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 <= 5000 <= rowSum[i], colSum[i] <= 108sum(rowSum) == sum(colSum)
Solutions
Solution 1: Greedy + Construction
We can first initialize an $m$ by $n$ answer matrix $ans$.
Next, we traverse each position $(i, j)$ in the matrix, set the element at this position to $x = \min(rowSum[i], colSum[j])$, and subtract $x$ from $rowSum[i]$ and $colSum[j]$ respectively. After traversing all positions, we can get a matrix $ans$ 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 $rowSum$ and $colSum$ is equal, so $rowSum[0]$ must be less than or equal to $\sum_{j = 0}^{n - 1} colSum[j]$. Therefore, after $n$ operations, $rowSum[0]$ can definitely be made $0$, and for any $j \in [0, n - 1]$, $colSum[j] \geq 0$ 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 $rowSum$ and $colSum$ are $0$, we can get a matrix $ans$ that meets the requirements of the problem.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the lengths of $rowSum$ and $colSum$ respectively.
Python3
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
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++
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
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
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
/**
* @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;
};