#2639

Easy

LeetCode Problem Workspace

Find the Width of Columns of a Grid

This problem asks you to compute the width of each column in a grid, where the width is defined by the length of the largest integer in each column.

array matrix

Problem Statement

You are given a 0-indexed matrix grid of size m x n. Each element in the matrix is an integer. The task is to compute the width of each column, where the width of a column is the maximum length of the integers in that column. The length of an integer is defined by the number of digits it has, with negative integers counted by one extra digit for the minus sign.

Return an integer array ans of size n, where ans[i] represents the width of the i-th column. You can find the length of a number by repeatedly dividing it by 10 and rounding it down until the number becomes 0. Ensure that your solution efficiently handles the matrix dimensions up to 100x100.

Examples

Example 1

Input: grid = [[1],[22],[333]]

Output: [3]

In the 0th column, 333 is of length 3.

Example 2

Input: grid = [[-15,1,3],[15,7,12],[5,6,-2]]

Output: [3,1,2]

In the 0th column, only -15 is of length 3. In the 1st column, all integers are of length 1. In the 2nd column, both 12 and -2 are of length 2.

Constraints

m == grid.length
n == grid[i].length
1 <= m, n <= 100
-109 <= grid[r][c] <= 109

Solution Approach

Iterate Through Columns

For each column, iterate through all rows, calculate the length of the integers in that column, and track the maximum length. This approach uses simple matrix traversal.

Determine Integer Length Efficiently

To find the length of an integer, repeatedly divide it by 10. For negative integers, add 1 to account for the negative sign. This avoids the need to convert integers to strings.

Optimize with Early Stopping

If the length of an integer in a column exceeds the length of all other integers in that column, you can stop checking further. This saves unnecessary iterations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(m * n), where m is the number of rows and n is the number of columns in the grid. For each column, we examine each row exactly once. The space complexity is O(n) due to the storage required for the result array.

What Interviewers Usually Probe

The candidate demonstrates a solid understanding of matrix manipulation and can compute the length of integers without using string conversions.
The candidate optimizes the solution by eliminating unnecessary checks in each column after the maximum width has been found.
The candidate is familiar with efficient iteration through matrices and can articulate the time and space complexity clearly.

Common Pitfalls or Variants

Common pitfalls

Forgetting to account for negative integers when calculating their length, which might lead to off-by-one errors.
Inefficiently converting numbers to strings to compute their length, leading to higher time complexity.
Not considering early stopping once the maximum length for a column has been determined, which can waste time.

Follow-up variants

Instead of returning a single integer array, return the width of each row alongside the column widths.
Extend the problem to handle floating-point numbers and account for decimal points when calculating the width.
Implement a solution that works with dynamically sized grids (e.g., not necessarily rectangular).

FAQ

How can I compute the width of a column without converting numbers to strings?

You can calculate the length of an integer by repeatedly dividing it by 10 until the number becomes 0. For negative integers, add 1 to account for the minus sign.

What is the time complexity of solving this problem?

The time complexity is O(m * n), where m is the number of rows and n is the number of columns in the grid. Each column requires examining all rows.

Can this problem be solved using a matrix-transposing approach?

While transposing the matrix could be one approach, it’s not necessary. Directly processing columns without transposing is more efficient for this problem.

How does the early stopping technique help optimize the solution?

Early stopping helps by eliminating unnecessary checks once the maximum width for a column is found, saving time and reducing redundant work.

What are common pitfalls when calculating the width of columns?

Common pitfalls include neglecting negative integers, using inefficient methods like string conversion, and not utilizing early stopping in column processing.

terminal

Solution

Solution 1: Simulation

We denote the number of columns in the matrix as $n$, and create an array $ans$ of length $n$, where $ans[i]$ represents the width of the $i$-th column. Initially, $ans[i] = 0$.

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class Solution:
    def findColumnWidth(self, grid: List[List[int]]) -> List[int]:
        return [max(len(str(x)) for x in col) for col in zip(*grid)]

Continue Practicing

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