2639. Find the Width of Columns of a Grid
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 is3
since-10
is of length3
.
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
}
}