1605

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:

• Reconstruct a 2-Row Binary Matrix (Medium)

Solution 1. Greedy

Fill row by row and column by column. For each cell (i, j), we fill ans[i][j] = min(rowSum[i], colSum[j]) and deduct ans[i][j] from rowSum[i] and colSum[j].

// OJ: https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/
// Author: github.com/lzl124631x
// 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;
    }
};

Solution 2. Greedy

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/
// Author: github.com/lzl124631x
// Time: O(MN) for initialization, O(M + N) for processing
// Space: O(1) extra space
// 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;
    }
};