Greedy Algorithm to 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 = 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)

Greedy Algorithm to Construct Valid Matrix Given Row and Column Sums

Although you can do a Backtracking algorithm to find such valid matrix, the most efficient algorithm is greedy in this case. We know that sum(colSums) = sum(rowSums) and we just need to greediy fill the element of the matrix by the minimal value of its rowSum and colSum and update the sum values accordingly.

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

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

Both time and space complexity is O(MN) where M is the number of rows and N is the number of the columns for the matrix.

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