Solution in Swift, DP
Approach\nWe begin by calculating the total sum of the arrays A and B as sa and sb respectively.\n\nIf no actions are taken, at i seconds, we would have a total
# Approach\nWe begin by calculating the total sum of the arrays A and B as sa and sb respectively.\n\nIf no actions are taken, at i seconds, we would have a total of sb * i + sa.\n\nDuring the t seconds, we select t elements. When we consider these selected elements, A[i] would be removed. The sum for these selected elements would be b1 * (t - 1) + b2 * (t - 2) + ... + bt * 0, where b1, b2, b3, ..., bt are arranged in increasing order.\n\nTo solve this problem, we sort all the pairs (B[i], A[i]) based on the value of B[i].\n\nWe then utilize dynamic programming (dp) with the following logic:\ndp[j][i] represents the maximum value we can reduce within i seconds, using j + 1 step-smallest integers.\n\nThe dp equation is as follows:\ndp[j][i] = max(dp[j][i], dp[j - 1][i - 1] + i * b + a)\n\nIn the end, we return the value of i seconds if sb * i + sa - dp[n - 1][i] is less than or equal to x. If not, we return -1.\n\nIt is possible to optimize the space complexity by storing only the first dimension of the dp array.\n\n# Complexity\n- Time complexity: `O(n^2)`\n- Space complexity: `O(n)`\n\n# Code\n```\nclass Solution {\n    func minimumTime(_ nums1: [Int], _ nums2: [Int], _ x: Int) -> Int {\n        let n = nums1.count\n        var dp = [Int](repeating: 0, count: n + 1)\n        \n        let sortedPairs = zip(nums2, nums1).sorted { $0.0 < $1.0 }\n        \n        for (j, (b, a)) in sortedPairs.enumerated() {\n            for i in stride(from: j + 1, through: 1, by: -1) {\n                dp[i] = max(dp[i], dp[i - 1] + i * b + a)\n            }\n        }\n        \n        let sa = nums1.reduce(0, +)\n        let sb = nums2.reduce(0, +)\n        \n        for i in 0...n {\n            if sb * i + sa - dp[i] <= x {\n                return i\n            }\n        }\n        \n        return -1\n    }\n}\n\n```
['Swift']