Minimum Time to Process Arrays with Constraints Beats 81.20%
Intuition\nThe problem seems to involve finding the minimum time required to perform certain operations on two arrays while satisfying certain conditions. \n\n#
# Intuition\nThe problem seems to involve finding the minimum time required to perform certain operations on two arrays while satisfying certain conditions. \n\n# Approach\n1. We start by calculating the total sums of both arrays, `nums1` and `nums2`, and storing the indices in a separate vector.\n2. If the sum of `nums1` is less than or equal to `x`, then we can return 0 as no operations are needed.\n3. Otherwise, we sort the indices based on the values of `nums2`.\n4. We then utilize dynamic programming to find the minimum time. We iterate over the sorted indices and calculate the time taken for each possible number of elements from `nums2` to be processed, considering the time it takes to process elements from both arrays.\n5. We keep track of the minimum time found so far, updating it whenever we find a new minimum time that satisfies the given condition.\n6. Finally, we return the minimum time found or -1 if it\'s not possible to satisfy the condition.\n\n# Complexity\n- Time complexity:\n  - Sorting the indices based on `nums2` takes O(n*log(n)) time.\n  - The dynamic programming approach takes O(n^2) time.\n  - Therefore, the overall time complexity is O(n^2).\n- Space complexity:\n  - We use additional space for the `ind` vector and the `dp` vector, both of size `n`, resulting in a space complexity of O(n).\n\n# Code\n``` C++ []\nclass Solution {\npublic:\n    int minimumTime(vector<int>& nums1, vector<int>& nums2, int x) {\n        const int n = nums1.size();\n        vector<int> ind(n);\n        int s = 0, d = 0;\n        for (int i = 0; i < n; ++i) {\n            ind[i] = i;\n            s += nums1[i];\n            d += nums2[i];\n        }\n        if (s <= x) {\n            return 0;\n        }\n        sort(ind.begin(), ind.end(), [&](const int a, const int b) {\n            return nums2[a] < nums2[b];\n        });\n        vector<int> dp(n + 1);\n        int r = n + 1;\n        for (int i = 1; i <= n; ++i) {\n            for (int j = min(i, r - 1); j; --j) {\n                dp[j] = max(dp[j], dp[j - 1] + nums2[ind[i - 1]] * j + nums1[ind[i - 1]]);\n                if (s + j * d - dp[j] <= x) {\n                    r = j;\n                }\n            }\n        }\n        return r <= n ? r : -1;\n    }\n};\n```\n```javascript []\nfunction minimumTime(nums1, nums2, x) {\n    const n = nums1.length;\n    let ind = Array.from({ length: n }, (_, i) => i);\n    let s = 0, d = 0;\n    \n    for (let i = 0; i < n; ++i) {\n        s += nums1[i];\n        d += nums2[i];\n    }\n    \n    if (s <= x) {\n        return 0;\n    }\n    \n    ind.sort((a, b) => nums2[a] - nums2[b]);\n    let dp = new Array(n + 1).fill(0);\n    let r = n + 1;\n    \n    for (let i = 1; i <= n; ++i) {\n        for (let j = Math.min(i, r - 1); j; --j) {\n            dp[j] = Math.max(dp[j], dp[j - 1] + nums2[ind[i - 1]] * j + nums1[ind[i - 1]]);\n            if (s + j * d - dp[j] <= x) {\n                r = j;\n            }\n        }\n    }\n    \n    return r <= n ? r : -1;\n}\n\n```\n```python []\nclass Solution:\n    def minimumTime(self, nums1: List[int], nums2: List[int], x: int) -> int:\n        n = len(nums1)\n        ind = list(range(n))\n        s, d = 0, 0\n        for i in range(n):\n            s += nums1[i]\n            d += nums2[i]\n        if s <= x:\n            return 0\n\n        # Custom comparator for sorting ind based on nums2 values\n        ind.sort(key=lambda i: nums2[i])\n\n        dp = [0] * (n + 1)\n        r = n + 1\n        for i in range(1, n + 1):\n            for j in range(min(i, r - 1), 0, -1):\n                dp[j] = max(dp[j], dp[j - 1] + nums2[ind[i - 1]] * j + nums1[ind[i - 1]])\n                if s + j * d - dp[j] <= x:\n                    r = j\n        return r if r <= n else -1\n```\n\n
['Array', 'Dynamic Programming', 'Sorting', 'Python', 'C++', 'Python3', 'JavaScript']