[LeetCode] 1605. Find Valid Matrix Given Row and Column Sums

2024-04-01 12:59:34

You are given two arrays rowSum and colSum of non-negative integers where rowSum[i] is the sum of the elements in the ith row and colSum[j] is the sum of the elements of the jth column of a 2D matrix. In other words, you do not know the elements of the matrix, but you do know the sums of each row and column.

Find any matrix of non-negative integers of size rowSum.length x colSum.length that satisfies the rowSum and colSum requirements.

Return a 2D array representing any matrix that fulfills the requirements. It's guaranteed that at least one matrix that fulfills the requirements exists.

Example 1:

Input: rowSum = [3,8], colSum = [4,7]
Output: [[3,0],
         [1,7]]
Explanation:
0th row: 3 + 0 = 0 == rowSum[0]
1st row: 1 + 7 = 8 == rowSum[1]
0th column: 3 + 1 = 4 == colSum[0]
1st column: 0 + 7 = 7 == colSum[1]
The row and column sums match, and all matrix elements are non-negative.
Another possible matrix is: [[1,2],
                             [3,5]]

Example 2:

Input: rowSum = [5,7,10], colSum = [8,6,8]
Output: [[0,5,0],
         [6,1,0],
         [2,0,8]]

Example 3:

Input: rowSum = [14,9], colSum = [6,9,8]
Output: [[0,9,5],
         [6,0,3]]

Example 4:

Input: rowSum = [1,0], colSum = [1]
Output: [[1],
         [0]]

Example 5:

Input: rowSum = [0], colSum = [0]
Output: [[0]]

Constraints:

1 <= rowSum.length, colSum.length <= 500
0 <= rowSum[i], colSum[i] <= 108
sum(rows) == sum(columns)

给定行和列的和求可行矩阵。

给你两个非负整数数组 rowSum 和 colSum ，其中 rowSum[i] 是二维矩阵中第 i 行元素的和， colSum[j] 是第 j 列元素的和。换言之你不知道矩阵里的每个元素，但是你知道每一行和每一列的和。

请找到大小为 rowSum.length x colSum.length 的任意非负整数矩阵，且该矩阵满足 rowSum 和 colSum 的要求。

请你返回任意一个满足题目要求的二维矩阵，题目保证存在至少一个可行矩阵。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-valid-matrix-given-row-and-column-sums
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是贪心。这道题跟1253题差不多但是个人感觉稍微难一点，尤其是证明贪心的正确性会是个难点。

我们首先创建一个二维数组 int[][] res 记录最后的结果，res的尺寸是 [rowSum.length][colSum.length]。接着我们遍历这个二维数组，对于每一个位置上的元素，我们写入 rowSum[i] 和 colSum[j] 的较小值，同时需要在 rowSum[i] 和 colSum[j]里面减去这个较小值。

证明：对于每一个位置 res[i][j]，我们一开始写入 Math.min(rowSum[i], colSum[j]) 应该没有什么疑问，这样操作之后，会使得 rowSum[i], colSum[j] 之中有一个变成0。如此操作直到最后一个位置上的时候，理应让 sum(rows) 和 sum(columns) 都变成0。

时间O(m + n)

空间O(mn)

Java实现

 1 class Solution {
 2     public int[][] restoreMatrix(int[] rowSum, int[] colSum) {
 3         int m = rowSum.length;
 4         int n = colSum.length;
 5         int[][] res = new int[m][n];
 6         for (int i = 0; i < m; i++) {
 7             for (int j = 0; j < n; j++) {
 8                 res[i][j] = Math.min(rowSum[i], colSum[j]);
 9                 rowSum[i] -= res[i][j];
10                 colSum[j] -= res[i][j];
11             }
12         }
13         return res;
14     }
15 }

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