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.

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?

Solution

func setZeroes(matrix [][]int) {
	m := len(matrix)
	n := len(matrix[0])
	row0 := false
	col0 := false

	// Check if 0th col needs to be marked all 0s in future
	for _, row := range matrix {
		if row[0] == 0 {
			col0 = true
			break
		}
	}

	// Check if 0th row needs to be marked all 0s in future
	for i := 0; i < n; i++ {
		if matrix[0][i] == 0 {
			row0 = true
			break
		}
	}

	// Store the signals in 0th row and column
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			if matrix[i][j] == 0 {
				matrix[i][0] = 0
				matrix[0][j] = 0
			}
		}
	}

	// Mark 0 for all cells based on signal from 0th row and 0th column
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			if matrix[i][0] == 0 || matrix[0][j] == 0 {
				matrix[i][j] = 0
			}
		}
	}

	// Set 0th column
	for i := 0; i < m; i++ {
		if col0 {
			matrix[i][0] = 0
		}
	}

	// Set 0th row
	for i := 0; i < n; i++ {
		if row0 {
			matrix[0][i] = 0
		}
	}
}

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