LeetCode 944 - Delete Columns to Make Sorted

Related:
LeetCode 944 - Delete Columns to Make Sorted
LeetCode 955 - Delete Columns to Make Sorted II
LeetCode 960 - Delete Columns to Make Sorted III
https://leetcode.com/problems/delete-columns-to-make-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.  The remaining rows of strings form columns when read north to south.

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: ["cba","daf","ghi"]
Output: 1
Explanation: 
After choosing D = {1}, each column ["c","d","g"] and ["a","f","i"] are in non-decreasing sorted order.
If we chose D = {}, then a column ["b","a","h"] would not be in non-decreasing sorted order.

Example 2:

Input: ["a","b"]
Output: 0
Explanation: D = {}

Example 3:

Input: ["zyx","wvu","tsr"]
Output: 3
Explanation: D = {0, 1, 2}

Note:

1. 1 <= A.length <= 100
2. 1 <= A[i].length <= 1000

For each column, check if its sorted. If it isn't, it must be deleted, so we add 1 to the final answer.

  public int minDeletionSize(String[] A) {

    int ans = 0;

    for (int c = 0; c < A[0].length(); ++c)

      for (int r = 0; r < A.length - 1; ++r)

        if (A[r].charAt(c) > A[r + 1].charAt(c)) {

          ans++;

          break;

        }

    return ans;

  }