The best solution!!! without native methods...
IntuitionThe problem requires finding the smallest possible (ans[i]) such that ((ans[i] | (ans[i] + 1)) = nums[i]). If no such (ans[i]) exists, it should return
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires finding the smallest possible \(ans[i]\) such that \((ans[i] | (ans[i] + 1)) = nums[i]\). If no such \(ans[i]\) exists, it should return \(-1\). The key insight is that for each \(nums[i]\), we can iterate through potential values for \(ans[i]\) starting from \(0\), and check the condition using bitwise operations.


# Approach
<!-- Describe your approach to solving the problem. -->
1. Use a loop to iterate through the \(nums\) array.
2. Use a variable \(i\) to represent potential values of \(ans[i]\). Start \(i\) from \(0\) and increment until the condition \((i | (i + 1)) == nums[j]\) is satisfied.
3. If the condition is met, append \(i\) to the answer and move to the next \(nums[j]\).
4. If \(i\) reaches \(nums[j]\) without finding a valid \(ans[i]\), append \(-1\) to the answer and move to the next element.
5. Reset \(i\) by assigning \(-1\) to start fresh for the next \(nums[j]\).

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
The time complexity is \(O(n \cdot m)\), where \(n\) is the length of the \(nums\) array, and \(m\) is the maximum number of iterations needed to find a valid \(ans[i]\). Each element of \(nums\) requires checking potential \(ans[i]\) values up to a maximum.

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
The space complexity is \(O(n)\), as the output array \(ans\) stores one value for each element in \(nums\).

# Code

```php []
class Solution {

    /**
     * @param Integer[] $nums
     * @return Integer[]
     */
    function minBitwiseArray($nums) {
        $ans = [];
        $j = 0;
        $c = count($nums);
        for($i = 0; $j < $c; $i++)
        {
            if(($i | ($i + 1)) == $nums[$j])
            {
                $ans[] = $i;
                $j++;
                $i = -1;
            }
            if($i == $nums[$j])
            {
                $ans[] = -1;
                $j++;
                $i = -1;
            }
        }
        return $ans;
    }
}

```
['PHP']