# 3033. Modify the Matrix

## Description

Given a 0-indexed m x n integer matrix matrix, create a new 0-indexed matrix called answer. Make answer equal to matrix, then replace each element with the value -1 with the maximum element in its respective column.

Return the matrix answer.

Example 1:

Input: matrix = [[1,2,-1],[4,-1,6],[7,8,9]]
Output: [[1,2,9],[4,8,6],[7,8,9]]
Explanation: The diagram above shows the elements that are changed (in blue).
- We replace the value in the cell [1][1] with the maximum value in the column 1, that is 8.
- We replace the value in the cell [0][2] with the maximum value in the column 2, that is 9.


Example 2:

Input: matrix = [[3,-1],[5,2]]
Output: [[3,2],[5,2]]
Explanation: The diagram above shows the elements that are changed (in blue).


Constraints:

• m == matrix.length
• n == matrix[i].length
• 2 <= m, n <= 50
• -1 <= matrix[i][j] <= 100
• The input is generated such that each column contains at least one non-negative integer.

## Solutions

### Solution 1: Simulation

We can follow the problem description, traverse each column, find the maximum value of each column, and then traverse each column again, replacing the elements with a value of -1 with the maximum value of that column.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively. The space complexity is $O(1)$.

• class Solution {
public int[][] modifiedMatrix(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
for (int j = 0; j < n; ++j) {
int mx = -1;
for (int i = 0; i < m; ++i) {
mx = Math.max(mx, matrix[i][j]);
}
for (int i = 0; i < m; ++i) {
if (matrix[i][j] == -1) {
matrix[i][j] = mx;
}
}
}
return matrix;
}
}

• class Solution {
public:
vector<vector<int>> modifiedMatrix(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size();
for (int j = 0; j < n; ++j) {
int mx = -1;
for (int i = 0; i < m; ++i) {
mx = max(mx, matrix[i][j]);
}
for (int i = 0; i < m; ++i) {
if (matrix[i][j] == -1) {
matrix[i][j] = mx;
}
}
}
return matrix;
}
};

• class Solution:
def modifiedMatrix(self, matrix: List[List[int]]) -> List[List[int]]:
m, n = len(matrix), len(matrix[0])
for j in range(n):
mx = max(matrix[i][j] for i in range(m))
for i in range(m):
if matrix[i][j] == -1:
matrix[i][j] = mx
return matrix


• func modifiedMatrix(matrix [][]int) [][]int {
m, n := len(matrix), len(matrix[0])
for j := 0; j < n; j++ {
mx := -1
for i := 0; i < m; i++ {
mx = max(mx, matrix[i][j])
}
for i := 0; i < m; i++ {
if matrix[i][j] == -1 {
matrix[i][j] = mx
}
}
}
return matrix
}


• function modifiedMatrix(matrix: number[][]): number[][] {
const [m, n] = [matrix.length, matrix[0].length];
for (let j = 0; j < n; ++j) {
let mx = -1;
for (let i = 0; i < m; ++i) {
mx = Math.max(mx, matrix[i][j]);
}
for (let i = 0; i < m; ++i) {
if (matrix[i][j] === -1) {
matrix[i][j] = mx;
}
}
}
return matrix;
}


• public class Solution {
public int[][] ModifiedMatrix(int[][] matrix) {
int m = matrix.Length, n = matrix[0].Length;
for (int j = 0; j < n; ++j) {
int mx = -1;
for (int i = 0; i < m; ++i) {
mx = Math.Max(mx, matrix[i][j]);
}
for (int i = 0; i < m; ++i) {
if (matrix[i][j] == -1) {
matrix[i][j] = mx;
}
}
}
return matrix;
}
}