💡✅🔥 Bitwise operations with detailed explanation🔥 C++ Simple Solution 🔥 🧠Optimized ✅
\n\n## Problem Overview\n\nThe goal of this problem is to find a value x for each number in the array nums such that:\n\n1. \( x \) is a non-negative integer le
\n\n## Problem Overview\n\nThe goal of this problem is to find a value `x` for each number in the array `nums` such that:\n\n1. \\( x \\) is a non-negative integer less than the current number \\( \\text{num} \\).\n2. \\( x \\) should satisfy the equation \\( x \\text{ OR } (x + 1) = \\text{num} \\), where "OR" represents the bitwise OR operation.\n\nIf such an `x` exists, we should return that value for the current number. If not, we return `-1` for that number.\n\n---\n\n## Intuition\n\nBefore diving into the code, let\'s first break down the key idea behind the problem. The task is to find a number `x` such that when we perform a bitwise OR operation between `x` and `x+1`, the result is equal to a given number from the array. Bitwise operations can be a little tricky, but with some experimentation, we can find a pattern.\n\n### What is Bitwise OR?\n\nBitwise OR (`|`) compares each corresponding bit of two numbers. If either of the bits is `1`, the result is `1`. Otherwise, the result is `0`. For example:\n\n- \\( 5 = 101_2 \\) (binary representation)\n- \\( 6 = 110_2 \\)\n- \\( 5 \\text{ OR } 6 = 111_2 = 7 \\)\n\nThus, performing a bitwise OR on 5 and 6 gives 7.\n\n### What We Are Looking For\n\nWe want to find a number `x` such that \\( x \\text{ OR } (x + 1) = \\text{num} \\). For example:\n\n- If \\( \\text{num} = 7 \\), we need to find `x` such that \\( x \\text{ OR } (x + 1) = 7 \\).\n- We can try \\( x = 5 \\):\n  - \\( 5 = 101_2 \\) (binary)\n  - \\( 6 = 110_2 \\)\n  - \\( 5 \\text{ OR } 6 = 111_2 = 7 \\), so `x = 5` works.\n\nOur task is to iterate over possible values of `x` for each number in the array `nums`, check the condition, and store the result.\n\n---\n\n## Approach\n\n### Step 1: Initialize a Result Vector\n\nWe need to return a vector of integers, where each entry corresponds to the smallest `x` for which \\( x \\text{ OR } (x + 1) = \\text{num} \\). So, we start by creating an empty vector `ans` that will hold these results.\n\n### Step 2: Iterate Over Each Number in `nums`\n\nWe need to process each number in the input array `nums`. For each number `num`, we try different values of `x` starting from `0` up to `num - 1`. We will check if \\( x \\text{ OR } (x + 1) \\) equals `num`.\n\n### Step 3: Find the Minimum Value of `x`\n\nFor each `num` in the array, we loop over all potential values of `x` from `0` to `num - 1`. We check if \\( x \\text{ OR } (x + 1) \\) equals `num`. If we find such an `x`, we store it in our result vector and stop the search for that `num`.\n\n### Step 4: Handle Cases Where No `x` is Found\n\nIf after trying all possible values of `x`, no suitable value is found, we append `-1` to the result vector. This indicates that no `x` satisfies the condition for that particular `num`.\n\n### Step 5: Return the Result\n\nAfter processing all numbers in `nums`, we return the result vector, which contains either the value of `x` or `-1` for each number.\n\n---\n\n## Code Explanation\n\n```cpp\n#include <vector>  // Include the vector library\n\nusing namespace std;  // Use the standard namespace\n\nclass Solution {\npublic:\n    vector<int> minBitwiseArray(vector<int>& nums) {\n        vector<int> ans;  // Create a vector to store the results\n        \n        // Iterate over each number in the input array \'nums\'\n        for (int num : nums) {\n            bool found = false;  // A flag to check if we find a valid \'x\'\n            \n            // Try different values of \'x\' starting from 0 up to \'num-1\'\n            for (int x = 0; x < num; ++x) {  // No need to check values greater than or equal to \'num\'\n                \n                // Check if the condition holds: x OR (x + 1) == num\n                if ((x | (x + 