leetcode-problemset/leetcode-cn/originData/earliest-second-to-mark-indices-ii.json

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        "question": {
            "questionId": "3289",
            "questionFrontendId": "3049",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2650520,
            "title": "Earliest Second to Mark Indices II",
            "titleSlug": "earliest-second-to-mark-indices-ii",
            "content": "<p>You are given two <strong>1-indexed</strong> integer arrays, <code>nums</code> and, <code>changeIndices</code>, having lengths <code>n</code> and <code>m</code>, respectively.</p>\n\n<p>Initially, all indices in <code>nums</code> are unmarked. Your task is to mark <strong>all</strong> indices in <code>nums</code>.</p>\n\n<p>In each second, <code>s</code>, in order from <code>1</code> to <code>m</code> (<strong>inclusive</strong>), you can perform <strong>one</strong> of the following operations:</p>\n\n<ul>\n\t<li>Choose an index <code>i</code> in the range <code>[1, n]</code> and <strong>decrement</strong> <code>nums[i]</code> by <code>1</code>.</li>\n\t<li>Set <code>nums[changeIndices[s]]</code> to any <strong>non-negative</strong> value.</li>\n\t<li>Choose an index <code>i</code> in the range <code>[1, n]</code>, where <code>nums[i]</code> is <strong>equal</strong> to <code>0</code>, and <strong>mark</strong> index <code>i</code>.</li>\n\t<li>Do nothing.</li>\n</ul>\n\n<p>Return <em>an integer denoting the <strong>earliest second</strong> in the range </em><code>[1, m]</code><em> when <strong>all</strong> indices in </em><code>nums</code><em> can be marked by choosing operations optimally, or </em><code>-1</code><em> if it is impossible.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,2,3], changeIndices = [1,3,2,2,2,2,3]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> In this example, we have 7 seconds. The following operations can be performed to mark all indices:\nSecond 1: Set nums[changeIndices[1]] to 0. nums becomes [0,2,3].\nSecond 2: Set nums[changeIndices[2]] to 0. nums becomes [0,2,0].\nSecond 3: Set nums[changeIndices[3]] to 0. nums becomes [0,0,0].\nSecond 4: Mark index 1, since nums[1] is equal to 0.\nSecond 5: Mark index 2, since nums[2] is equal to 0.\nSecond 6: Mark index 3, since nums[3] is equal to 0.\nNow all indices have been marked.\nIt can be shown that it is not possible to mark all indices earlier than the 6th second.\nHence, the answer is 6.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,0,1,2], changeIndices = [1,2,1,2,1,2,1,2]\n<strong>Output:</strong> 7\n<strong>Explanation:</strong> In this example, we have 8 seconds. The following operations can be performed to mark all indices:\nSecond 1: Mark index 1, since nums[1] is equal to 0.\nSecond 2: Mark index 2, since nums[2] is equal to 0.\nSecond 3: Decrement index 4 by one. nums becomes [0,0,1,1].\nSecond 4: Decrement index 4 by one. nums becomes [0,0,1,0].\nSecond 5: Decrement index 3 by one. nums becomes [0,0,0,0].\nSecond 6: Mark index 3, since nums[3] is equal to 0.\nSecond 7: Mark index 4, since nums[4] is equal to 0.\nNow all indices have been marked.\nIt can be shown that it is not possible to mark all indices earlier than the 7th second.\nHence, the answer is 7.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3], changeIndices = [1,2,3]\n<strong>Output:</strong> -1\n<strong>Explanation: </strong>In this example, it can be shown that it is impossible to mark all indices, as we don&#39;t have enough seconds. \nHence, the answer is -1.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == nums.length &lt;= 5000</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= m == changeIndices.length &lt;= 5000</code></li>\n\t<li><code>1 &lt;= changeIndices[i] &lt;= n</code></li>\n</ul>\n",
            "translatedTitle": "标记所有下标的最早秒数 II",
