README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2600-2699/2639.Find%20the%20Width%20of%20Columns%20of%20a%20Grid/README_EN.md	1282	Biweekly Contest 102 Q1	Array Matrix

2639. Find the Width of Columns of a Grid

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Description

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.

 

Example 1:

Input: grid = [[1],[22],[333]]
Output: [3]
Explanation: 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]
Explanation: 
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

Solutions

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, $ans[i] = 0$.

We traverse each row in the matrix. For each element in each row, we calculate its string length $w$, and update the value of $ans[j]$ to be $\max(ans[j], w)$.

After traversing all rows, each element in the array $ans$ is the width of the corresponding column.

The time complexity is $O(m \times n)$, and the space complexity is $O(\log M)$. Where $m$ and $n$ are the number of rows and columns in the matrix respectively, and $M$ is the absolute value of the maximum element in the matrix.

Python3

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)]

Java

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;
    }
}

C++

class Solution {
public:
    vector<int> findColumnWidth(vector<vector<int>>& grid) {
        int n = grid[0].size();
        vector<int> ans(n);
        for (auto& row : grid) {
            for (int j = 0; j < n; ++j) {
                int w = to_string(row[j]).size();
                ans[j] = max(ans[j], w);
            }
        }
        return ans;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}