Problem 1605 Find Valid Matrix Given Row and Column Sums
Table of Contents
Problem Statement
Link - Problem 1605
Question
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] <= 10^8`
- `sum(rowSum) == sum(colSum)`
Solution
class Solution {
public:
vector<vector<int>> restoreMatrix(vector<int>& rowSum, vector<int>& colSum) {
int rows = rowSum.size(), cols = colSum.size();
vector<vector<int>>res(rows,vector<int>(cols,0));
for(int i = 0; i<rows; i++){
for(int j = 0; j<cols; j++){
res[i][j] = min(rowSum[i],colSum[j]);
rowSum[i]-=res[i][j];
colSum[j]-=res[i][j];
}
}
return res;
}
};
Complexity Analysis
| Algorithm | Time Complexity | Space Complexity |
| ---------------- | --------------- | ---------------- |
| Matrix Traversal | O(mn) | O(mn) |
Explanation
1. Intuition
- Seeing the problem statement it might seem its very hard.
- We can use greedy method to solve this question.
- Greedy works because of the following mathematical fact
> If given a m variable system of equations and n constraints, if m>n, then the system is underdetermined and has infinite solutions.
- We can use this fact to solve the problem.
- If we can fix the value of one cell, we can derive the values of other cells.
- First set the value of the cell to be the minimum of the rowSum and colSum.
- Then subtract the value from the rowSum and colSum.
- Continue this process until all the cells are filled.
2. Implementation
- Initialize a result matrix `res` of size `rows x cols` with all values as 0.
- Traverse the matrix and set the value of the cell to be the minimum of the rowSum and colSum.
- Subtract the value from the rowSum and colSum.
- Continue this process until all the cells are filled.