leetcode-problemset/leetcode-cn/originData/mark-elements-on-array-by-performing-queries.json

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        "question": {
            "questionId": "3306",
            "questionFrontendId": "3080",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2688104,
            "title": "Mark Elements on Array by Performing Queries",
            "titleSlug": "mark-elements-on-array-by-performing-queries",
            "content": "<p>You are given a <strong>0-indexed</strong> array <code>nums</code> of size <code>n</code> consisting of positive integers.</p>\n\n<p>You are also given a 2D array <code>queries</code> of size <code>m</code> where <code>queries[i] = [index<sub>i</sub>, k<sub>i</sub>]</code>.</p>\n\n<p>Initially all elements of the array are <strong>unmarked</strong>.</p>\n\n<p>You need to apply <code>m</code> queries on the array in order, where on the <code>i<sup>th</sup></code> query you do the following:</p>\n\n<ul>\n\t<li>Mark the element at index <code>index<sub>i</sub></code> if it is not already marked.</li>\n\t<li>Then mark <code>k<sub>i</sub></code> unmarked elements in the array with the <strong>smallest</strong> values. If multiple such elements exist, mark the ones with the smallest indices. And if less than <code>k<sub>i</sub></code> unmarked elements exist, then mark all of them.</li>\n</ul>\n\n<p>Return <em>an array answer of size </em><code>m</code><em> where </em><code>answer[i]</code><em> is the <strong>sum</strong> of unmarked elements in the array after the </em><code>i<sup>th</sup></code><em> query</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\" style=\"border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;\">\n<p><strong>Input: </strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\">nums = [1,2,2,1,2,3,1], queries = [[1,2],[3,3],[4,2]]</span></p>\n\n<p><strong>Output: </strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\">[8,3,0]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>We do the following queries on the array:</p>\n\n<ul>\n\t<li>Mark the element at index <code>1</code>, and <code>2</code> of the smallest unmarked elements with the smallest indices if they exist, the marked elements now are <code>nums = [<strong><u>1</u></strong>,<u><strong>2</strong></u>,2,<u><strong>1</strong></u>,2,3,1]</code>. The sum of unmarked elements is <code>2 + 2 + 3 + 1 = 8</code>.</li>\n\t<li>Mark the element at index <code>3</code>, since it is already marked we skip it. Then we mark <code>3</code> of the smallest unmarked elements with the smallest indices, the marked elements now are <code>nums = [<strong><u>1</u></strong>,<u><strong>2</strong></u>,<u><strong>2</strong></u>,<u><strong>1</strong></u>,<u><strong>2</strong></u>,3,<strong><u>1</u></strong>]</code>. The sum of unmarked elements is <code>3</code>.</li>\n\t<li>Mark the element at index <code>4</code>, since it is already marked we skip it. Then we mark <code>2</code> of the smallest unmarked elements with the smallest indices if they exist, the marked elements now are <code>nums = [<strong><u>1</u></strong>,<u><strong>2</strong></u>,<u><strong>2</strong></u>,<u><strong>1</strong></u>,<u><strong>2</strong></u>,<strong><u>3</u></strong>,<u><strong>1</strong></u>]</code>. The sum of unmarked elements is <code>0</code>.</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\" style=\"border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;\">\n<p><strong>Input: </strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\">nums = [1,4,2,3], queries = [[0,1]]</span></p>\n\n<p><strong>Output: </strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\">[7]</span></p>\n\n<p><strong>Explanation: </strong> We do one query which is mark the element at index <code>0</code> and mark the smallest element among unmarked elements. The marked elements will be <code>nums = [<strong><u>1</u></strong>,4,<u><strong>2</strong></u>,3]</code>, and the sum of unmarked elements is <code>4 + 3 = 7</code>.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>m == queries.length</code></li>\n\t<li><code>1 &lt;= m &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>queries[i].length == 2</code></li>\n\t<li><code>0 &lt;= index<sub>i</sub>, k<sub>i</sub> &lt;= n - 1</code></li>\n</ul>\n",
