leetcode-problemset/leetcode-cn/originData/minimum-time-to-make-array-sum-at-most-x.json

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            "title": "Minimum Time to Make Array Sum At Most x",
            "titleSlug": "minimum-time-to-make-array-sum-at-most-x",
            "content": "<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code> of equal length. Every second, for all indices <code>0 &lt;= i &lt; nums1.length</code>, value of <code>nums1[i]</code> is incremented by <code>nums2[i]</code>. <strong>After</strong> this is done, you can do the following operation:</p>\n\n<ul>\n\t<li>Choose an index <code>0 &lt;= i &lt; nums1.length</code> and make <code>nums1[i] = 0</code>.</li>\n</ul>\n\n<p>You are also given an integer <code>x</code>.</p>\n\n<p>Return <em>the <strong>minimum</strong> time in which you can make the sum of all elements of </em><code>nums1</code><em> to be<strong> less than or equal</strong> to </em><code>x</code>, <em>or </em><code>-1</code><em> if this is not possible.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,2,3], nums2 = [1,2,3], x = 4\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> \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</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,2,3], nums2 = [3,3,3], x = 4\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> It can be shown that the sum of nums1 will always be greater than x, no matter which operations are performed.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code><font face=\"monospace\">1 &lt;= nums1.length &lt;= 10<sup>3</sup></font></code></li>\n\t<li><code>1 &lt;= nums1[i] &lt;= 10<sup>3</sup></code></li>\n\t<li><code>0 &lt;= nums2[i] &lt;= 10<sup>3</sup></code></li>\n\t<li><code>nums1.length == nums2.length</code></li>\n\t<li><code>0 &lt;= x &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
            "translatedTitle": "使数组和小于等于 x 的最少时间",
            "translatedContent": "<p>给你两个长度相等下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;。每一秒，对于所有下标&nbsp;<code>0 &lt;= i &lt; nums1.length</code>&nbsp;，<code>nums1[i]</code>&nbsp;的值都增加&nbsp;<code>nums2[i]</code>&nbsp;。操作&nbsp;<strong>完成后</strong>&nbsp;，你可以进行如下操作：</p>\n\n<ul>\n\t<li>选择任一满足&nbsp;<code>0 &lt;= i &lt; nums1.length</code>&nbsp;的下标 <code>i</code>&nbsp;，并使&nbsp;<code>nums1[i] = 0</code>&nbsp;。</li>\n</ul>\n\n<p>同时给你一个整数&nbsp;<code>x</code>&nbsp;。</p>\n\n<p>请你返回使&nbsp;<code>nums1</code>&nbsp;中所有元素之和 <strong>小于等于</strong>&nbsp;<code>x</code>&nbsp;所需要的 <strong>最少</strong>&nbsp;时间，如果无法实现，那么返回 <code>-1</code>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>nums1 = [1,2,3], nums2 = [1,2,3], x = 4\n<b>输出：</b>3\n<b>解释：</b>\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</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums1 = [1,2,3], nums2 = [3,3,3], x = 4\n<b>输出：</b>-1\n<b>解释：</b>不管如何操作，nums1 的和总是会超过 x 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums1.length &lt;= 10<sup>3</sup></code></li>\n\t<li><code>1 &lt;= nums1[i] &lt;= 10<sup>3</sup></code></li>\n\t<li><code>0 &lt;= nums2[i] &lt;= 10<sup>3</sup></code></li>\n\t<li><code>nums1.length == nums2.length</code></li>\n\t<li><code>0 &lt;= x &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
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                "<div class=\"_1l1MA\">It can be proven that in the optimal solution, for each index <code>i</code>, we only need to set <code>nums1[i]</code> to <code>0</code> at most once. (If we have to set it twice, we can simply remove the earlier set and all the operations “shift left” by <code>1</code>.)</div>",
                "<div class=\"_1l1MA\">It can also be proven that if we select several indexes <code>i<sub>1</sub>, i<sub>2</sub>, ..., i<sub>k</sub></code> and set <code>nums1[i<sub>1</sub>], nums1[i<sub>2</sub>], ..., nums1[i<sub>k</sub>]</code> to <code>0</code>, it’s always optimal to set them in the order of <code>nums2[i<sub>1</sub>] <= nums2[i<sub>2</sub>] <= ... <= nums2[i<sub>k</sub>]</code> (the larger the increase is, the later we should set it to <code>0</code>).</div>",
                "<div class=\"_1l1MA\">Let’s sort all the values by <code>nums2</code> (in non-decreasing order). Let <code>dp[i][j]</code> represent the maximum total value that can be reduced if we do <code>j</code> operations on the first <code>i</code> elements. Then we have <code>dp[i][0] = 0</code> (for all <code>i = 0, 1, ..., n</code>) and <code>dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + nums2[i - 1] * j + nums1[i - 1])</code> (for <code>1 <= i <= n</code> and <code>1 <= j <= i</code>).</div>",
                "<div class=\"_1l1MA\">The answer is the minimum value of <code>t</code>, such that <code>0 <= t <= n</code> and <code>sum(nums1) + sum(nums2) * t - dp[n][t] <= x</code>, or <code>-1</code> if it doesn’t exist.</div>"
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