Leetcode 1605 find valid matrix given row and Column Sums

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 = 3 == 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]]

Constraints:

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

Topic link ：https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/

The main idea of the topic ： Tell you the sum of elements in each row and column of a matrix , Find an original matrix that satisfies the condition , Make sure there's a solution

Topic analysis ： Inspired by the previous question （https://leetcode.com/problems/two-city-scheduling/）, Also wrote Answer key , But it's not the best practice , The best thing to do is to let everyone go to one city first and then calculate the cost of moving someone to another city . Back to the subject , It's a little hard to start by considering the ranks at the same time , You might as well fill in the line first , That is, fill in all the values of row sum in the first column , Then the answer is constructed by modifying the column sum , From the top down , Enumerate from left to right , If the prefix sum of the row in the front row has exceeded the expected sum of the column , Then subtract the accumulated part from the row value , The rest goes on to the right column

8ms, Time beats 63.5%

class Solution {
    public int[][] restoreMatrix(int[] rowSum, int[] colSum) {
        int n = rowSum.length;
        int m = colSum.length;
        int[][] ans = new int[n][m];

        for (int j = 0; j < m; j++) {
            int sum = 0;
            for (int i = 0; i < n; i++) {
                if (sum + rowSum[i] >= colSum[j]) {
                    ans[i][j] = colSum[j] - sum;
                    rowSum[i] -= ans[i][j];
                    break;
                }
                sum += rowSum[i];
                ans[i][j] = rowSum[i];
                rowSum[i] = 0;
            }
        }
        return ans;
    }
}