities.  \n\n# Complexity  \n- **Time complexity:**  \n  $$O(n \\cdot m)$$  \n  Where \\( n \\) is the length of the input list and \\( m \\) is the value of the largest element in the list.  \n\n- **Space complexity:**  \n  $$O(n)$$  \n\n# Code  \n```python3\nclass Solution:\n    def minBitwiseArray(self, nums: List[int]) -> List[int]:\n        ans = []\n        for i in nums:\n            for j in range(i):\n                if j | (j + 1) == i:\n                    ans.append(j)\n                    break\n            else:\n                ans.append(-1)\n        return ans\n
['Array', 'Bit Manipulation', 'Python3']