algoadvance

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.

Clarifying Questions:

1. Are there any constraints on the size of the matrix (m and n)?
   • Typical constraints for LeetCode problems are decent sized matrices (e.g., up to 200x200).
2. Is there any restriction on the auxiliary space usage?
   • Given the requirement to do it “in place,” it is implied that we should aim to minimize extra space usage.
3. Are there specific constraints on matrix elements (range of values)?
   • The problem is typically about integers and we should assume all elements can be any integer, including negatives.

Strategy:

1. Identify Zero Entries:
   • Traverse the matrix and identify the rows and columns that need to be zeroed out.
2. Mark Rows and Columns:
   • Use the first row and the first column as markers to store this information.
3. Zero Out Marked Rows and Columns:
   • Use the markers to update the matrix elements to zero accordingly.
4. Edge Cases to Consider:
   • Handle the zeroing of the first row and first column separately since they are used as markers.

Code:

def setZeroes(matrix):
    if not matrix or not matrix[0]:
        return

    m, n = matrix.length, matrix[0].length
    first_row_has_zero = any(matrix[0][j] == 0 for j in range(n))
    first_col_has_zero = any(matrix[i][0] == 0 for i in range(m))
    
    # Use first row and column to mark zeros
    for i in range(m):
        for j in range(n):
            if matrix[i][j] == 0:
                matrix[i][0] = 0
                matrix[0][j] = 0
    
    # Zero out cells based on markers except the first row and first column
    for i in range(1, m):
        for j in range(1, n):
            if matrix[i][0] == 0 or matrix[0][j] == 0:
                matrix[i][j] = 0

    # Zero out the first row if needed
    if first_row_has_zero:
        for j in range(n):
            matrix[0][j] = 0
    
    # Zero out the first column if needed
    if first_col_has_zero:
        for i in range(m):
            matrix[i][0] = 0

Time Complexity:

• Time Complexity: O(m * n)
  • We only traverse the matrix a couple of times, leading to a linear scan of all elements.
• Space Complexity: O(1)
  • We use the matrix’s first row and first column as storage for zeroing instructions, hence no extra space proportional to the input size is used except for a few variables.

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