1605. Find Valid Matrix Given Row and Column Sums

Problem Link

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)

Solution

---

C++

class Solution {
public:
    vector<vector<int>> restoreMatrix(vector<int>& rowSum, vector<int>& colSum) {
        int rows = rowSum.size(), cols = colSum.size();
        vector<vector<int>> res(rows, vector<int>(cols, 0));
 
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                int currMin = min(rowSum[i], colSum[j]);
                res[i][j] = currMin;
                rowSum[i] -= currMin;
                colSum[j] -= currMin;
            }
        }
 
        return res;
    }
};

Python3

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