leetcode-problemset/leetcode-cn/originData/paint-house-iii.json

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        "question": {
            "questionId": "1583",
            "questionFrontendId": "1473",
            "categoryTitle": "Algorithms",
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            "title": "Paint House III",
            "titleSlug": "paint-house-iii",
            "content": "<p>There is a row of <code>m</code> houses in a small city, each house must be painted with one of the <code>n</code> colors (labeled from <code>1</code> to <code>n</code>), some houses that have been painted last summer should not be painted again.</p>\n\n<p>A neighborhood is a maximal group of continuous houses that are painted with the same color.</p>\n\n<ul>\n\t<li>For example: <code>houses = [1,2,2,3,3,2,1,1]</code> contains <code>5</code> neighborhoods <code>[{1}, {2,2}, {3,3}, {2}, {1,1}]</code>.</li>\n</ul>\n\n<p>Given an array <code>houses</code>, an <code>m x n</code> matrix <code>cost</code> and an integer <code>target</code> where:</p>\n\n<ul>\n\t<li><code>houses[i]</code>: is the color of the house <code>i</code>, and <code>0</code> if the house is not painted yet.</li>\n\t<li><code>cost[i][j]</code>: is the cost of paint the house <code>i</code> with the color <code>j + 1</code>.</li>\n</ul>\n\n<p>Return <em>the minimum cost of painting all the remaining houses in such a way that there are exactly</em> <code>target</code> <em>neighborhoods</em>. If it is not possible, return <code>-1</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3\n<strong>Output:</strong> 9\n<strong>Explanation:</strong> Paint houses of this way [1,2,2,1,1]\nThis array contains target = 3 neighborhoods, [{1}, {2,2}, {1,1}].\nCost of paint all houses (1 + 1 + 1 + 1 + 5) = 9.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3\n<strong>Output:</strong> 11\n<strong>Explanation:</strong> Some houses are already painted, Paint the houses of this way [2,2,1,2,2]\nThis array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}]. \nCost of paint the first and last house (10 + 1) = 11.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> Houses are already painted with a total of 4 neighborhoods [{3},{1},{2},{3}] different of target = 3.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == houses.length == cost.length</code></li>\n\t<li><code>n == cost[i].length</code></li>\n\t<li><code>1 &lt;= m &lt;= 100</code></li>\n\t<li><code>1 &lt;= n &lt;= 20</code></li>\n\t<li><code>1 &lt;= target &lt;= m</code></li>\n\t<li><code>0 &lt;= houses[i] &lt;= n</code></li>\n\t<li><code>1 &lt;= cost[i][j] &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "粉刷房子 III",
            "translatedContent": "<p>在一个小城市里，有 <code>m</code> 个房子排成一排，你需要给每个房子涂上 <code>n</code> 种颜色之一（颜色编号为 <code>1</code> 到 <code>n</code> ）。有的房子去年夏天已经涂过颜色了，所以这些房子不可以被重新涂色。</p>\n\n<p>我们将连续相同颜色尽可能多的房子称为一个街区。（比方说 <code>houses = [1,2,2,3,3,2,1,1]</code> ，它包含 5 个街区 <code> [{1}, {2,2}, {3,3}, {2}, {1,1}]</code> 。）</p>\n\n<p>给你一个数组 <code>houses</code> ，一个 <code>m * n</code> 的矩阵 <code>cost</code> 和一个整数 <code>target</code> ，其中：</p>\n\n<ul>\n\t<li><code>houses[i]</code>：是第 <code>i</code> 个房子的颜色，<strong>0</strong> 表示这个房子还没有被涂色。</li>\n\t<li><code>cost[i][j]</code>：是将第 <code>i</code> 个房子涂成颜色 <code>j+1</code> 的花费。</li>\n</ul>\n\n<p>请你返回房子涂色方案的最小总花费，使得每个房子都被涂色后，恰好组成 <code>target</code> 个街区。如果没有可用的涂色方案，请返回 <strong>-1</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3\n<strong>输出：</strong>9\n<strong>解释：</strong>房子涂色方案为 [1,2,2,1,1]\n此方案包含 target = 3 个街区，分别是 [{1}, {2,2}, {1,1}]。\n涂色的总花费为 (1 + 1 + 1 + 1 + 5) = 9。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3\n<strong>输出：</strong>11\n<strong>解释：</strong>有的房子已经被涂色了，在此基础上涂色方案为 [2,2,1,2,2]\n此方案包含 target = 3 个街区，分别是 [{2,2}, {1}, {2,2}]。\n给第一个和最后一个房子涂色的花费为 (10 + 1) = 11。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>houses = [0,0,0,0,0], cost = [[1,10],[10,1],[1,10],[10,1],[1,10]], m = 5, n = 2, target = 5\n<strong>输出：</strong>5\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<strong>输入：</strong>houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3\n<strong>输出：</strong>-1\n<strong>解释：</strong>房子已经被涂色并组成了 4 个街区，分别是 [{3},{1},{2},{3}] ，无法形成 target = 3 个街区。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>m == houses.length == cost.length</code></li>\n\t<li><code>n == cost[i].length</code></li>\n\t<li><code>1 <= m <= 100</code></li>\n\t<li><code>1 <= n <= 20</code></li>\n\t<li><code>1 <= target <= m</code></li>\n\t<li><code>0 <= houses[i] <= n</code></li>\n\t<li><code>1 <= cost[i][j] <= 10^4</code></li>\n</ul>\n",
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href=\\\"https:\\/\\/github.com\\/emirpasic\\/gods\\/tree\\/v1.18.1\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.10<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap 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