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.

Submitted Code

class Solution:
    def minDeletionSize(self, strs: List[str]) -> int:
        len_list, len_str = len(strs), len(strs[0])
        delete_col = 0

        for j in range(len_str):                # 열
            for i in range(1, len_list):        # 1행부터 시작해서 이전 행과 비교
                if strs[i][j] < strs[i-1][j]:
                    delete_col += 1
                    break
                    
        return delete_col

Runtime: 63 ms | Beats 59.42%
Memory: 18.22 MB | Beats 71.66%

각 열을 세로로 읽으면서 사전순대로 정렬되어있는지 확인하면 된다. 0행은 시작부분이기 때문에 이전 값과 비교할 필요가 없어서 1행부터 순회했다.

Other Solutions

1st

    def minDeletionSize(self, A):
        return sum(list(col) != sorted(col) for col in zip(*A))

time complexity: 𝑂(𝑛𝑚log𝑚)
space complexity: 𝑂(𝑚)

    def minDeletionSize(self, A):
        return sum(any(a > b for a, b in zip(col, col[1:])) for col in zip(*A))

time complexity: 𝑂(𝑛𝑚)
space complexity: 𝑂(1)

zip(*A)로 간단하게 열 단위 튜플들을 생성할 수도 있다.