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Formatted question description: https://leetcode.ca/all/1605.html

# 1605. Find Valid Matrix Given Row and Column Sums (Medium)

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)

Related Topics:
Greedy

Similar Questions:

## Solution 1.

// OJ: https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/
// Time: O(MN)
// Space: O(1)
class Solution {
public:
vector<vector<int>> restoreMatrix(vector<int>& R, vector<int>& C) {
int M = R.size(), N = C.size(), d;
vector<vector<int>> ans(M, vector<int>(N));
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
d = ans[i][j] = min(R[i], C[j]);
R[i] -= d;
C[j] -= d;
}
}
return ans;
}
};


## Solution 2.

We just need to go from the top-left to the bottom-right. Once R[i] or C[j] becomes 0, the rest of the elements in row i or column j must be 0s, so we can increment i or j.

// OJ: https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/
// Time: O(MN) for initialization, O(M + N) for processing
// Space: O(MN)
// Ref: https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/876845/JavaC%2B%2BPython-Easy-and-Concise-with-Prove
class Solution {
public:
vector<vector<int>> restoreMatrix(vector<int>& R, vector<int>& C) {
int M = R.size(), N = C.size(), i = 0, j = 0, d;
vector<vector<int>> ans(M, vector<int>(N));
while (i < M && j < N) {
d = ans[i][j] = min(R[i], C[j]);
if ((R[i] -= d) == 0) ++i;
if ((C[j] -= d) == 0) ++j;
}
return ans;
}
};

• class Solution {
public int[][] restoreMatrix(int[] rowSum, int[] colSum) {
int rows = rowSum.length, columns = colSum.length;
int[][] matrix = new int[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
int num = Math.min(rowSum[i], colSum[j]);
matrix[i][j] = num;
rowSum[i] -= num;
colSum[j] -= num;
}
}
return matrix;
}
}

############

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;
}
}

• // OJ: https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/
// Time: O(MN)
// Space: O(1) extra space
class Solution {
public:
vector<vector<int>> restoreMatrix(vector<int>& R, vector<int>& C) {
int M = R.size(), N = C.size(), d;
vector<vector<int>> ans(M, vector<int>(N));
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
d = ans[i][j] = min(R[i], C[j]);
R[i] -= d;
C[j] -= d;
}
}
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

############

# 1605. Find Valid Matrix Given Row and Column Sums
# https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/

class Solution:
def restoreMatrix(self, rs: List[int], cs: List[int]) -> List[List[int]]:
r = len(rs)
c = len(cs)

lst = [[0]*c for _ in range(r)]

for i in range(r):
for j in range(c):

cur = min(rs[i], cs[j])

lst[i][j] = cur
rs[i] -= cur
cs[j] -= cur

return lst


• 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
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

• 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;
};