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Abhishek Chaudhary
Abhishek Chaudhary

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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:


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:
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:
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:
All 3 columns are not sorted, so you will delete all 3.


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


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
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