            "translatedContent": "<p>给你两个下标从 <strong>1</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code> 和&nbsp;<code>changeIndices</code>&nbsp;，数组的长度分别为&nbsp;<code>n</code> 和&nbsp;<code>m</code>&nbsp;。</p>\n\n<p>一开始，<code>nums</code>&nbsp;中所有下标都是未标记的，你的任务是标记 <code>nums</code>&nbsp;中 <strong>所有</strong>&nbsp;下标。</p>\n\n<p>从第 <code>1</code>&nbsp;秒到第 <code>m</code>&nbsp;秒（<b>包括&nbsp;</b>第&nbsp;<code>m</code>&nbsp;秒），对于每一秒 <code>s</code>&nbsp;，你可以执行以下操作 <strong>之一</strong>&nbsp;：</p>\n\n<ul>\n\t<li>选择范围&nbsp;<code>[1, n]</code>&nbsp;中的一个下标 <code>i</code>&nbsp;，并且将&nbsp;<code>nums[i]</code> <strong>减少</strong>&nbsp;<code>1</code>&nbsp;。</li>\n\t<li>将&nbsp;<code>nums[changeIndices[s]]</code>&nbsp;设置成任意的 <strong>非负</strong>&nbsp;整数。</li>\n\t<li>选择范围&nbsp;<code>[1, n]</code>&nbsp;中的一个下标&nbsp;<code>i</code>&nbsp;， 满足&nbsp;<code>nums[i]</code> <strong>等于</strong> <code>0</code>, 并 <strong>标记</strong>&nbsp;下标&nbsp;<code>i</code> 。</li>\n\t<li>什么也不做。</li>\n</ul>\n\n<p>请你返回范围 <code>[1, m]</code>&nbsp;中的一个整数，表示最优操作下，标记&nbsp;<code>nums</code>&nbsp;中 <strong>所有</strong>&nbsp;下标的 <strong>最早秒数</strong>&nbsp;，如果无法标记所有下标，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>nums = [3,2,3], changeIndices = [1,3,2,2,2,2,3]\n<b>输出：</b>6\n<b>解释：</b>这个例子中，我们总共有 7 秒。按照以下操作标记所有下标：\n第 1 秒：将 nums[changeIndices[1]] 变为 0 。nums 变为 [0,2,3] 。\n第 2 秒：将 nums[changeIndices[2]] 变为 0 。nums 变为 [0,2,0] 。\n第 3 秒：将 nums[changeIndices[3]] 变为 0 。nums 变为 [0,0,0] 。\n第 4 秒：标记下标 1 ，因为 nums[1] 等于 0 。\n第 5 秒：标记下标 2 ，因为 nums[2] 等于 0 。\n第 6 秒：标记下标 3 ，因为 nums[3] 等于 0 。\n现在所有下标已被标记。\n最早可以在第 6 秒标记所有下标。\n所以答案是 6 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [0,0,1,2], changeIndices = [1,2,1,2,1,2,1,2]\n<b>输出：</b>7\n<b>解释：</b>这个例子中，我们总共有 8 秒。按照以下操作标记所有下标：\n第 1 秒：标记下标 1 ，因为 nums[1] 等于 0 。\n第 2 秒：标记下标 2 ，因为 nums[2] 等于 0 。\n第 3 秒：将 nums[4] 减少 1 。nums 变为 [0,0,1,1] 。\n第 4 秒：将 nums[4] 减少 1 。nums 变为 [0,0,1,0] 。\n第 5 秒：将 nums[3] 减少 1 。nums 变为 [0,0,0,0] 。\n第 6 秒：标记下标 3 ，因为 nums[3] 等于 0 。\n第 7 秒：标记下标 4 ，因为 nums[4] 等于 0 。\n现在所有下标已被标记。\n最早可以在第 7 秒标记所有下标。\n所以答案是 7 。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [1,2,3], changeIndices = [1,2,3]\n<b>输出：</b>-1\n<strong>解释：</strong>这个例子中，无法标记所有下标，因为我们没有足够的秒数。\n所以答案是 -1 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == nums.length &lt;= 5000</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= m == changeIndices.length &lt;= 5000</code></li>\n\t<li><code>1 &lt;= changeIndices[i] &lt;= n</code></li>\n</ul>\n",
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            "hints": [
                "We need at least <code>n</code> seconds, and at most <code>sum(nums[i]) + n</code> seconds.",
                "We can binary search the earliest second where all indices can be marked.",
                "If there is an operation where we change <code>nums[changeIndices[i]]</code> to a non-negative value, it is best for it to satisfy the following constraints:<ul>\r\n<li><code>nums[changeIndices[i]]</code> should not be equal to <code>0</code>.</li>\r\n<li><code>nums[changeIndices[i]]</code> should be changed to <code>0</code>.</li>\r\n<li>It should be the first position where <code>changeIndices[i]</code> occurs in <code>changeIndices</code>.</li>\r\n<li>There should be another second, <code>j</code>, where <code>changeIndices[i]</code> will be marked. <code>j</code> is in the range <code>[i + 1, m]</code>.</li>\r\n</ul>",
                "Let <code>time_needed = sum(nums[i]) + n</code>. To check if we can mark all indices at some second <code>x</code>, we need to make <code>time_needed <= x</code>, using non-negative change operations as described previously.",
                "Using a non-negative change operation on some <code>nums[changeIndices[i]]</code> that satisfies the constraints described previously reduces <code>time_needed</code> by <code>nums[changeIndices[i]] - 1</code>. So, we need to maximize the sum of <code>(nums[changeIndices[i]] - 1)</code> while ensuring that the non-negative change operations still satisfy the constraints.",
                "Maximizing the sum of <code>(nums[changeIndices[i]] - 1)</code> can be done greedily using a min-priority queue and going in reverse starting from second <code>x</code> to second <code>1</code>, maximizing the sum of the values in the priority queue and ensuring that for every non-negative change operation on <code>nums[changeIndices[i]]</code> chosen, there is another second <code>j</code> in the range <code>[i + 1, x]</code> where <code>changeIndices[i]</code> can be marked.",
                "The answer is the first value of <code>x</code> in the range <code>[1, m]</code> where it is possible to make <code>time_needed <= x</code>, or <code>-1</code> if there is no such second."
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