TS Solution
Create an array ans of the same length as nums, initialized with -1 (default value if no valid ans[i] is found).\n\nMain Loop:\n\nFor each element in nums, we t
Create an array ans of the same length as nums, initialized with -1 (default value if no valid ans[i] is found).\n\nMain Loop:\n\nFor each element in nums, we try to find the smallest x such that x | (x + 1) == nums[i].\nIf we find such a value, we update ans[i] and break out of the loop to move to the next index.\n\nThe condition \n\nx | (x + 1) == nums[i] checks if the bitwise OR of x and x + 1 equals the current number in nums.\n\n# Complexity\n- Time complexity: 6ms\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: 52.76MB\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```typescript []\nfunction minBitwiseArray(nums: number[]): number[] {\n    let ans = new Array(nums.length).fill(-1)\n    \n    for (let i = 0; i < nums.length; i++) {\n      for (let x = 0; x < nums[i]; x++) {\n        if ((x | (x + 1)) == nums[i]) {\n          ans[i] = x\n          break\n        }\n      }\n    }\n  return ans\n};\n\n```
['TypeScript']