1)) == num) {\n                    ans.push_back(x);  // If condition is met, store \'x\' in the result vector\n                    found = true;  // Set the flag to true as we found a valid \'x\'\n                    break;  // Stop searching as we found the minimum \'x\'\n                }\n            }\n            \n            // If no such \'x\' was found, append -1 to the result vector\n            if (!found) {\n                ans.push_back(-1);\n            }\n        }\n        \n        return ans;  // Return the result vector\n    }\n};\n```\n\n---\n\n## Detailed Explanation\n\n1. **Include Necessary Libraries**:\n   - We include the `vector` library since we need to return a vector containing the results.\n\n2. **Class Definition**:\n   - The `Solution` class encapsulates our solution.\n\n3. **minBitwiseArray Function**:\n   - **Input**: The function takes a vector of integers (`nums`) as input.\n   - **Output**: It returns a vector of integers where each element corresponds to the minimum value of `x` (or `-1` if no valid `x` is found).\n\n4. **Vector Initialization**:\n   - We create an empty vector `ans` that will store the result for each number in `nums`.\n\n5. **Loop Through Each Number**:\n   - We iterate over the numbers in the array `nums`. For each number `num`, we initialize a flag `found` to `false`, indicating that we haven\'t found a valid `x` yet.\n   \n6. **Inner Loop to Find `x`**:\n   - For each `num`, we try different values of `x` from `0` to `num - 1`. If we find that \\( x \\text{ OR } (x + 1) = \\text{num} \\), we append `x` to `ans` and break out of the loop. If no such `x` is found after checking all possibilities, we append `-1` to `ans`.\n\n7. **Return the Result**:\n   - After processing all the numbers, we return the vector `ans`, which contains either the minimum `x` or `-1` for each number in `nums`.\n\n---\n\n## Complexity Analysis\n\n### Time Complexity\n\n- The outer loop runs once for each element in the `nums` array. If there are `n` elements, this takes \\( O(n) \\) time.\n- For each number `num`, the inner loop runs at most `num` times, where `num` is the value of the current number. In the worst case, this could be \\( O(\\text{max}(nums)) \\) for each number.\n\nThus, the overall time complexity is:\n\n\\[ O(n \\cdot \\text{max}(nums)) \\]\n\nWhere `n` is the number of elements in `nums`, and `max(nums)` is the maximum value in the array.\n\n### Space Complexity\n\n- We use an additional vector `ans` to store the results, which has the same size as the input `nums` array. Therefore, the space complexity is \\( O(n) \\), where `n` is the number of elements in `nums`.\n\n---\n\n## Example Walkthrough\n\nLet\'s consider an example to understand the execution of the code.\n\n### Input:\n```cpp\nnums = [7, 5, 10]\n```\n\n### Process:\n\n1. **For `num = 7`:**\n   - We try `x = 0`: \\( 0 \\text{ OR } 1 = 1 \\neq 7 \\).\n   - We try `x = 1`: \\( 1 \\text{ OR } 2 = 3 \\neq 7 \\).\n   - We try `x = 2`: \\( 2 \\text{ OR } 3 = 3 \\neq 7 \\).\n   - We try `x = 3`: \\( 3 \\text{ OR } 4 = 7 = 7 \\). Success! So, we append `3` to `ans`.\n\n2. **For `num = 5`:**\n   - We try `x = 0`: \\( 0 \\text{ OR } 1 = 1 \\neq 5 \\).\n   - We try `x = 1`: \\( 1 \\text{ OR } 2 = 3 \\neq 5 \\).\n   - We try `x = 2`: \\( 2 \\text{ OR } 3 = 3 \\neq 5 \\).\n   - We try `x = 3`: \\( 3 \\text{ OR }\n\n 4 = 7 \\neq 5 \\). No valid `x`, so we append `-1`.\n\n3. **For `num = 10`:**\n   - We try `x = 0` to `x = 8`: None of these work.\n   - We try `x = 9`: \\( 9 \\text{ OR } 10 = 10 = 10 \\). Success! So, we append `9` to `ans`.\n\n### Output:\n```cpp\nans = [3, -1, 9]\n```\n\n---\n\n## Conclusion\n\nThis solution uses bitwise operations to solve the problem by finding the smallest `x` for which \\( x \\text{ OR } (x + 1) = \\text{num} \\). By iterating through possible values of `x`, we can efficiently find a result or determine if no solution exists for each number.
['Array', 'Bit Manipulation', 'C++']