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955. Delete Columns to Make Sorted II
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

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; } }