Efficient Binary Search on Answer Solution
Hey everyone,\n\nI wanted to share my approach to solving the "3048. Earliest Second to Mark Indices I" problem, which I found quite interesting and challenging
Hey everyone,\n\nI wanted to share my approach to solving the "3048. Earliest Second to Mark Indices I" problem, which I found quite interesting and challenging. The problem essentially asks for the earliest second when all indices in a given array nums can be marked, given certain constraints and operations that can be performed in each second.\n\nHere\'s a quick recap of the problem statement:\n\nYou have an integer array nums and another array changeIndices.\nInitially, all indices in nums are unmarked.\nYou can perform certain operations in each second to mark indices in nums.\nThe goal is to find the earliest second when all indices can be marked, or return -1 if it\'s impossible.\nApproach:\nMy solution revolves around using a binary search on the seconds to find the minimum time required to mark all indices. Here\'s a breakdown of my approach:\n\nBinary Search on Time: Since the problem asks for the "earliest second," binary search is a perfect fit to minimize the time complexity. We search over the range of seconds [1, m] where m is the length of the changeIndices array.\n\nSimulating the Process: For each mid-point in the binary search, we simulate the marking process up to that second. We maintain a power counter, which represents our ability to decrement nums[i] by 1, and a mark counter to keep track of the indices we have successfully marked.\n\nHashMap Optimization: To efficiently check if we can mark an index at a given second, I use a hashmap that maps each index in changeIndices to its latest occurrence up to the current second. This helps to quickly determine if an index is eligible for marking.\n\nEarly Termination: If at any point our power goes negative (meaning we don\'t have enough capacity to decrement a nums[i]), or if we cannot mark all indices by the end of our simulation, we know that the current second is not feasible.\n\nSolution:\n```\nclass Solution:\n    def earliestSecondToMarkIndices(self, nums: List[int], idx: List[int]) -> int:\n        def binary_search__on_answer(t):\n            hashmap = {idx[i]: i for i in range(t + 1)} \n            power,mark = 0,0\n            for i in range(t + 1):\n                    if hashmap[idx[i]] != i:\n                        power += 1\n                    else:\n                        power -= nums[idx[i]-1]\n                        mark += 1\n                        if power < 0:\n                            return False\n            return mark == len(nums)\n\n        result = -1\n        l, h = 0, len(idx) - 1\n        while l <= h:\n            mid = (l + h) // 2 \n            if binary_search__on_answer(mid):\n                result = mid + 1\n                h = mid - 1\n            else:\n                l = mid + 1\n        return result\n```\nComplexity Analysis:\nTime Complexity: O(log m * t), where m is the length of the changeIndices array, and t is the time taken to simulate the process for each mid-point in the binary search.\nSpace Complexity: O(t), primarily due to the hashmap used to store the indices up to the current second.\nFinal Thoughts:\nThis problem was a great exercise in applying binary search to a scenario that isn\'t straightforward sorting or searching. It required careful consideration of how to simulate the process efficiently and how to utilize additional data structures like hashmaps for optimization.\n\nWould love to hear any feedback or alternative approaches!\n
['Binary Tree']