3033. Modify the Matrix LeetCode Solution

In this guide, you will get 3033. Modify the Matrix LeetCode Solution with the best time and space complexity. The solution to Modify the Matrix problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Modify the Matrix solution in C++
Modify the Matrix solution in Java
Modify the Matrix solution in Python
Additional Resources

3033. Modify the Matrix LeetCode Solution image

Problem Statement of Modify the Matrix

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.

Complexity Analysis

Time Complexity: O(mn)
Space Complexity: O(mn)

3033. Modify the Matrix LeetCode Solution in C++

class Solution {
 public:
  vector<vector<int>> modifiedMatrix(vector<vector<int>>& matrix) {
    const int m = matrix.size();
    const int n = matrix[0].size();
    vector<vector<int>> ans = matrix;

    for (int j = 0; j < n; ++j) {
      int mx = 0;
      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)
          ans[i][j] = mx;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

3033. Modify the Matrix LeetCode Solution in Java

class Solution {
  public int[][] modifiedMatrix(int[][] matrix) {
    final int m = matrix.length;
    final int n = matrix[0].length;
    int[][] ans = matrix.clone();

    for (int j = 0; j < n; ++j) {
      int mx = 0;
      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)
          ans[i][j] = mx;
    }

    return ans;
  }
}
// code provided by PROGIEZ

3033. Modify the Matrix LeetCode Solution in Python

class Solution:
  def modifiedMatrix(self, matrix: list[list[int]]) -> list[list[int]]:
    m = len(matrix)
    n = len(matrix[0])
    ans = matrix.copy()

    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:
          ans[i][j] = mx

    return ans
 # code by PROGIEZ