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1605. Find Valid Matrix Given Row and Column Sums

SourceBiweekly Contest 36 Q3DifficultyMediumRating1867

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: Greedy + Construction

Thinking

We must fill a nonnegative matrix whose row and column sums match the given arrays, with \(\sum \textit{rowSum} = \sum \textit{colSum}\). Searching cell by cell is impossible for \(m,n \le 500\).

Putting \(x=\min(\textit{rowSum}[i],\textit{colSum}[j])\) at \((i,j)\) and subtracting from both remainders leaves a smaller instance that is still consistent, so the greedy choice is safe.

A single row-major scan constructs a valid matrix without backtracking.

We can first initialize an \(m\) by \(n\) answer matrix \(\textit{ans}\).

Next, we traverse each position \((i, j)\) in the matrix, set the element at this position to \(x = \min(\textit{rowSum}[i], \textit{colSum}[j])\), and subtract \(x\) from \(\textit{rowSum}[i]\) and \(\textit{colSum}[j]\) respectively. After traversing all positions, we can get a matrix \(\textit{ans}\) that meets the requirements of the problem.

The correctness of the above strategy is explained as follows:

According to the requirements of the problem, we know that the sum of \(\textit{rowSum}\) and \(\textit{colSum}\) is equal, so \(\textit{rowSum}[0]\) must be less than or equal to \(\sum_{j = 0}^{n - 1} \textit{colSum}[j]\). Therefore, after \(n\) operations, \(\textit{rowSum}[0]\) can definitely be made \(0\), and for any \(j \in [0, n - 1]\), \(\textit{colSum}[j] \geq 0\) is guaranteed.

Therefore, we reduce the original problem to a subproblem with \(m-1\) rows and \(n\) columns, continue the above operations, until all elements in \(\textit{rowSum}\) and \(\textit{colSum}\) are \(0\), we can get a matrix \(\textit{ans}\) that meets the requirements of the problem.

The time complexity is \(O(mn)\), and the space complexity is \(O(mn)\), where \(m\) and \(n\) are the lengths of \(\textit{rowSum}\) and \(\textit{colSum}\) respectively.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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;
};

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