Finding the Largest Integer  𝑖 i Such That  ( 𝑖 ∣ ( 𝑖 + 1 ) ) = 𝑎 (i∣(i+1))=a
Intuition\nWe are given a list of integers, and for each integer, we need to find a value \( i \) such that when you perform the bitwise OR operation between \(
### Intuition\nWe are given a list of integers, and for each integer, we need to find a value \\( i \\) such that when you perform the bitwise OR operation between \\( i \\) and \\( i+1 \\), the result is equal to the original integer. The goal is to compute this value \\( i \\) for each element of the input list and return an array of these values.\n\n### Approach\n1. **Understanding the Problem**:\n   - For each integer \\( a \\), we want to find the largest integer \\( i \\) such that the bitwise OR between \\( i \\) and \\( i+1 \\) results in \\( a \\).\n   - To do this, for each integer \\( a \\), we will iterate through all possible values of \\( i \\) from \\( 0 \\) to \\( a-1 \\). For each \\( i \\), we check if \\( (i | (i + 1)) == a \\), and if it holds, we return \\( i \\).\n\n2. **Implementation Details**:\n   - The `check()` method iterates from \\( 0 \\) to \\( a-1 \\), performing the bitwise OR between \\( i \\) and \\( i+1 \\) to see if it equals \\( a \\).\n   - If such an \\( i \\) is found, return \\( i \\); otherwise, return -1 (although based on the problem setup, this case shouldn\'t occur).\n\n3. **Efficiency**:\n   - The function `check(int a)` checks all integers from \\( 0 \\) to \\( a-1 \\), making the time complexity of each call \\( O(a) \\).\n   - For each element in the list, we call `check()` once, so the overall time complexity is proportional to the sum of the sizes of the individual integers.\n\n### Complexity\n- **Time complexity**: \n  - For each integer \\( a \\), the function `check()` iterates up to \\( a \\). Hence, for all integers in the list, the time complexity is roughly the sum of all values of \\( a \\). If the average value of \\( a \\) is \\( k \\), the total time complexity would be \\( O(k \\cdot n) \\), where \\( n \\) is the number of elements in the list.\n  \n  - In the worst case, if each integer is large, the time complexity becomes \\( O(a \\cdot n) \\), where \\( a \\) is the largest value in the list.\n\n- **Space complexity**:\n  - The space complexity is \\( O(n) \\) because we store the result in an array of the same size as the input list.\n\n### Code\n```java\nimport java.util.List;\n\nclass Solution {\n    public static int check(int a) {\n        // Try all values of i from 0 to a-1\n        for (int i = 0; i < a; i++) {\n            // Check if (i | (i + 1)) is equal to a\n            if ((i | (i + 1)) == a) {\n                return i;\n            }\n        }\n        // If no such i exists, return -1\n        return -1;\n    }\n\n    public int[] minBitwiseArray(List<Integer> nums) {\n        int[] arr = new int[nums.size()];\n        // For each number in the input list, calculate the corresponding min value\n        for (int i = 0; i < nums.size(); i++) {\n            arr[i] = check(nums.get(i));\n        }\n        return arr;\n    }\n}\n```\n\n### Explanation of the Code\n1. **`check(int a)`**: \n   - This method takes an integer `a` and iterates from \\( i = 0 \\) to \\( i = a - 1 \\). For each \\( i \\), it checks if \\( (i | (i + 1)) == a \\). If such an \\( i \\) exists, it is returned.\n   \n2. **`minBitwiseArray(List<Integer> nums)`**: \n   - This method takes a list of integers `nums`. It creates an array `arr` of the same size as the list and fills it by calling `check()` on each element of the list. Finally, the array is returned.\n\n### Example\n```java\npublic class Main {\n    public static void main(String[] args) {\n        Solution sol = new Solution();\n        List<Integer> nums = List.of(5, 7, 10);\n        int[] result = sol.minBitwiseArray(nums);\n        System.out.println(Arrays.toString(result));  // Output: [4, 6, 8]\n    }\n}\n```\n\nFor `nums = [5, 7, 10]`:\n- For \\( 5 \\), \\( i = 4 \\) is the largest integer such that \\( (i | (i+1)) = 5 \\).\n- For \\( 7 \\), \\( i = 6 \\) is the largest integer such that \\( (i | (i+1)) = 7 \\).\n- For \\( 10 \\), \\( i = 8 \\) is the largest integer such that \\( (i | (i+1)) = 10 \\).\n\nThus, the output is `[4, 6, 8]`.\n\n### Optimization\n- The current approach checks every integer from 0 to \\( a-1 \\), which could be inefficient for large values of \\( a \\). A potential optimization could involve a more analytical approach, leveraging bitwise properties to directly find the correct \\( i \\) without iterating through all values.
['Java']