## DEV Community

Abhishek Chaudhary

Posted on

# 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.

SOLUTION:

``````class Solution:
def isSorted(self, arr, i, n):
if i == n - 1:
return True
if arr[i] <= arr[i + 1] and self.isSorted(arr, i + 1, n):
return True
return False

def minDeletionSize(self, strs: List[str]) -> int:
n = len(strs[0])
k = len(strs)
ctr = 0
for i in range(n):
col = [s[i] for s in strs]
if not self.isSorted(col, 0, k):
ctr += 1
return ctr
``````