1605. Find Valid Matrix Given Row and Column Sums

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)

Solutions

Solution 1

Python Code
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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

Java Code
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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;
    }
}

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

Go Code
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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
}

TypeScript Code
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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;
}

JavaScript Code
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/**
 * @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;
};