LeetCode-in-All

73. Set Matrix Zeroes

Medium

Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0’s, and return the matrix.

You must do it in place.

Example 1:

Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]

Output: [[1,0,1],[0,0,0],[1,0,1]]

Example 2:

Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]

Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]

Constraints:

• m == matrix.length
• n == matrix[0].length
• 1 <= m, n <= 200
• -231 <= matrix[i][j] <= 231 - 1

Follow up:

• A straightforward solution using O(mn) space is probably a bad idea.
• A simple improvement uses O(m + n) space, but still not the best solution.
• Could you devise a constant space solution?

To solve the “Set Matrix Zeroes” problem in Swift with the Solution class, follow these steps:

1. Define a method setZeroes in the Solution class that takes a 2D integer matrix matrix as input and modifies it in place to set the entire row and column to zeros if an element is zero.
2. Initialize two boolean arrays rowZero and colZero of size m and n respectively, where m is the number of rows in the matrix and n is the number of columns. These arrays will track whether a row or column needs to be set to zero.
3. Iterate over the matrix to mark the rows and columns that contain zeros:
   • If matrix[i][j] is zero, set rowZero[i] = true and colZero[j] = true.
4. Iterate over the matrix again and set the entire row to zeros if rowZero[i] = true or the entire column to zeros if colZero[j] = true.
5. Return the modified matrix.

Here’s the implementation of the setZeroes method in Swift:

class Solution {
    func setZeroes(_ matrix: inout [[Int]]) {
        var rows = Set<Int>()
        var columns = Set<Int>()

        //traverse the matrix
        for i in 0..<matrix.count {
            for j in 0..<matrix[i].count {
                if matrix[i][j] == 0 {
                    rows.insert(i)
                    columns.insert(j)
                }
            }
        }

        //change rows
        for row in rows {
            matrix[row] = Array(repeating: 0, count: matrix[row].count)
        }

        //change columns
        for column in columns {
            for row in 0..<matrix.count {
                matrix[row][column] = 0
            }
        }
    }
}

This implementation modifies the matrix in place to set entire rows and columns to zeros, with a time complexity of O(m * n) and a space complexity of O(m + n), where m is the number of rows and n is the number of columns in the matrix.

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