Actually Helpful Explanation with Proof and Not Just Code translated in English
We will find a value MaxRemove[i], this tells us the maximum sum of values we can remove if we can find the answer in i seconds/moves.\n\nIf we can find this Ma
We will find a value MaxRemove[i], this tells us the maximum sum of values we can remove if we can find the answer in i seconds/moves.\n\nIf we can find this MaxRemove[i] for i = 0 to N, then we can just check at every position if the total would be less than X, then we are done.\n\nLet\'s try to find this now.\n\n- There is no point in making one element zero more than once. Because last time when it is made zero, it is zero now, choosing this in any move before this doesn\'t makes sense. This tells us that the maximum number of seconds required is N, otherwise not possible.\n\n- If we have a set of K indexes which we are choosing to make zero at some point, then it is always better to choose the index with the larger num2 value in the end because this will contribute most to the answer otherwise, then second largest in second last position and so on. This tells us that if we try to update our MaxRemove state, it would make sense to iterate from smaller num2 values, as we would know which position to put the latest index in.\n\nNow the idea is initially our set of used indexes is empty. MaxRemove is 0 for all i = 0 to i = N.\nNow we would iterate through the index in increasing order of (num2[i], nums1[i]) and try to update the MaxRemove.\n\nThe state transition is - \nEither take the current element or not.\nIf taken it will present in the last position, so update accordingly.\nMaxRemove[i] = max(MaxRemove[i], MaxRemove[i - 1] + nums1[ind] + nums2[ind] * i);\n\nRefer to my code if something is unclear.\n\n```\n#define all(c) (c).begin(), (c).end()\n\nclass Solution {\npublic:\n    int minimumTime(vector<int>& nums1, vector<int>& nums2, int x) {\n        int n = nums1.size();\n        vector<int> Ind(n, -1);\n        iota(all(Ind), 0);\n        sort(all(Ind), [&](int i, int j) {\n            return make_pair(nums2[i], nums1[i]) < make_pair(nums2[j], nums1[j]);\n        });\n        vector<int> MaxRemove(n + 1, 0);\n        for(auto ind : Ind) {\n            for(int i = n; i >= 1; --i) {\n                MaxRemove[i] = max(MaxRemove[i], MaxRemove[i - 1] + nums1[ind] + nums2[ind] * i);\n            }\n        }\n        int tot = accumulate(all(nums1), 0);\n        int perTurn = accumulate(all(nums2), 0);\n        for(int i = 0; i <= n; i++) {\n            if(tot + perTurn * i - MaxRemove[i] <= x) return i;\n        }\n        return -1;\n    }\n};\n```
['C']