"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 <= i < 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 <= i < 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> </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> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code><font face=\"monospace\">1 <= nums1.length <= 10<sup>3</sup></font></code></li>\n\t<li><code>1 <= nums1[i] <= 10<sup>3</sup></code></li>\n\t<li><code>0 <= nums2[i] <= 10<sup>3</sup></code></li>\n\t<li><code>nums1.length == nums2.length</code></li>\n\t<li><code>0 <= x <= 10<sup>6</sup></code></li>\n</ul>\n",
"<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>"
"envInfo":"{\"cpp\": [\"C++\", \"<p>Compiled with <code> clang 11 </code> using the latest C++ 20 standard.</p>\\r\\n\\r\\n<p>Your code is compiled with level two optimization (<code>-O2</code>). <a href=\\\"https://github.com/google/sanitizers/wiki/AddressSanitizer\\\" target=\\\"_blank\\\">AddressSanitizer</a> is also enabled to help detect out-of-bounds and use-after-free bugs.</p>\\r\\n\\r\\n<p>Most standard library headers are already included automatically for your convenience.</p>\"], \"java\": [\"Java\", \"<p><code>OpenJDK 17</code>. Java 8 features such as lambda expressions and stream API can be used. </p>\\r\\n\\r\\n<p>Most standard library headers are already included automatically for your convenience.</p>\\r\\n<p>Includes <code>Pair</code> class from https://docs.oracle.com/javase/8/javafx/api/javafx/util/Pair.html.</p>\"], \"python\": [\"Python\", \"<p><code>Python 2.7.12</code>.</p>\\r\\n\\r\\n<p>Most libraries are already imported automatically for your convenience, such as <a href=\\\"https://docs.python.org/2/library/array.html\\\" target=\\\"_blank\\\">array</a>, <a href=\\\"https://docs.python.org/2/library/bisect.html\\\" target=\\\"_blank\\\">bisect</a>, <a href=\\\"https://docs.python.org/2/library/collections.html\\\" target=\\\"_blank\\\">collections</a>. If you need more libraries, you can import it yourself.</p>\\r\\n\\r\\n<p>For Map/TreeMap data structure, you may use <a href=\\\"http://www.grantjenks.com/docs/sortedcontainers/\\\" target=\\\"_blank\\\">sortedcontainers</a> library.</p>\\r\\n\\r\\n<p>Note that Python 2.7 <a href=\\\"https://www.python.org/dev/peps/pep-0373/\\\" target=\\\"_blank\\\">will not be maintained past 2020</a>. For the latest Python, please choose Python3 instead.</p>\"], \"c\": [\"C\", \"<p>Compiled with <code>gcc 8.2</code> using the gnu11 standard.</p>\\r\\n\\r\\n<p>Your code is compiled with level one optimization (<code>-O1</code>). <a href=\\\"https://github.com/google/sanitizers/wiki/AddressSanitizer\\\" target=\\\"_blank\\\">AddressSanitizer</a> is also enabled to help detect out-of-bounds and use-after-free bugs.</p>\\r\\n\\r\\n<p>Most standard library headers are already included automatically for your convenience.</p>\\r\\n\\r\\n<p>For hash table operations, you may use <a href=\\\"https://troydhanson.github.io/uthash/\\\" target=\\\"_blank\\\">uthash</a>. \\\"uthash.h\\\" is included by default. Below are some examples:</p>\\r\\n\\r\\n<p><b>1. Adding an item to a hash.</b>\\r\\n<pre>\\r\\nstruct hash_entry {\\r\\n int id; /* we'll use this field as the key */\\r\\n char name[10];\\r\\n UT_hash_handle hh; /* makes this structure hashable */\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n</pre>\\r\\n</p>\\r\\n\\r\\n<p><b>2. Looking up an item in a hash:</b>\\r\\n<pre>\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n struct hash_entry *s;\\r\\n HASH_FIND_INT(users, &user_id, s);\\r\\n return s;\\r\\n}\\r\\n</pre>\\r\\n</p>\\r\\n\\r\\n<p><b>3. Deleting an item in a hash:</b>\\r\\n<pre>\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n HASH_DEL(users, user); \\r\\n}\\r\\n</pre>\\r\\n</p>\"], \"csharp\": [\"C#\", \"<p><a href=\\\"https://learn.microsoft.com/en-us/dotnet/csharp/whats-new/csharp-10\\\" target=\\\"_blank\\\">C# 10 with .NET 6 runtime</a></p>\"], \"javascript\": [\"JavaScript\", \"<p><code>Node.js 16.13.2</code>.</p>\\r\\n\\r\\n<p>Your code is run with <code>--harmony</code> flag, enabling <a href=\\\"http://node.green/\\\" target=\\\"_blank\\\">new ES6 features</a>.</p>\\r\\n\\r\\n<p><a href=\\\"https://lodash.com\\\" target=\\\"_blank\\\">lodash.js</a> library is included by default.</p>\\r\\n\\r\\n<p>For Priority Queue / Queue data structures, you may use 5.3.0 version of <a href=\\\"https://github.com/datastructures-js/priority-queue/tree/fb4fdb984834421279aeb081df7af624d17c2a03\\\" target=\\\"_blank\\\">datastructures-js/priority-queue</a> and 4.2.1 version of <a href=\\\"https://githu
