5 hours, mega failure
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nimport java.util.*;\n\n\npublic class Solution {\n    // sum2 - max(A2) > X   => IMPOSSIBLE  = -1\n    // sum2 is CONSTANT !!!!\n\n    // so we are looking for steps using i,j...\n    // sum1 + (sum2 - A2[i] - A1[i]) + (sum2 - A2[j] - A1[j]) .... <= X\n    // sum1 + numSteps * sum2 - (A2[i] + A1[i] + A2[j] + A1[j]...) <= X\n\n\n    // Flipping an index sets it to 0 = MIN possible value\n    // its best to let it accumulate and flip it ONCE at the end, make it 0; no point flipping it twice...\n    // nums1 = [1000,   1]    x = 900\n    // nums2 = [900 ,1000]\n    //          1st ,  2nd\n\n    // REALIZATION 1: NEVER FLIP TWICE !!!!!! so the problem becomes about WHICH INDEXES and WHICH ORDER !!!!!!!\n    // REALIZATION 2: ORDER OF FLIPS: ALWAYS FLIP smallest gains A2[i] first !!!!!!! they are slower to accumulate\n    // while we flip bigger ones later, allowing them to accumulate and kill them off immediately; OR NO FLIP AT ALL\n\n    // IDEA: sort by the gain; then we have only 2 choices: flip or not !!!\n\n    // initial DP approach: can be also iterative; works on the SUM as dimension; but the sum range is 0..1,000,000!!!\n\n    // Convert the problem into something else to extract the sum OUT OF THE DP DIMENSIONS !!!!!\n    // the other possible dimensions left are: INDEX and NUM_STEPS which go 0...N !!!! so:  O(N^2)\n    // redefine DP:  dp[i][steps] = MIN sum possible using indexes 0..i and this num steps\n    // NOTICE: when we flip A[i] totalSum INCREASES but prefixSum DECREASES !!!!\n    // so do DP on the prefixSum !!!!\n\n\n    public int minimumTime(List<Integer> nums1, List<Integer> nums2, int x) {\n        this.n = nums1.size();\n        this.x = x;\n        this.A = new int[n][2];\n        int minSteps = Integer.MAX_VALUE;\n        for (int i = 0; i < n; i++) {\n            sum1 += A[i][0] = nums1.get(i);\n            sum2 += A[i][1] = nums2.get(i);\n        }\n        if (sum1 <= x) return 0;\n        Arrays.sort(A, (a, b) -> Integer.compare(a[1], b[1])); // sort by gain\n\n        int[] prefA1 = new int[n + 1]; // prefSum of A1\n        int[] prefA2 = new int[n + 1]; // prefixSum of A2 = prefixGain\n        for (int i = 1; i <= n; i++) {\n            prefA1[i] = prefA1[i - 1] + A[i - 1][0];\n            prefA2[i] = prefA2[i - 1] + A[i - 1][1];\n        }\n        int[][] dp = new int[n + 1][n + 1];\n        for (int i = 1; i <= n; i++) {\n            dp[i][0] = prefA1[i];\n            for (int j = 1; j <= i; j++) { // stepsTaken\n                int take = dp[i - 1][j - 1] + prefA2[i - 1]; // take;  make A1[i] = 0\n                int notTake = dp[i - 1][j] + A[i - 1][0] + j * A[i - 1][1];  // NOT take\n\n                dp[i][j] = take;\n                if (i > j) dp[i][j] = Math.min(dp[i][j], notTake);\n                if (i == n && dp[i][j] <= x) minSteps = Math.min(minSteps, j);\n            }\n        }\n\n        return minSteps == Integer.MAX_VALUE ? -1 : minSteps;\n//        int res = recMemo(0, sum1, 0);\n//        return res >= Integer.MAX_VALUE / 2 ? -1 : res;\n    }\n\n    // [9,10,4,4]   x = 16\n    // [1, 3,4,4]\n\n\n    // initial DP approach: can be also iterative; works on the SUM as dimension; but the sum range is 0..1,000,000!!!\n    // DIMENSION is TOO BIG !!!!!!!\n//    int recMemo(int i, int sum, int stepsTaken) { // adding cache is too much memory; O(n*n*1 000 000)\n//        if (sum <= x) return 0;\n//        if (i == n) return Integer.MAX_VALUE / 2;\n//\n//        int gainRest = sum2 - A[i][1];\n//        int currVal = A[i][0] + stepsTaken * A[i][1];\n//        int take = recMemo(i + 1, sum + gainRest - currVal, stepsTaken + 1);\n//        int notTake = recMemo(i + 1, sum, stepsTaken);\n//\n//        return Math.min(1 + take, notTake);\n//    }\n\n    int sum1, sum2, n, x;\n    int[][] A;\n\n\n}\n\n\n```
['Java']