# 1605. Find Valid Matrix Given Row and Column Sums

## Description

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)

## Solutions

• class Solution {
public int[][] restoreMatrix(int[] rowSum, int[] colSum) {
int m = rowSum.length;
int n = colSum.length;
int[][] ans = new int[m][n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int x = Math.min(rowSum[i], colSum[j]);
ans[i][j] = x;
rowSum[i] -= x;
colSum[j] -= x;
}
}
return ans;
}
}

• class Solution {
public:
vector<vector<int>> restoreMatrix(vector<int>& rowSum, vector<int>& colSum) {
int m = rowSum.size(), n = colSum.size();
vector<vector<int>> ans(m, vector<int>(n));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int x = min(rowSum[i], colSum[j]);
ans[i][j] = x;
rowSum[i] -= x;
colSum[j] -= x;
}
}
return ans;
}
};

• class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
m, n = len(rowSum), len(colSum)
ans = [[0] * n for _ in range(m)]
for i in range(m):
for j in range(n):
x = min(rowSum[i], colSum[j])
ans[i][j] = x
rowSum[i] -= x
colSum[j] -= x
return ans


• func restoreMatrix(rowSum []int, colSum []int) [][]int {
m, n := len(rowSum), len(colSum)
ans := make([][]int, m)
for i := range ans {
ans[i] = make([]int, n)
}
for i := range rowSum {
for j := range colSum {
x := min(rowSum[i], colSum[j])
ans[i][j] = x
rowSum[i] -= x
colSum[j] -= x
}
}
return ans
}

• function restoreMatrix(rowSum: number[], colSum: number[]): number[][] {
const m = rowSum.length;
const n = colSum.length;
const ans = Array.from(new Array(m), () => new Array(n).fill(0));
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
const x = Math.min(rowSum[i], colSum[j]);
ans[i][j] = x;
rowSum[i] -= x;
colSum[j] -= x;
}
}
return ans;
}


• /**
* @param {number[]} rowSum
* @param {number[]} colSum
* @return {number[][]}
*/
var restoreMatrix = function (rowSum, colSum) {
const m = rowSum.length;
const n = colSum.length;
const ans = Array.from(new Array(m), () => new Array(n).fill(0));
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
const x = Math.min(rowSum[i], colSum[j]);
ans[i][j] = x;
rowSum[i] -= x;
colSum[j] -= x;
}
}
return ans;
};