# 944. Delete Columns to Make Sorted

## Description

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 follows:
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.

## Solutions

• class Solution {
public int minDeletionSize(String[] strs) {
int m = strs[0].length(), n = strs.length;
int ans = 0;
for (int j = 0; j < m; ++j) {
for (int i = 1; i < n; ++i) {
if (strs[i].charAt(j) < strs[i - 1].charAt(j)) {
++ans;
break;
}
}
}
return ans;
}
}

• class Solution {
public:
int minDeletionSize(vector<string>& strs) {
int n = strs.size();
int m = strs[0].size();
int res = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n - 1; ++j) {
if (strs[j][i] > strs[j + 1][i]) {
res++;
break;
}
}
}
return res;
}
};


• class Solution:
def minDeletionSize(self, strs: List[str]) -> int:
m, n = len(strs[0]), len(strs)
ans = 0
for j in range(m):
for i in range(1, n):
if strs[i][j] < strs[i - 1][j]:
ans += 1
break
return ans


• func minDeletionSize(strs []string) int {
m, n := len(strs[0]), len(strs)
ans := 0
for j := 0; j < m; j++ {
for i := 1; i < n; i++ {
if strs[i][j] < strs[i-1][j] {
ans++
break
}
}
}
return ans
}

• impl Solution {
pub fn min_deletion_size(strs: Vec<String>) -> i32 {
let n = strs.len();
let m = strs[0].len();
let mut res = 0;
for i in 0..m {
for j in 1..n {
if strs[j - 1].as_bytes()[i] > strs[j].as_bytes()[i] {
res += 1;
break;
}
}
}
res
}
}


• function minDeletionSize(strs: string[]): number {
const [m, n] = [strs[0].length, strs.length];
let ans = 0;
for (let j = 0; j < m; ++j) {
for (let i = 1; i < n; ++i) {
if (strs[i][j] < strs[i - 1][j]) {
++ans;
break;
}
}
}
return ans;
}