C Solution || Beats 100% || Bit Manipulation Approach || With Explanations
Intuition\nThe goal is to modify each number in the array such that for odd numbers, we find the minimum bitwise transformation that satisfies a given condition
# Intuition\nThe goal is to modify each number in the array such that for odd numbers, we find the minimum bitwise transformation that satisfies a given condition. For even numbers, a predetermined result is returned. The key observation is that for odd numbers, we can use bitwise operations to manipulate individual bits in a controlled manner.\n\n# Approach\n\nWe need to modify each number in the input array based on specific rules. If the number is even, we simply set the result for that number to `-1`. If the number is odd, we need to find the smallest number that can be created using a bitwise XOR (`^`) operation with a bitmask.\n\nThe approach can be broken down into the following steps:\n\n1. **Even Numbers**: If the current number is even, the result is automatically set to `-1`. This simplifies handling for even numbers and avoids unnecessary computation.\n\n2. **Odd Numbers**: For odd numbers, we need to find the smallest possible result by applying a bitwise operation. We do this by:\n   - Starting with a bitmask of `1` (which corresponds to the lowest bit).\n   - For each bitmask, we calculate the result of XOR between the number and the bitmask (`num ^ mask`).\n   - After that, we check if the condition `(val | (val + 1)) == num` is satisfied. This condition ensures that the transformed number still retains a valid bit structure similar to the original odd number.\n   - We repeat this process by shifting the bitmask to the left (doubling it) and checking different bits of the number until we find a valid result.\n\n# Code\n```c []\n/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* minBitwiseArray(int* nums, int numsSize, int* returnSize) {\n    int *result = calloc(numsSize, sizeof(int));\n    *returnSize = numsSize;\n\n    for (int i = 0; i < numsSize; i++) {\n        int num = nums[i];\n\n        if (num % 2 == 0) {\n            result[i] = -1;\n            continue;\n        }\n\n        int mask = 1;\n\n        while (mask <= num) {\n            int val = mask ^ num;\n\n            if ((val | (val + 1)) == num) result[i] = val;\n\n            mask <<= 1;\n        }\n    }\n\n    return result;\n}\n```
['Array', 'Bit Manipulation', 'C']