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.
Rust Solution
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[0].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);
}
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