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Modify the Matrix

Difficulty: Easy

Problem 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.

Key Insights

  • We need to traverse each column to find the maximum value.
  • We will create a new matrix based on the original matrix.
  • For each element in the original matrix, if it is -1, we will replace it with the maximum value found in its column.
  • The constraints ensure that each column contains at least one non-negative integer.

Space and Time Complexity

Time Complexity: O(m * n) - We traverse the entire matrix twice: once to find max values and once to create the answer matrix. Space Complexity: O(m * n) - We create a new matrix to store the result.

Solution

The approach involves two main steps:

  1. Identify the maximum value in each column of the matrix.
  2. Construct the answer matrix by copying the values from the original matrix and replacing any occurrence of -1 with the corresponding maximum value from its column.

We will use arrays to store the maximum values for each column and then iterate through the matrix to build the answer.

Code Solutions

def modifyMatrix(matrix):
    # Get the dimensions of the matrix
    m, n = len(matrix), len(matrix[0])
    # Initialize an array to hold the maximum values for each column
    max_in_columns = [float('-inf')] * n

    # First pass: find the maximum value in each column
    for i in range(m):
        for j in range(n):
            if matrix[i][j] != -1:
                max_in_columns[j] = max(max_in_columns[j], matrix[i][j])

    # Create a new answer matrix
    answer = [[0] * n for _ in range(m)]

    # Second pass: build the answer matrix
    for i in range(m):
        for j in range(n):
            if matrix[i][j] == -1:
                answer[i][j] = max_in_columns[j]  # Replace -1 with the max value
            else:
                answer[i][j] = matrix[i][j]  # Keep the original value

    return answer
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