# LeetCode 944. Delete Columns to Make Sorted

## Description

https://leetcode.com/problems/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.

## Explanation

Check column by column if sorted.

## Python Solution

``````class Solution:
def minDeletionSize(self, strs: List[str]) -> int:
columns = defaultdict(list)

for word in strs:
for i in range(len(word)):
ch = word[i]
columns[i].append(ch)

count = 0
for key, value in columns.items():
if value != sorted(value):
count += 1

return count
``````
• Time Complexity: O(N))
• Space Complexity: O(N)