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+---
+id: modify-the-matrix
+title: Modify the Matrix (LeetCode)
+sidebar_label: 3033-ModifyTheMatrix
+tags:
+  - Matrix
+  - Array
+description: Modify a matrix by replacing each element with the value -1 with the maximum element in its respective column.
+sidebar_position: 3033
+---
+
+## Problem Description
+
+| Problem Statement | Solution Link | LeetCode Profile |
+| :---------------- | :------------ | :--------------- |
+| [Modify the Matrix](https://leetcode.com/problems/modify-the-matrix/) | [Modify the Matrix Solution on LeetCode](https://leetcode.com/problems/modify-the-matrix/solutions/) | [vaishu_1904](https://leetcode.com/u/vaishu_1904/) |
+
+## Problem 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.
+
+## Approach
+
+To solve this problem, we need to replace each occurrence of `-1` in the matrix with the maximum value in its respective column. Here are the steps:
+
+1. Traverse each column of the matrix to find the maximum value.
+2. Traverse the matrix again to replace each `-1` with the corresponding column maximum value.
+
+### Solution Code
+
+#### Python
+
+```python
+class Solution:
+    def modifiedMatrix(self, matrix: List[List[int]]) -> List[List[int]]:
+        m, n = len(matrix), len(matrix[0])
+        col_max = [max(matrix[i][j] for i in range(m) if matrix[i][j] != -1) for j in range(n)]
+        
+        for i in range(m):
+            for j in range(n):
+                if matrix[i][j] == -1:
+                    matrix[i][j] = col_max[j]
+        
+        return matrix
+```
+
+#### C++
+```c++
+#include <vector>
+#include <algorithm>
+using namespace std;
+
+class Solution {
+public:
+    vector<vector<int>> modifiedMatrix(vector<vector<int>>& matrix) {
+        int m = matrix.size(), n = matrix[0].size();
+        vector<int> colMax(n, INT_MIN);
+        
+        for (int j = 0; j < n; ++j) {
+            for (int i = 0; i < m; ++i) {
+                if (matrix[i][j] != -1) {
+                    colMax[j] = max(colMax[j], matrix[i][j]);
+                }
+            }
+        }
+        
+        for (int i = 0; i < m; ++i) {
+            for (int j = 0; j < n; ++j) {
+                if (matrix[i][j] == -1) {
+                    matrix[i][j] = colMax[j];
+                }
+            }
+        }
+        
+        return matrix;
+    }
+};
+
+```
+
+#### Java
+```java
+class Solution {
+    public int[][] modifiedMatrix(int[][] matrix) {
+        int m = matrix.length, n = matrix[0].length;
+        int[] colMax = new int[n];
+        Arrays.fill(colMax, Integer.MIN_VALUE);
+        
+        for (int j = 0; j < n; ++j) {
+            for (int i = 0; i < m; ++i) {
+                if (matrix[i][j] != -1) {
+                    colMax[j] = Math.max(colMax[j], matrix[i][j]);
+                }
+            }
+        }
+        
+        for (int i = 0; i < m; ++i) {
+            for (int j = 0; j < n; ++j) {
+                if (matrix[i][j] == -1) {
+                    matrix[i][j] = colMax[j];
+                }
+            }
+        }
+        
+        return matrix;
+    }
+}
+
+```
+
+### Conclusion
+The solutions traverse the matrix twice to efficiently find the maximum values for each column and 
+then replace the -1 values. This approach ensures that the problem is solved in a straightforward 
+and efficient manner across different programming languages.
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