mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-09-05 23:41:41 +08:00
190 lines
24 KiB
JSON
190 lines
24 KiB
JSON
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"title": "Sorting Three Groups",
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"content": "<p>You are given an integer array <code>nums</code>. Each element in <code>nums</code> is 1, 2 or 3. In each operation, you can remove an element from <code>nums</code>. Return the <strong>minimum</strong> number of operations to make <code>nums</code> <strong>non-decreasing</strong>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [2,1,3,2,1]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">3</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>One of the optimal solutions is to remove <code>nums[0]</code>, <code>nums[2]</code> and <code>nums[3]</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [1,3,2,1,3,3]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">2</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>One of the optimal solutions is to remove <code>nums[1]</code> and <code>nums[2]</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [2,2,2,2,3,3]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">0</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p><code>nums</code> is already non-decreasing.</p>\n</div>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 100</code></li>\n\t<li><code>1 <= nums[i] <= 3</code></li>\n</ul>\n\n<p> </p>\n<strong>Follow-up:</strong> Can you come up with an algorithm that runs in <code>O(n)</code> time complexity?",
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"translatedContent": "<p>给你一个整数数组 <code>nums</code> 。<code>nums</code> 的每个元素是 1,2 或 3。在每次操作中,你可以删除 <code>nums</code> 中的一个元素。返回使 nums 成为 <strong>非递减</strong> 顺序所需操作数的 <strong>最小值</strong>。</p>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1:</strong></p>\n\n<pre>\n<b>输入:</b>nums = [2,1,3,2,1]\n<b>输出:</b>3\n<b>解释:</b>\n其中一个最优方案是删除 nums[0],nums[2] 和 nums[3]。\n</pre>\n\n<p><strong class=\"example\">示例 2:</strong></p>\n\n<pre>\n<b>输入:</b>nums = [1,3,2,1,3,3]\n<b>输出:</b>2\n<b>解释:</b>\n其中一个最优方案是删除 nums[1] 和 nums[2]。\n</pre>\n\n<p><strong class=\"example\">示例 3:</strong></p>\n\n<pre>\n<b>输入:</b>nums = [2,2,2,2,3,3]\n<b>输出:</b>0\n<b>解释:</b>\nnums 已是非递减顺序的。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 100</code></li>\n\t<li><code>1 <= nums[i] <= 3</code></li>\n</ul>\n\n<p><strong>进阶:</strong>你可以使用 <code>O(n)</code> 时间复杂度以内的算法解决吗?</p>\n",
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"The problem asks to change the array nums to make it sorted (i.e., all the 1s are on the left of 2s, and all the 2s are on the left of 3s.).",
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"We can try all the possibilities to make nums indices range in [0, i) to 0 and [i, j) to 1 and [j, n) to 2. Note the ranges are left-close and right-open; each might be empty. Namely, 0 <= i <= j <= n.",
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