leetcode-problemset/leetcode-cn/originData/reconstruct-a-2-row-binary-matrix.json

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        "question": {
            "questionId": "1379",
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            "categoryTitle": "Algorithms",
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            "title": "Reconstruct a 2-Row Binary Matrix",
            "titleSlug": "reconstruct-a-2-row-binary-matrix",
            "content": "<p>Given the following details of a matrix with <code>n</code> columns and <code>2</code> rows :</p>\n\n<ul>\n\t<li>The matrix is a binary matrix, which means each element in the matrix can be <code>0</code> or <code>1</code>.</li>\n\t<li>The sum of elements of the 0-th(upper) row is given as <code>upper</code>.</li>\n\t<li>The sum of elements of the 1-st(lower) row is given as <code>lower</code>.</li>\n\t<li>The sum of elements in the i-th column(0-indexed) is <code>colsum[i]</code>, where <code>colsum</code> is given as an integer array with length <code>n</code>.</li>\n</ul>\n\n<p>Your task is to reconstruct the matrix with <code>upper</code>, <code>lower</code> and <code>colsum</code>.</p>\n\n<p>Return it as a 2-D integer array.</p>\n\n<p>If there are more than one valid solution, any of them will be accepted.</p>\n\n<p>If no valid solution exists, return an empty 2-D array.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> upper = 2, lower = 1, colsum = [1,1,1]\n<strong>Output:</strong> [[1,1,0],[0,0,1]]\n<strong>Explanation: </strong>[[1,0,1],[0,1,0]], and [[0,1,1],[1,0,0]] are also correct answers.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> upper = 2, lower = 3, colsum = [2,2,1,1]\n<strong>Output:</strong> []\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> upper = 5, lower = 5, colsum = [2,1,2,0,1,0,1,2,0,1]\n<strong>Output:</strong> [[1,1,1,0,1,0,0,1,0,0],[1,0,1,0,0,0,1,1,0,1]]\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= colsum.length &lt;= 10^5</code></li>\n\t<li><code>0 &lt;= upper, lower &lt;= colsum.length</code></li>\n\t<li><code>0 &lt;= colsum[i] &lt;= 2</code></li>\n</ul>\n",
            "translatedTitle": "重构 2 行二进制矩阵",
            "translatedContent": "<p>给你一个&nbsp;<code>2</code>&nbsp;行 <code>n</code> 列的二进制数组：</p>\n\n<ul>\n\t<li>矩阵是一个二进制矩阵，这意味着矩阵中的每个元素不是&nbsp;<code>0</code>&nbsp;就是&nbsp;<code>1</code>。</li>\n\t<li>第 <code>0</code> 行的元素之和为&nbsp;<code>upper</code>。</li>\n\t<li>第 <code>1</code> 行的元素之和为 <code>lower</code>。</li>\n\t<li>第 <code>i</code> 列（从 <code>0</code> 开始编号）的元素之和为&nbsp;<code>colsum[i]</code>，<code>colsum</code>&nbsp;是一个长度为&nbsp;<code>n</code>&nbsp;的整数数组。</li>\n</ul>\n\n<p>你需要利用&nbsp;<code>upper</code>，<code>lower</code>&nbsp;和&nbsp;<code>colsum</code>&nbsp;来重构这个矩阵，并以二维整数数组的形式返回它。</p>\n\n<p>如果有多个不同的答案，那么任意一个都可以通过本题。</p>\n\n<p>如果不存在符合要求的答案，就请返回一个空的二维数组。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>upper = 2, lower = 1, colsum = [1,1,1]\n<strong>输出：</strong>[[1,1,0],[0,0,1]]\n<strong>解释：</strong>[[1,0,1],[0,1,0]] 和 [[0,1,1],[1,0,0]] 也是正确答案。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>upper = 2, lower = 3, colsum = [2,2,1,1]\n<strong>输出：</strong>[]\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>upper = 5, lower = 5, colsum = [2,1,2,0,1,0,1,2,0,1]\n<strong>输出：</strong>[[1,1,1,0,1,0,0,1,0,0],[1,0,1,0,0,0,1,1,0,1]]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= colsum.length &lt;= 10^5</code></li>\n\t<li><code>0 &lt;= upper, lower &lt;= colsum.length</code></li>\n\t<li><code>0 &lt;= colsum[i] &lt;= 2</code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    vector<vector<int>> reconstructMatrix(int upper, int lower, vector<int>& colsum) {\n\n    }\n};",
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                    "code": "class Solution(object):\n    def reconstructMatrix(self, upper, lower, colsum):\n        \"\"\"\n        :type upper: int\n        :type lower: int\n        :type colsum: List[int]\n        :rtype: List[List[int]]\n        \"\"\"",
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                "You cannot do anything about colsum[i] = 2 case or colsum[i] = 0 case. Then you put colsum[i] = 1 case to the upper row until upper has reached. Then put the rest into lower row.",
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