A Middle-Earth Tale of Smallest Subarrays and Maximum Bitwise OR
Solution\n\n\nfrom typing import List\n\nclass Solution:\n    def smallestSubarrays(self, nums: List[int]) -> List[int]:\n\n        def update_last_seen_bits(nu
**Solution**\n\n```\nfrom typing import List\n\nclass Solution:\n    def smallestSubarrays(self, nums: List[int]) -> List[int]:\n\n        def update_last_seen_bits(num, index, last_seen_bits):\n            """Updates the last seen indices of each bit in \'num\'."""\n            bit_index = 0\n            while (num >> bit_index):\n                if (num >> bit_index) & 1:\n                    last_seen_bits[bit_index] = index\n                bit_index += 1\n\n        # Calculate the maximum bit length to cover all bits in \'nums\'.\n        max_bit_length = max(1, max(n.bit_length() for n in nums))\n\n        # Initialize the last seen bits array.\n        last_seen_bits = [0] * max_bit_length\n\n        # Initialize the result array.\n        result = [0] * len(nums)\n\n        # Iterate over \'nums\' in reverse order.\n        for i in range(len(nums) - 1, -1, -1):\n            update_last_seen_bits(nums[i], i, last_seen_bits)\n            # The size of the smallest subarray with maximum bitwise OR \n            # starting at \'i\' is the maximum last seen bit index - \'i\' + 1.\n            result[i] = max(1, max(last_seen_bits) - i + 1)\n\n        return result\n\n```\n\n\nIn the land of Middle-Earth, where shadows lie and algorithms thrive, we find ourselves on an epic quest. A quest not for the One Ring, but for an understanding of a powerful algorithmic problem - finding the smallest subarrays with maximum bitwise OR.\n\n## Setting Forth on the Quest\n\nThe brave fellowship of coders are given an array of non-negative integers, `nums`. Their task is to find, for each starting point in the array, the length of the smallest subarray which yields the maximum bitwise OR. The bitwise OR of an array is simply the bitwise OR of all its numbers. The output of their journey will be an array of integers, each representing the smallest subarray length with maximum bitwise OR, for each starting point in `nums`.\n\n## The Secret of the Last Seen Bits\n\nThe wise wizards of the fellowship devised a cunning strategy. They knew that to compute the maximum bitwise OR, they would need to understand the layout of the bits in each number. But they also knew that the bit length of any number, on most mortal and immortal computers, is limited to either 32 or 64 bits. Hence, they could create an array, `last_seen_bits`, to keep track of the last occurrence of each bit in the numbers from `nums`.\n\n```python\nmax_bit_length = max(1, max(n.bit_length() for n in nums))\nlast_seen_bits = [0] * max_bit_length\n```\nIn this part of the journey, the `max_bit_length` is calculated. This ensures that the `last_seen_bits` array covers all possible bits in `nums`. The `max(1, ...)` construction is used to handle the case where `nums` is an empty array.\n\n## The Journey Backwards\n\nThe wise wizards also knew that the smallest subarray with maximum bitwise OR, for each starting point in the array, can be found by traversing the array in reverse order. As they traverse `nums`, for each number, they go through its bits. For each bit that is set (1), they update the corresponding element in `last_seen_bits` to the current index. This enchantment is performed by the `update_last_seen_bits` function.\n\n```python\ndef update_last_seen_bits(num, index, last_seen_bits):\n\tbit_index = 0\n\twhile (num >> bit_index):\n\t\tif (num >> bit_index) & 1:\n\t\t\tlast_seen_bits[bit_index] = index\n\t\tbit_index += 1\n```\n\n## One Subarray to Rule Them All\n\nThe fellowship now understood that for each starting index `i`, the length of the smallest subarray with maximum bitwise OR is given by the maximum last seen bit index - `i` + 1. This is because the maximum bitwise OR starting at `i` must include all bits set in `nums[i]`, and hence must include all numbers up to the maximum last seen bit index. \n\n```python\nresult = [0] * len(nums)\nfor i in range(len(nums) - 1, -1, -1):\n\tupdate_last_seen_bits(nums[i], i, last_seen_bits)\n\tresult[i] = max(1, max(last_seen_bits) - i + 1)\nreturn result\n```\n\n## The Path to Mount Doom\n\nThis valiant strategy led the fellowship through their quest with both speed and precision. The time complexity of their journey is O(nm), where `n` is the number of elements in `nums` and `m` is the maximum bit length in `nums`. However, given that `m` is limited to either 32 or 64 on most machines, it can be treated as a constant factor, making the overall time complexity effectively O(n).\n\nTheir journey also left a small footprint. The space complexity of the solution is O(m), which arises from the need to store the result array and the `last_seen_bits` array. Again, given the constraints on `m`, it can be treated as a constant factor, yielding a space complexity of O(1).\n\n## Return of the King\n\nWith their quest completed, the fellowship has illuminated the path to understanding the smallest subarrays with maximum bitwise OR. Their tale serves as a reminder that even in the deepest darkness, there is a light that never goes out - the light of knowledge, of understanding, and of the joy of coding.
['Two Pointers', 'Python']