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{
"data": {
"question": {
"questionId": "3470",
"questionFrontendId": "3225",
"categoryTitle": "Algorithms",
"boundTopicId": 2849095,
"title": "Maximum Score From Grid Operations",
"titleSlug": "maximum-score-from-grid-operations",
"content": "<p>You are given a 2D matrix <code>grid</code> of size <code>n x n</code>. Initially, all cells of the grid are colored white. In one operation, you can select any cell of indices <code>(i, j)</code>, and color black all the cells of the <code>j<sup>th</sup></code> column starting from the top row down to the <code>i<sup>th</sup></code> row.</p>\n\n<p>The grid score is the sum of all <code>grid[i][j]</code> such that cell <code>(i, j)</code> is white and it has a horizontally adjacent black cell.</p>\n\n<p>Return the <strong>maximum</strong> score that can be achieved after some number of operations.</p>\n\n<p>&nbsp;</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\">grid = [[0,0,0,0,0],[0,0,3,0,0],[0,1,0,0,0],[5,0,0,3,0],[0,0,0,0,2]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">11</span></p>\n\n<p><strong>Explanation:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/05/11/one.png\" style=\"width: 300px; height: 200px;\" />\n<p>In the first operation, we color all cells in column 1 down to row 3, and in the second operation, we color all cells in column 4 down to the last row. The score of the resulting grid is <code>grid[3][0] + grid[1][2] + grid[3][3]</code> which is equal to 11.</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\">grid = [[10,9,0,0,15],[7,1,0,8,0],[5,20,0,11,0],[0,0,0,1,2],[8,12,1,10,3]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">94</span></p>\n\n<p><strong>Explanation:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/05/11/two-1.png\" style=\"width: 300px; height: 200px;\" />\n<p>We perform operations on 1, 2, and 3 down to rows 1, 4, and 0, respectively. The score of the resulting grid is <code>grid[0][0] + grid[1][0] + grid[2][1] + grid[4][1] + grid[1][3] + grid[2][3] + grid[3][3] + grid[4][3] + grid[0][4]</code> which is equal to 94.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;=&nbsp;n == grid.length &lt;= 100</code></li>\n\t<li><code>n == grid[i].length</code></li>\n\t<li><code>0 &lt;= grid[i][j] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
"translatedTitle": "网格图操作后的最大分数",
"translatedContent": "<p>给你一个大小为 <code>n x n</code>&nbsp;的二维矩阵&nbsp;<code>grid</code>&nbsp;,一开始所有格子都是白色的。一次操作中,你可以选择任意下标为&nbsp;<code>(i, j)</code>&nbsp;的格子,并将第&nbsp;<code>j</code>&nbsp;列中从最上面到第&nbsp;<code>i</code>&nbsp;行所有格子改成黑色。</p>\n\n<p>如果格子 <code>(i, j)</code>&nbsp;为白色,且左边或者右边的格子至少一个格子为黑色,那么我们将 <code>grid[i][j]</code>&nbsp;加到最后网格图的总分中去。</p>\n\n<p>请你返回执行任意次操作以后,最终网格图的 <strong>最大</strong>&nbsp;总分数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入:</b>grid = [[0,0,0,0,0],[0,0,3,0,0],[0,1,0,0,0],[5,0,0,3,0],[0,0,0,0,2]]</span></p>\n\n<p><span class=\"example-io\"><b>输出:</b>11</span></p>\n\n<p><strong>解释:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/05/11/one.png\" style=\"width: 300px; height: 200px;\" />\n<p>第一次操作中,我们将第 1 列中,最上面的格子到第 3 行的格子染成黑色。第二次操作中,我们将第 4 列中,最上面的格子到最后一行的格子染成黑色。最后网格图总分为&nbsp;<code>grid[3][0] + grid[1][2] + grid[3][3]</code>&nbsp;等于 11 。</p>\n</div>\n\n<p><strong class=\"example\">示例 2</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入:</b>grid = [[10,9,0,0,15],[7,1,0,8,0],[5,20,0,11,0],[0,0,0,1,2],[8,12,1,10,3]]</span></p>\n\n<p><span class=\"example-io\"><b>输出:</b>94</span></p>\n\n<p><strong>解释:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/05/11/two-1.png\" style=\"width: 300px; height: 200px;\" />\n<p>我们对第 1 2 3 列分别从上往下染黑色到第 1 4 0 行。最后网格图总分为&nbsp;<code>grid[0][0] + grid[1][0] + grid[2][1] + grid[4][1] + grid[1][3] + grid[2][3] + grid[3][3] + grid[4][3] + grid[0][4]</code>&nbsp;等于 94 。</p>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;=&nbsp;n == grid.length &lt;= 100</code></li>\n\t<li><code>n == grid[i].length</code></li>\n\t<li><code>0 &lt;= grid[i][j] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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"Use dynamic programming.",
"Solve the problem in O(N^4) using a 3-states dp.",
"Let <code>dp[i][lastHeight][beforeLastHeight]</code> denote the maximum score if the grid was limited to column <code>i</code>, and the height of column <code>i - 1</code> is <code>lastHeight</code> and the height of column <code>i - 2</code> is <code>beforeLastHeight</code>.",
"The third state, <code>beforeLastHeight</code>, is used to determine which values of column <code>i - 1</code> will be added to the score. We can replace this state with another state that only takes two values 0 or 1.",
"Let <code>dp[i][lastHeight][isBigger]</code> denote the maximum score if the grid was limited to column <code>i</code>, and where the height of column <code>i - 1</code> is <code>lastHeight</code>. Additionally, if <code>isBigger == 1</code>, the number of black cells in column <code>i</code> is assumed to be larger than the number of black cells in column <code>i - 2</code>, and vice versa. Note that if our assumption is wrong, it would lead to a suboptimal score and, therefore, it would not be considered as the final answer."
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