Problem Description

You are given a 2D integer matrix grid with dimensions m x n (m rows and n columns), where the matrix is 0-indexed.

The problem asks you to find the width of each column in the matrix. The width of a column is defined as the maximum length among all integers in that column.

To determine the length of an integer:

• For non-negative integers, the length equals the number of digits
• For negative integers, the length equals the number of digits plus 1 (to account for the negative sign)

For example:

• The integer 123 has length 3
• The integer -10 has length 3 (2 digits + 1 for the negative sign)
• The integer 5 has length 1

Given a matrix like grid = [[-10], [3], [12]]:

• The only column contains values: -10, 3, and 12
• Their lengths are: 3, 1, and 2 respectively
• The width of this column is the maximum: 3

Your task is to return an integer array ans of size n (where n is the number of columns), where ans[i] represents the width of the i-th column.

The solution uses zip(*grid) to transpose the matrix and access columns directly, then calculates the maximum string length for each column using max(len(str(x)) for x in col). This efficiently computes the width by converting each number to a string and finding the longest one in each column.

Intuition

The key insight is that we need to process the matrix column by column rather than row by row, since we're looking for the maximum width within each column.

When we think about finding the "width" of a column, we're essentially asking: "What's the longest string representation among all numbers in this column?" This naturally leads us to consider converting each number to a string and measuring its length.

The string conversion approach len(str(x)) elegantly handles both positive and negative numbers. Python's str() function automatically includes the negative sign for negative numbers, so -10 becomes "-10" with length 3, while 10 becomes "10" with length 2. This perfectly aligns with the problem's definition of length.

To work with columns efficiently, we can use Python's zip(*grid) trick. The * operator unpacks the rows of the grid, and zip then groups elements by their column index. For example, if grid = [[1, 2], [3, 4], [5, 6]], then zip(*grid) gives us [(1, 3, 5), (2, 4, 6)] - effectively transposing the matrix and giving us direct access to columns.

Once we have each column as a sequence, finding the maximum width becomes straightforward: for each column, convert all numbers to strings, measure their lengths, and take the maximum. This can be expressed concisely with a list comprehension: [max(len(str(x)) for x in col) for col in zip(*grid)].

This approach is both intuitive and efficient - we visit each element exactly once, perform a simple string conversion, and maintain the maximum for each column.

Solution Approach

The solution follows a simulation approach where we calculate the width of each column by examining all elements within it.

Step 1: Matrix Transposition

We use zip(*grid) to transpose the matrix and access columns directly. The * operator unpacks the rows, and zip groups elements by their position:

• Original matrix: rows are [[a, b], [c, d], [e, f]]
• After zip(*grid): columns become [(a, c, e), (b, d, f)]

Step 2: Width Calculation for Each Column

For each column obtained from the transposition, we:

1. Convert each integer x to its string representation using str(x)
2. Calculate the length of each string using len(str(x))
3. Find the maximum length within the column using max(...)

Step 3: List Comprehension

The entire process is wrapped in a list comprehension:

[max(len(str(x)) for x in col) for col in zip(*grid)]

This creates a result array where each element represents the width of the corresponding column.

Example Walkthrough:

Consider grid = [[1, -22], [333, 4]]:

1. zip(*grid) gives us columns: [(1, 333), (-22, 4)]
2. For first column (1, 333):
   • str(1) = "1", length = 1
   • str(333) = "333", length = 3
   • Maximum = 3
3. For second column (-22, 4):
   • str(-22) = "-22", length = 3
   • str(4) = "4", length = 1
   • Maximum = 3
4. Result: [3, 3]

The time complexity is O(m × n) where we visit each element once, and the space complexity is O(n) for storing the result array.

Example Walkthrough

Let's walk through the solution with a concrete example to see how it works step by step.

