# [LeetCode]#944. Delete Columns to Make Sorted

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Environment: Python 3.7

Key technique: zip(*), sorted

We are given an array `A` of `N` lowercase letter strings, all of the same length.

Now, we may choose any set of deletion indices, and for each string, we delete all the characters in those indices.

For example, if we have an array `A = ["abcdef","uvwxyz"]` and deletion indices `{0, 2, 3}`, then the final array after deletions is `["bef", "vyz"]`, and the remaining columns of `A` are `["b","v"]`, `["e","y"]`, and `["f","z"]`. (Formally, the `c`-th column is `[A[0][c], A[1][c], ..., A[A.length-1][c]]`).

Suppose we chose a set of deletion indices `D` such that after deletions, each remaining column in A is in non-decreasing sorted order.

Return the minimum possible value of `D.length`.

Example 1:

`Input: A = ["cba","daf","ghi"]Output: 1`

Analysis:

1. Use zip(*A) and get (“c”,”d”,”g”), (“b”,”a”,”h”), (“a”,”f”,”i”)
2. Use sorted and compare original list.
3. If they are not match, add 1.
4. Summarize all mismatch pair.

Solution:

`class Solution:    def minDeletionSize(self, A):        cnt=0        for i in zip(*A):            if list(i) !=sorted(i):                cnt+=1        return cnt`

Submissions:

Reference:

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