Easy to UnderStand Beginner Friendly
To address this problem, here\u2019s the breakdown:\n\n# Intuition\nThe goal is to transform each element of the array v by minimizing its binary representation
To address this problem, here\u2019s the breakdown:\n\n# Intuition\nThe goal is to transform each element of the array `v` by minimizing its binary representation. For each element `v[i]`, if it is odd, we attempt to adjust it such that the resulting value retains as few high-value bits as possible while being slightly modified if necessary. This may involve clearing specific bits in a controlled way to achieve the minimal required configuration.\n\n# Approach\n1. **Helper Function `f`**:  \n   - This function minimizes a given integer `n` by selectively clearing one of the bits to create the smallest possible number that still closely resembles `n` in its binary structure.\n   - We iterate through each bit of `n`. When we encounter the first unset bit (`0`), we adjust the last set bit (`1`) we added, which helps minimize the integer.\n  \n2. **Main Transformation in `minBitwiseArray`**:\n   - **Check Each Element**:\n     - If `v[i]` is even, we skip it, as we only want to modify odd numbers.\n     - For odd numbers:\n       - If `v[i]` is a power of two minus one (e.g., `1, 3, 7, 15...`), its minimized value should be half of `v[i]` minus one (a straightforward case).\n       - Otherwise, we use the `f` function to get a minimized version.\n   - **Store Result**: We create an array `ans` to store the minimized results for each `v[i]`.\n\n# Complexity\n- **Time Complexity**:  \n  - Let `n` be the length of the input array `v`. For each odd element, we process its bits (approximately `O(log k)` for integer `k`).\n  - Overall complexity is `O(nlog k)` in the worst case.\n\n- **Space Complexity**:  \n  - `O(n)` for storing the resulting minimized array.\n\n# Beats\n![image.png](https://assets.leetcode.com/users/images/def2d063-d956-4101-ac46-a2fbbf6fd4ba_1731318664.2222877.png)\n\n\n# Code\n```cpp\n#include <vector>\nusing namespace std;\n\nclass Solution {\npublic:\n    // Helper function to minimize binary form of odd number `n`\n    int f(int n) {\n        int ans = 0, i = 0;\n        bool flag = true;\n        \n        // Iterate through bits of `n` to minimize its value\n        while(n) {\n            if(n % 2) ans += 1 << i;   // If bit is set, add it to `ans`\n            else if(flag) {\n                ans -= 1 << (i - 1);   // Clear the last added bit to minimize\n                flag = false;\n            }\n            i++;\n            n /= 2;\n        }\n        return ans;\n    }\n    \n    // Main function to transform array based on minimization rules\n    vector<int> minBitwiseArray(vector<int>& v) {\n        int n = v.size();\n        vector<int> ans(n, -1);  // Initialize results with `-1`\n        \n        // Process each element in the input vector\n        for(int i = 0; i < n; i++) {\n            if(v[i] % 2 == 0) continue;  // Skip even numbers\n            else if(((v[i] + 1) & (v[i])) == 0)  // Special case for `2^k - 1`\n                ans[i] = (v[i] - 1) / 2;\n            else ans[i] = f(v[i]);  // Use helper function for other odd numbers\n        }\n        return ans;\n    }\n};\n```
['Array', 'Bit Manipulation', 'C++']