Iterative Bitwise Search | Dart Solution | Time O(n · m) | Space O(n)
ApproachThe solution processes each element in the input array and, for each element, finds the smallest candidate integer that satisfies a specific bitwise pro
## Approach
The solution processes each element in the input array and, for each element, finds the smallest candidate integer that satisfies a specific bitwise property:

1. **Iterate Over the Input Array**:  
  For every number in the list, the algorithm attempts to find the smallest integer `j` such that the bitwise OR of `j` and `j + 1` equals that number.

2. **Bitwise Condition Check**:  
  For each candidate `j` from 1 up to the current number, the solution checks if  
  $$j \;OR\; (j+1) == \text{current number}$$
  If the condition is met, `j` is added to the result, and the search continues with the next element.

3. **Handling No Valid Candidate**:  
  If no candidate satisfies the condition for an element, the solution appends `-1` to the result list.

4. **Final Result**:  
  Once all elements have been processed, the constructed list is returned.

## Complexity

- **Time complexity**: $$O(n \times m)$$  
  where $$ n $$ is the number of elements in the input array and $$ m $$ is the value of the current element in the worst case, since for each element the inner loop may iterate up to that number.

- **Space complexity**: $$O(n)$$  
  The algorithm uses additional space for the result list, which stores one integer per element in the input array.

# Code
```dart []
class Solution {
  List<int> minBitwiseArray(List<int> nums) {
    List<int> ans = [];
    main:
    for (var i = 0; i < nums.length; i++) {
      for (var j = 1; j <= nums[i]; j++) {
        if (j | (j + 1) == nums[i]) {
          ans.add(j);
          continue main;
        }
      }
      ans.add(-1);
    }
    return ans;
  }
}

```
['Dart']