944. Delete Columns to Make Sorted

You are given an array of `n` strings `strs`, all of the same length.

The strings can be arranged such that there is one on each line, making a grid. For example, `strs = ["abc", "bce", "cae"]` can be arranged as:

```abc
bce
cae
```

You want to delete the columns that are not sorted lexicographically. In the above example (0-indexed), columns 0 (`'a'`, `'b'`, `'c'`) and 2 (`'c'`, `'e'`, `'e'`) are sorted while column 1 (`'b'`, `'c'`, `'a'`) is not, so you would delete column 1.

Return the number of columns that you will delete.

Example 1:

```Input: strs = ["cba","daf","ghi"]
Output: 1
Explanation: The grid looks as follows:
cba
daf
ghi
Columns 0 and 2 are sorted, but column 1 is not, so you only need to delete 1 column.
```

Example 2:

```Input: strs = ["a","b"]
Output: 0
Explanation: The grid looks as follows:
a
b
Column 0 is the only column and is sorted, so you will not delete any columns.
```

Example 3:

```Input: strs = ["zyx","wvu","tsr"]
Output: 3
Explanation: The grid looks as follows:
zyx
wvu
tsr
All 3 columns are not sorted, so you will delete all 3.
```

Constraints:

• `n == strs.length`
• `1 <= n <= 100`
• `1 <= strs[i].length <= 1000`
• `strs[i]` consists of lowercase English letters.

944. Delete Columns to Make Sorted
``````struct Solution;

impl Solution {
fn min_deletion_size(a: Vec<String>) -> i32 {
let mut d = 0;
let a: Vec<&[u8]> = a.iter().map(|s| s.as_bytes()).collect();
let n = a.len();
let m = a.len();
for i in 0..m {
for j in 1..n {
if a[j][i] < a[j - 1][i] {
d += 1;
break;
}
}
}
d
}
}

#[test]
fn test() {
let a: Vec<String> = vec_string!["cba", "daf", "ghi"];
assert_eq!(Solution::min_deletion_size(a), 1);
let a: Vec<String> = vec_string!["a", "b"];
assert_eq!(Solution::min_deletion_size(a), 0);
let a: Vec<String> = vec_string!["zyx", "wvu", "tsr"];
assert_eq!(Solution::min_deletion_size(a), 3);
}
``````