Unraveling the Optimal Time Scheduler
Intuition\nImagine you have two lists of tasks with respective times required. You have a goal to optimize the sum of these tasks such that, at every step, you 
# Intuition\nImagine you have two lists of tasks with respective times required. You have a goal to optimize the sum of these tasks such that, at every step, you can either use the time from one list or swap it with a time from the other list. The problem might sound tricky, but a methodical approach can reveal patterns that help us arrive at an optimal solution. Our immediate thought here is to prioritize tasks based on which has the most significant benefit when swapped, setting the stage for a dynamic programming solution.\n\n# Approach\n    Pairing & Prioritizing Tasks: We start by pairing each task from nums1 with its corresponding task in nums2. The main idea is to figure out which task is most advantageous to swap earlier. This decision is made by sorting these pairs based on the values of nums2. Why? Because when we swap out a value in nums1, the penalty (or addition) we get is from nums2 multiplied by its position. Hence, larger values in nums2 are beneficial to swap out later.\n\n    Dynamic Programming Table: We initialize a table, dp, with dimensions (n+1) x (n+1), where n is the number of tasks. This table will help us determine the maximum time reduction possible for any given number of tasks with a set number of operations.\n\n    Populating the Table: For each task (in the sorted order), we consider two options:\n    a. Not using this task and hence retaining the previous maximum reduction.\n    b. Using this task, which means we take the value from nums1 and add a penalty based on its position and value in nums2.\n\n    Finding the Minimum Time: Once our table is populated, we go through each time from 0 to n, and check if the total time minus the reduction obtained from our dp table is less than or equal to x (our target). If yes, we return that time.\n\n# Complexity\n- Time complexity:\nO(n^2). The major time consumption happens during the DP table population, which involves nested loops running through the number of tasks.\n\n- Space complexity:\nO(n^2). Primarily due to the space occupied by the DP table. We\'ve also optimized space by utilizing the array data structure, making it more memory-efficient.\n\n# Code\n```\nfrom array import array\n\nclass Solution:\n    def minimumTime(self, nums1, nums2, x):\n        n = len(nums1)\n        \n        indices = list(range(n))\n        indices.sort(key=lambda i: nums2[i])\n        \n        dp = [array(\'I\', (0 for _ in range(n + 1))) for _ in range(n + 1)]\n        \n        for idx in range(1, n + 1):\n            a, b = nums1[indices[idx-1]], nums2[indices[idx-1]]\n            for j in range(1, n + 1):\n                dp[idx][j] = max(dp[idx-1][j], dp[idx-1][j-1] + a + b * j)\n        \n        sum1 = sum(nums1)\n        sum2 = sum(nums2)\n        \n        for t in range(n + 1):\n            prod = sum2 * t\n            if sum1 + prod - dp[n][t] <= x:\n                return t\n        return -1\n\n```
['Array', 'Dynamic Programming', 'Sorting', 'Python']