You are given two 0-indexed integer arrays nums1 and nums2 of equal length. Every second, for all indices 0 <= i < nums1.length, value of nums1[i] is incremented by nums2[i]. After this is done, you can do the following operation:
0 <= i < nums1.length and make nums1[i] = 0.You are also given an integer x.
Return the minimum time in which you can make the sum of all elements of nums1 to be less than or equal to x, or -1 if this is not possible.
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Example 1:
\n\n\nInput: nums1 = [1,2,3], nums2 = [1,2,3], x = 4\nOutput: 3\nExplanation: \nFor the 1st second, we apply the operation on i = 0. Therefore nums1 = [0,2+2,3+3] = [0,4,6]. \nFor the 2nd second, we apply the operation on i = 1. Therefore nums1 = [0+1,0,6+3] = [1,0,9]. \nFor the 3rd second, we apply the operation on i = 2. Therefore nums1 = [1+1,0+2,0] = [2,2,0]. \nNow sum of nums1 = 4. It can be shown that these operations are optimal, so we return 3.\n\n\n\n
Example 2:
\n\n\nInput: nums1 = [1,2,3], nums2 = [3,3,3], x = 4\nOutput: -1\nExplanation: It can be shown that the sum of nums1 will always be greater than x, no matter which operations are performed.\n\n\n
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Constraints:
\n\n1 <= nums1.length <= 1031 <= nums1[i] <= 1030 <= nums2[i] <= 103nums1.length == nums2.length0 <= x <= 106i, we only need to set nums1[i] to 0 at most once. (If we have to set it twice, we can simply remove the earlier set and all the operations “shift left” by 1.)i1, i2, ..., ik and set nums1[i1], nums1[i2], ..., nums1[ik] to 0, it’s always optimal to set them in the order of nums2[i1] <= nums2[i2] <= ... <= nums2[ik] (the larger the increase is, the later we should set it to 0).nums2 (in non-decreasing order). Let dp[i][j] represent the maximum total value that can be reduced if we do j operations on the first i elements. Then we have dp[i][0] = 0 (for all i = 0, 1, ..., n) and dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + nums2[i - 1] * j + nums1[i - 1]) (for 1 <= i <= n and 1 <= j <= i).t, such that 0 <= t <= n and sum(nums1) + sum(nums2) * t - dp[n][t] <= x, or -1 if it doesn’t exist.Compiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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