README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0900-0999/0955.Delete%20Columns%20to%20Make%20Sorted%20II/README_EN.md	Greedy Array String

955. Delete Columns to Make Sorted II

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Description

You are given an array of n strings strs, all of the same length.

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

For example, if we have strs = ["abcdef","uvwxyz"] and deletion indices {0, 2, 3}, then the final array after deletions is ["bef", "vyz"].

Suppose we chose a set of deletion indices answer such that after deletions, the final array has its elements in lexicographic order (i.e., strs[0] <= strs[1] <= strs[2] <= ... <= strs[n - 1]). Return the minimum possible value of answer.length.

 

Example 1:

Input: strs = ["ca","bb","ac"]
Output: 1
Explanation: 
After deleting the first column, strs = ["a", "b", "c"].
Now strs is in lexicographic order (ie. strs[0] <= strs[1] <= strs[2]).
We require at least 1 deletion since initially strs was not in lexicographic order, so the answer is 1.

Example 2:

Input: strs = ["xc","yb","za"]
Output: 0
Explanation: 
strs is already in lexicographic order, so we do not need to delete anything.
Note that the rows of strs are not necessarily in lexicographic order:
i.e., it is NOT necessarily true that (strs[0][0] <= strs[0][1] <= ...)

Example 3:

Input: strs = ["zyx","wvu","tsr"]
Output: 3
Explanation: We have to delete every column.

 

Constraints:

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

Solutions

Solution 1

Java

class Solution {
    public int minDeletionSize(String[] A) {
        if (A == null || A.length <= 1) {
            return 0;
        }
        int len = A.length, wordLen = A[0].length(), res = 0;
        boolean[] cut = new boolean[len];
    search:
        for (int j = 0; j < wordLen; j++) {
            // 判断第 j 列是否应当保留
            for (int i = 0; i < len - 1; i++) {
                if (!cut[i] && A[i].charAt(j) > A[i + 1].charAt(j)) {
                    res += 1;
                    continue search;
                }
            }
            // 更新 cut 的信息
            for (int i = 0; i < len - 1; i++) {
                if (A[i].charAt(j) < A[i + 1].charAt(j)) {
                    cut[i] = true;
                }
            }
        }
        return res;
    }
}