BEATS 100% || O(N) EASY SOLUTION :-
IntuitionApproachCheck for even numbers:The condition ans[i] OR (ans[i] + 1) = nums[i] is impossible to satisfy for even numbers because the result of ans[i] OR
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->

# Approach
Check for even numbers:

The condition ans[i] OR (ans[i] + 1) = nums[i] is impossible to satisfy for even numbers because the result of ans[i] OR (ans[i] + 1) is always an odd number, while nums[i] is even.
Therefore, for any even nums[i], you directly set nums[i] = -1 and move to the next element.
For odd numbers:

Convert each nums[i] to a bitset<30>. This allows you to work with the binary representation of nums[i] easily, as each bitset will represent a 30-bit binary number (enough to handle prime numbers up to 10^9).
Find the smallest ans[i]:

Bit Manipulation:
Iterate through the bits of nums[i] (from the least significant bit to the most significant bit).
If a bit is 0, toggle it to 0 (this is done in the inner loop).
Then, compute the corresponding value x from the modified bitset.
Check the condition:

Once the modified x is computed, check if x | (x + 1) == nums[i]. If this condition is satisfied, then x is the answer.
If the condition is not met, increment x and check again until you find a valid x.

# Complexity
- Time complexity:
- O(N)
- Space complexity:
- O(1)

# Code
```cpp []
class Solution {
public:
    vector<int> minBitwiseArray(vector<int>& nums) {
        int n = nums.size();
        for(int i = 0; i < n; i++) {
            // Step 1: Check if the number is even
            if(nums[i] % 2 == 0) {
                nums[i] = -1;  // For even numbers, it's not possible
                continue;
            }
            
            // Step 2: Convert the number to a bitset to work with its binary representation
            bitset<30> b(nums[i]);
            
            // Step 3: Modify the bitset to find the smallest valid number
            for(int j = 0; j < 30; j++) {
                if(b[j] == 0) {
                    j--;  // Rewind to the previous bit
                    b[j] = 0;  // Reset the bit to 0
                    break;
                }
            }
            
            // Step 4: Compute the modified number from the bitset
            int x = 0, a = 1;
            for(int j = 0; j < 30; j++) {
                x += (a * b[j]);
                a *= 2;
            }
            
            // Step 5: Check if this x satisfies the condition or increment until it does
            while(x <= nums[i] - 1) {
                if((x | (x + 1)) == nums[i]) {
                    nums[i] = x;
                    break;
                }
                x++;  // Increment x until we find a valid answer
            }
        }
        
        // Return the modified array
        return nums;
    }
};

```
['C++']