            "translatedTitle": "执行操作标记数组中的元素",
            "translatedContent": "<p>给你一个长度为 <code>n</code>&nbsp;下标从 <strong>0</strong>&nbsp;开始的正整数数组&nbsp;<code>nums</code>&nbsp;。</p>\n\n<p>同时给你一个长度为 <code>m</code>&nbsp;的二维操作数组&nbsp;<code>queries</code>&nbsp;，其中&nbsp;<code>queries[i] = [index<sub>i</sub>, k<sub>i</sub>]</code>&nbsp;。</p>\n\n<p>一开始，数组中的所有元素都 <strong>未标记</strong>&nbsp;。</p>\n\n<p>你需要依次对数组执行 <code>m</code>&nbsp;次操作，第 <code>i</code>&nbsp;次操作中，你需要执行：</p>\n\n<ul>\n\t<li>如果下标&nbsp;<code>index<sub>i</sub></code>&nbsp;对应的元素还没标记，那么标记这个元素。</li>\n\t<li>然后标记&nbsp;<code>k<sub>i</sub></code>&nbsp;个数组中还没有标记的&nbsp;<strong>最小</strong>&nbsp;元素。如果有元素的值相等，那么优先标记它们中下标较小的。如果少于&nbsp;<code>k<sub>i</sub></code>&nbsp;个未标记元素存在，那么将它们全部标记。</li>\n</ul>\n\n<p>请你返回一个长度为 <code>m</code>&nbsp;的数组 <code>answer</code>&nbsp;，其中<em>&nbsp;</em><code>answer[i]</code>是第&nbsp;<code>i</code>&nbsp;次操作后数组中还没标记元素的&nbsp;<strong>和</strong>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\" style=\"border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;\">\n<p><strong>输入：</strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\">nums = [1,2,2,1,2,3,1], queries = [[1,2],[3,3],[4,2]]</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\">[8,3,0]</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>我们依次对数组做以下操作：</p>\n\n<ul>\n\t<li>标记下标为&nbsp;<code>1</code>&nbsp;的元素，同时标记&nbsp;<code>2</code>&nbsp;个未标记的最小元素。标记完后数组为&nbsp;<code>nums = [<em><strong>1</strong></em>,<em><strong>2</strong></em>,2,<em><strong>1</strong></em>,2,3,1]</code>&nbsp;。未标记元素的和为&nbsp;<code>2 + 2 + 3 + 1 = 8</code>&nbsp;。</li>\n\t<li>标记下标为&nbsp;<code>3</code>&nbsp;的元素，由于它已经被标记过了，所以我们忽略这次标记，同时标记最靠前的&nbsp;<code>3</code>&nbsp;个未标记的最小元素。标记完后数组为&nbsp;<code>nums = [<em><strong>1</strong></em>,<em><strong>2</strong></em>,<em><strong>2</strong></em>,<em><strong>1</strong></em>,<em><strong>2</strong></em>,3,<em><strong>1</strong></em>]</code>&nbsp;。未标记元素的和为&nbsp;<code>3</code>&nbsp;。</li>\n\t<li>标记下标为 <code>4</code>&nbsp;的元素，由于它已经被标记过了，所以我们忽略这次标记，同时标记最靠前的 <code>2</code>&nbsp;个未标记的最小元素。标记完后数组为&nbsp;<code>nums = [<em><strong>1</strong></em>,<em><strong>2</strong></em>,<em><strong>2</strong></em>,<em><strong>1</strong></em>,<em><strong>2</strong></em>,<em><strong>3</strong></em>,<em><strong>1</strong></em>]</code>&nbsp;。未标记元素的和为&nbsp;<code>0</code>&nbsp;。</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\" style=\"border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;\">\n<p><strong>输入：</strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\">nums = [1,4,2,3], queries = [[0,1]]</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\">[7]</span></p>\n\n<p><strong>解释：</strong>我们执行一次操作，将下标为&nbsp;<code>0</code>&nbsp;处的元素标记，并且标记最靠前的&nbsp;<code>1</code>&nbsp;个未标记的最小元素。标记完后数组为&nbsp;<code>nums = [<em><strong>1</strong></em>,4,<em><strong>2</strong></em>,3]</code>&nbsp;。未标记元素的和为&nbsp;<code>4 + 3 = 7</code>&nbsp;。</p>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>m == queries.length</code></li>\n\t<li><code>1 &lt;= m &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>queries[i].length == 2</code></li>\n\t<li><code>0 &lt;= index<sub>i</sub>, k<sub>i</sub> &lt;= n - 1</code></li>\n</ul>\n",
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