Easy Solution In Java ✅ | 9 Lines | Beats 100% ✅✅✅ In Both Space And Time Complexity!!!!
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires generating an array based on bitwise operations. For each element 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires generating an array based on bitwise operations. For each element in the input list, the task is to find the smallest number j such that modifying its bits in a specific way results in the original value. To do this, bitwise operations are useful for checking and toggling bits.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Initial Setup:** Create an array ans initialized with -1 values, as placeholders for values that are either invalid or don\'t meet the condition.\n2. **Iterate through the input list nums:**\n- For each value val in nums, we check if it is equal to 2. If so, skip processing as 2 doesn\u2019t need further evaluation.\n- Otherwise, iterate through the bits of val.\n3. **Bitwise Check and Toggle:**\n- Use a bitmask (1 << j) to check the jth bit of val. If it\'s 0, this is where we can potentially toggle the bits to form a smaller number.\n- When a valid number is found (using XOR to flip the bits), it is stored in the ans array.\n4. **Return the result:** After iterating through all elements, return the modified ans array.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n    O(nlog(val))\n    \n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n    O(n)\n\n# Code\n```java []\nclass Solution {\n    public int[] minBitwiseArray(List<Integer> nums) {\n        int [] ans = new int[nums.size()];\n        Arrays.fill(ans, -1);\n        for (int i = 0; i < nums.size(); i++) \n        {\n            int val = nums.get(i);\n            if(val == 2) continue;\n            for (int j = 0; j < val; j++) \n            {\n                // if ((j | (j + 1)) == val) \n                // {\n                //     ans[i] = j; \n                //     break;\n                // }\n\n                if((val & (1 << j)) == 0)\n                {\n                    ans[i] = val ^ (1 << (j - 1));\n                    break;\n                }\n            }\n        }\n        return ans;\n    }\n}\n```
['Java']