Given: grid = [[10, -5, 0], [8, 234, -99]]

Step 1: Identify what we need

• We have a 2×3 matrix (2 rows, 3 columns)
• We need to find the width of each of the 3 columns

Step 2: Transpose the matrix using zip(*grid)

• Original rows: [10, -5, 0] and [8, 234, -99]
• After zip(*grid):
  • Column 0: (10, 8)
  • Column 1: (-5, 234)
  • Column 2: (0, -99)

Step 3: Calculate width for each column

Column 0: (10, 8)

• 10 → str(10) = "10" → length = 2
• 8 → str(8) = "8" → length = 1
• Maximum length = 2

Column 1: (-5, 234)

• -5 → str(-5) = "-5" → length = 2 (1 digit + 1 for negative sign)
• 234 → str(234) = "234" → length = 3
• Maximum length = 3

Column 2: (0, -99)

• 0 → str(0) = "0" → length = 1
• -99 → str(-99) = "-99" → length = 3 (2 digits + 1 for negative sign)
• Maximum length = 3

Step 4: Compile the result

• Column widths: [2, 3, 3]

Solution code execution:

[max(len(str(x)) for x in col) for col in zip(*grid)]

This produces [2, 3, 3], which matches our manual calculation.

The key insight is that string conversion automatically handles the negative sign, making our width calculation straightforward and accurate.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m × n × log M)

The code iterates through all elements in the grid using zip(*grid) which transposes the matrix, visiting each of the m × n elements once. For each element x, we convert it to a string using str(x), which takes O(log |x|) time (the number of digits in the absolute value of x). In the worst case, this is O(log M) where M is the absolute value of the maximum element. Therefore, the total time complexity is O(m × n × log M).

Space Complexity: O(m × n + log M)

The space complexity consists of:

• O(m × n) for storing the transposed matrix created by zip(*grid), which creates a new collection containing all elements
• O(log M) for the temporary string representation of each number during conversion
• The output list of size O(n) is not counted as it's the required output

The dominant term is O(m × n), but if we only consider auxiliary space excluding the transposed matrix (treating the zip operation as iteration without storage), then the space complexity would be O(log M) as stated in the reference answer, accounting only for the string conversion buffer.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Empty Grid Handling

The code will fail if the input grid is empty (either [] or contains empty rows like [[]]). When zip(*grid) is called on an empty grid, it returns an empty iterator, which could lead to unexpected behavior.

Problem Example:

grid = []  # or grid = [[]]
# zip(*grid) returns empty, causing issues

Solution: Add an edge case check at the beginning:

def findColumnWidth(self, grid: List[List[int]]) -> List[int]:
    if not grid or not grid[0]:
        return []
  
    return [max(len(str(element)) for element in column) 
            for column in zip(*grid)]

2. Confusion About Integer Zero

Developers might mistakenly think that 0 has a special case or no width, but 0 has a length of 1 just like any other single digit.

Clarification:

• str(0) = "0", length = 1
• str(-0) = "0", length = 1 (Python treats -0 as 0)

3. Memory Overhead with Large Grids

Using zip(*grid) creates tuples for all columns in memory simultaneously. For very large grids, this could cause memory issues.

Alternative Approach for Memory Optimization:

def findColumnWidth(self, grid: List[List[int]]) -> List[int]:
    m, n = len(grid), len(grid[0])
    result = [0] * n
  
    for col in range(n):
        for row in range(m):
            result[col] = max(result[col], len(str(grid[row][col])))
  
    return result

4. Assuming Regular Matrix Dimensions

The code assumes all rows have the same number of columns. If the grid is jagged (irregular), zip will truncate to the shortest row length.

Problem Example:

grid = [[1, 2, 3], [4, 5]]  # Irregular grid
# zip(*grid) will only process first 2 columns

Solution: Validate input or handle irregular grids explicitly:

def findColumnWidth(self, grid: List[List[int]]) -> List[int]:
    # Validate that all rows have the same length
    if grid and len(set(len(row) for row in grid)) > 1:
        raise ValueError("Irregular grid dimensions")
  
    return [max(len(str(element)) for element in column) 
            for column in zip(*grid)]