LeetCode #2639 — EASY

Find the Width of Columns of a Grid

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed m x n integer matrix grid. The width of a column is the maximum length of its integers.

For example, if grid = [[-10], [3], [12]], the width of the only column is 3 since -10 is of length 3.

Return an integer array ans of size n where ans[i] is the width of the i^th column.

The length of an integer x with len digits is equal to len if x is non-negative, and len + 1 otherwise.

Example 1:

Input: grid = [[1],[22],[333]]
Output: [3]
Explanation: In the 0^th column, 333 is of length 3.

Example 2:

Input: grid = [[-15,1,3],[15,7,12],[5,6,-2]]
Output: [3,1,2]
Explanation: 
In the 0^th column, only -15 is of length 3.
In the 1^st column, all integers are of length 1. 
In the 2^nd 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

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed m x n integer matrix grid. The width of a column is the maximum length of its integers. For example, if grid = [[-10], [3], [12]], the width of the only column is 3 since -10 is of length 3. Return an integer array ans of size n where ans[i] is the width of the ith column. The length of an integer x with len digits is equal to len if x is non-negative, and len + 1 otherwise.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[1],[22],[333]]

Example 2

[[-15,1,3],[15,7,12],[5,6,-2]]

Core Insight

What unlocks the optimal approach

You can find the length of a number by dividing it by 10 and then rounding it down again and again until this number becomes equal to 0. Add 1 if this number is negative.
Traverse the matrix column-wise to find the maximum length in each column.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2639: Find the Width of Columns of a Grid
class Solution {
    public int[] findColumnWidth(int[][] grid) {
        int n = grid[0].length;
        int[] ans = new int[n];
        for (var row : grid) {
            for (int j = 0; j < n; ++j) {
                int w = String.valueOf(row[j]).length();
                ans[j] = Math.max(ans[j], w);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2639: Find the Width of Columns of a Grid
func findColumnWidth(grid [][]int) []int {
	ans := make([]int, len(grid[0]))
	for _, row := range grid {
		for j, x := range row {
			w := len(strconv.Itoa(x))
			ans[j] = max(ans[j], w)
		}
	}
	return ans
}

# Accepted solution for LeetCode #2639: Find the Width of Columns of a Grid
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)]

// Accepted solution for LeetCode #2639: Find the Width of Columns of a Grid
impl Solution {
    pub fn find_column_width(grid: Vec<Vec<i32>>) -> Vec<i32> {
        let mut ans = vec![0; grid[0].len()];

        for row in grid.iter() {
            for (j, num) in row.iter().enumerate() {
                let width = num.to_string().len() as i32;
                ans[j] = std::cmp::max(ans[j], width);
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #2639: Find the Width of Columns of a Grid
function findColumnWidth(grid: number[][]): number[] {
    const n = grid[0].length;
    const ans: number[] = new Array(n).fill(0);
    for (const row of grid) {
        for (let j = 0; j < n; ++j) {
            const w: number = String(row[j]).length;
            ans[j] = Math.max(ans[j], w);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(log M)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.