carloscn
commented
1 year ago Analysis
pub fn find_column_width(grid: Vec<Vec<i32>>) -> Vec<i32>
{
let mut ret:Vec<i32> = vec![];
if grid.len() < 1 {
return ret;
}
let len = grid[0].len();
for i in 0..len {
let mut max_len:usize = usize::MIN;
let mut col_len:usize;
for j in 0..grid.len() {
col_len = grid[j][i].to_string().len();
max_len = max_len.max(col_len);
}
ret.push(max_len as i32);
}
return ret;
}
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