1605. Find Valid Matrix Given Row and Column Sums

Problem

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(rowSum) == sum(colSum)

Solution

/**
 * @param {number[]} rowSum
 * @param {number[]} colSum
 * @return {number[][]}
 */
var restoreMatrix = function(rowSum, colSum) {
    var m = rowSum.length;
    var n = colSum.length;
    var res = Array(m).fill(0).map(() => Array(n).fill(0));
    for (var i = 0; i < m; i++) {
        res[i][0] = rowSum[i];
    }
    for (var i = 0; i < n - 1; i++) {
        for (var j = 0; j < m; j++) {
            var num = Math.min(res[j][i], colSum[i]);
            res[j][i + 1] = res[j][i] - num;
            res[j][i] = num;
            colSum[i] -= num;
        }
    }
    return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n * m).
• Space complexity : O(n * m).