# Set matrix to zero-coding question asked by Atlassian, Intuit

# Problem statement:

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.

**Input Format:**

`Matrix`

**Output Format:**

`Matrix with rows and columns having an element zero set to zero.`

**Sample Input 1:**

`matrix = [[1,1,1],[1,0,1],[1,1,1]]`

**Sample Output 1:**

`[[1,0,1],[0,0,0],[1,0,1]]`

**Sample Input 2:**

`matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]`

**Sample Output 2:**

`[[0,0,0,0],[0,4,5,0],[0,3,1,0]]`

**Constraint:**

m == matrix.lengthn == matrix[0].length1 <= m, n <= 200-2³¹ <= matrix[i][j] <= 2³¹ — 1

# Approach:

If any cell of the matrix has a zero we can record its row and column number. All the cells of this recorded row and column can be marked zero in the next iteration.

We make a pass over our original array and look for zero entries.If we find that an entry at `[i, j]`

is 0, then we need to record somewhere the row `i`

and column `j`

.So, we use two `sets`

, one for the rows and one for the columns.

`if cell[i][j] == 0 { row_set.add(i) column_set.add(j) }`

Finally, we iterate over the original matrix. For every cell we check if the row `r`

or column `c`

had been marked earlier. If any of them was marked, we set the value in the cell to 0.

`if r in row_set or c in column_set { cell[r][c] = 0 }`

Time complexity:0(m*n)

Space complexity: O(m+n)

# Code:

# Thanks for Reading

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