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 <= 106给你两个长度相等下标从 0 开始的整数数组 nums1 和 nums2 。每一秒，对于所有下标 0 <= i < nums1.length ，nums1[i] 的值都增加 nums2[i] 。操作 完成后 ，你可以进行如下操作：
0 <= i < nums1.length 的下标 i ，并使 nums1[i] = 0 。同时给你一个整数 x 。
请你返回使 nums1 中所有元素之和 小于等于 x 所需要的 最少 时间，如果无法实现，那么返回 -1 。
\n\n
示例 1：
\n\n\n输入：nums1 = [1,2,3], nums2 = [1,2,3], x = 4\n输出：3\n解释：\n第 1 秒，我们对 i = 0 进行操作，得到 nums1 = [0,2+2,3+3] = [0,4,6] 。\n第 2 秒，我们对 i = 1 进行操作，得到 nums1 = [0+1,0,6+3] = [1,0,9] 。\n第 3 秒，我们对 i = 2 进行操作，得到 nums1 = [1+1,0+2,0] = [2,2,0] 。\n现在 nums1 的和为 4 。不存在更少次数的操作，所以我们返回 3 。\n\n\n
示例 2：
\n\n\n输入：nums1 = [1,2,3], nums2 = [3,3,3], x = 4\n输出：-1\n解释：不管如何操作，nums1 的和总是会超过 x 。\n\n\n
\n\n
提示：
\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.\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a 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