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(rowSum) == sum(colSum)

Approach 01

#include <bits/stdc++.h>

using namespace std;

class Solution {
public:
    vector<vector<int>> restoreMatrix(vector<int>& rowSums, vector<int>& colSums) {
        const int numRows = rowSums.size();
        const int numCols = colSums.size();
        vector<vector<int>> resultMatrix(numRows, vector<int>(numCols));

        for (int row = 0; row < numRows; ++row)
            for (int col = 0; col < numCols; ++col) {
                resultMatrix[row][col] = min(rowSums[row], colSums[col]);
                rowSums[row] -= resultMatrix[row][col];
                colSums[col] -= resultMatrix[row][col];
            }

        return resultMatrix;
    }
};

from typing import List

class Solution:
    def restoreMatrix(self, rowSums: List[int], colSums: List[int]) -> List[List[int]]:
        numRows = len(rowSums)
        numCols = len(colSums)
        resultMatrix = [[0] * numCols for _ in range(numRows)]

        for row in range(numRows):
            for col in range(numCols):
                resultMatrix[row][col] = min(rowSums[row], colSums[col])
                rowSums[row] -= resultMatrix[row][col]
                colSums[col] -= resultMatrix[row][col]

        return resultMatrix

Time Complexity

• Initialization:

  Creating the resultMatrix takes \( O(m \cdot n) \) time, where \( m \) is the number of rows and \( n \) is the number of columns.

• Main Loop:

  The nested loops iterate over each cell of the matrix once, which takes \( O(m \cdot n) \) time.

• Overall Time Complexity:

  The overall time complexity is \( O(m \cdot n) \).

Space Complexity

• Auxiliary Space:

  The algorithm uses additional space for the resultMatrix, which stores \( m \times n \) elements, resulting in \( O(m \cdot n) \) space usage.

• Overall Space Complexity:

  The overall space complexity is \( O(m \cdot n) \).