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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;
    }
}

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