The Jedi Path to Unraveling the Smallest Subarrays with Maximum Bitwise OR
Solution\n\nfrom collections import defaultdict\nfrom typing import List\n\nclass Solution:\n    def smallestSubarrays(self, nums: List[int]) -> List[int]:\n\n 
**Solution**\n```\nfrom collections import defaultdict\nfrom typing import List\n\nclass Solution:\n    def smallestSubarrays(self, nums: List[int]) -> List[int]:\n\n        def add_to_dict(num, bit_dict):\n            """Adds the bits of \'num\' to the dictionary \'bit_dict\'."""\n            bit_index = 0\n            while (num >> bit_index):\n                if (num >> bit_index) & 1:\n                    bit_dict[bit_index] += 1\n                bit_index += 1\n\n        def remove_from_dict(num, bit_dict):\n            """Removes the bits of \'num\' from the dictionary \'bit_dict\'."""\n            bit_index = 0\n            while (num >> bit_index):\n                if (num >> bit_index) & 1:\n                    bit_dict[bit_index] -= 1\n                    if bit_dict[bit_index] == 0:\n                        del bit_dict[bit_index]\n                bit_index += 1\n\n        bit_counts = defaultdict(int)\n        for num in nums:\n            add_to_dict(num, bit_counts)\n\n        current_bit_counts = defaultdict(int)\n        right_ptr = 0\n        ans = [0] * len(nums)\n        \n        for left_ptr, num in enumerate(nums):\n            if len(bit_counts) == 0:\n                for j in range(left_ptr, len(nums)):\n                    ans[j] = 1\n                break\n            \n            while len(current_bit_counts) != len(bit_counts):\n                add_to_dict(nums[right_ptr], current_bit_counts)\n                right_ptr += 1\n                \n            ans[left_ptr] = right_ptr - left_ptr\n            \n            remove_from_dict(num, bit_counts)\n            remove_from_dict(num, current_bit_counts)\n        \n        return ans\n\n```\n\nIn a galaxy far, far away, the Jedi Council of Data Structures and Algorithms has uncovered a new challenge, one that requires a deep understanding of the Force, also known as Bit Manipulation and Sliding Window techniques. This complex task involves finding the smallest subarrays within a given array, such that the bitwise OR of each subarray is maximized. Worry not, young padawans. The path may be treacherous, but the Force is with us. We shall embark on this quest and solve this enigma together.\n\n## The Task at Hand\n\nOur mission is to determine the size of the smallest subarray with maximum possible bitwise OR starting at each index i in a given array, `nums`. The bitwise OR of an array is the bitwise OR of all numbers within it. Our output shall be an array of equal size, where each element corresponds to the length of the smallest subarray starting at the respective index in `nums` that achieves this maximum bitwise OR.\n\n## The Force of Sliding Windows and Bit Manipulation\n\nOur Jedi solution calls upon two powerful techniques in the coding universe - the Sliding Window and Bit Manipulation. The Sliding Window approach harnesses the powers of two pointers, `left_ptr` and `right_ptr`, that define our current window within the array. As these pointers traverse the galaxy of our array, they uncover the smallest subarray size needed to reach the maximum bitwise OR at each index. \n\nBit Manipulation, our second tool, comes into play when we\'re calculating the bitwise OR for the numbers in our current window. We use it to track and manipulate the individual bits of our numbers, allowing us to reach the maximum bitwise OR with Jedi-like precision and efficiency.\n\n## The Path to the Solution\n\nNow, let\'s embark on our coding journey together. Remember, a Jedi\'s strength flows from the Force. \n\n```python\nfrom collections import defaultdict\nfrom typing import List\n\nclass Solution:\n    def smallestSubarrays(self, nums: List[int]) -> List[int]:\n```\n\nThe first step in our journey is the creation of two helper functions, `add_to_dict` and `remove_from_dict`. These two functions use the power of Bit Manipulation to respectively add and remove the bits of a number to and from a dictionary:\n\n```python\n        def add_to_dict(num, bit_dict):\n            """Adds the bits of \'num\' to the dictionary \'bit_dict\'."""\n            bit_index = 0\n            while (num >> bit_index):\n                if (num >> bit_index) & 1:\n                    bit_dict[bit_index] += 1\n                bit_index += 1\n\n        def remove_from_dict(num, bit_dict):\n            """Removes the bits of \'num\' from the dictionary \'bit_dict\'."""\n            bit_index = 0\n            while (num >> bit_index):\n                if (num >> bit_index) & 1:\n                    bit_dict[bit_index] -= 1\n                    if bit_dict[bit_index] == 0:\n                        del bit_dict[bit_index]\n                bit_index += 1\n```\n\nHaving constructed our helper functions, we now create the `bit_counts` dictionary. This Jedi tool stores the count of each bit present in the bitwise OR of all numbers in `nums`:\n\n```python\n        bit_counts = defaultdict(int)\n        for num in nums:\n            add_to_dict(num, bit_counts)\n```\n\nNext, we initialize `current_bit_counts`, a dictionary that will keep track of the bits in the bitwise OR of the numbers within our current window. We also set up `right_ptr`, the pointer that will slide across `nums` to extend our\n\n current window:\n\n```python\n        current_bit_counts = defaultdict(int)\n        right_ptr = 0\n        ans = [0] * len(nums)\n```\n\nOur next task is to traverse the universe of our array with the `left_ptr` pointer. The current window starts at the `left_ptr` index, and we extend this window by moving `right_ptr` until the bitwise OR of the numbers in the window equals the maximum bitwise OR. This equality is checked by comparing `current_bit_counts` with `bit_counts`. \n\n```python\n        for left_ptr, num in enumerate(nums):\n            if len(bit_counts) == 0:\n                for j in range(left_ptr, len(nums)):\n                    ans[j] = 1\n                break\n            \n            while len(current_bit_counts) != len(bit_counts):\n                add_to_dict(nums[right_ptr], current_bit_counts)\n                right_ptr += 1\n                \n            ans[left_ptr] = right_ptr - left_ptr\n```\n\nThe last part of our journey involves moving the `left_ptr` to the right, effectively reducing the window size by one. The `num` at this index is removed from both `bit_counts` and `current_bit_counts` to reflect this change. This process continues until the entire array has been traversed:\n\n```python\n            remove_from_dict(num, bit_counts)\n            remove_from_dict(num, current_bit_counts)\n        \n        return ans\n```\n\n## The Computational Complexity of the Force\n\nThe path we have tread together is not just the way of the Jedi, but also the way of Optimal Algorithm Design. Our approach here is efficient and elegant, boasting a time complexity of O(n). Each number in the array is processed exactly once due to the nature of the sliding window technique, which results in a linear time complexity. \n\nIn terms of space complexity, our solution sits at O(n). This arises from the need to store the output array `ans` and the dictionaries for bit counts, which in the worst-case scenario could be as large as the input array.\n\n## In a Galaxy not so far away\u2026\n\nOur journey has now come to an end. We have successfully solved the problem with the wisdom of the Jedi, unraveling the secrets of Bit Manipulation and the Sliding Window technique. Remember, the Force will be with you. Always.\n\n
['Two Pointers', 'Python']