Delete Columns to Make Sorted-优快云博客

本文链接：https://blog.youkuaiyun.com/m0_37889928/article/details/88029636

本文探讨了一种算法问题，即如何在给定的字符串数组构成的字符矩阵中，通过删除某些列使剩余列保持非递减顺序，目标是最小化删除列的数量。文章提供了具体的例子和解决方案。

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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: ["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 <= A.length <= 100
1 <= A[i].length <= 1000

题目理解：

给定一个字符串数组，用这些字符串构成一个字符矩阵，矩阵的每一行是字符串数组中的一个字符。删除矩阵中的一些列，可以使得矩阵中剩下的列都是非递减（递增或者相等）序列，问最少删除多少列。

解题思路：

找到所有不是非递减序列的列，返回其数量

class Solution {
    public int minDeletionSize(String[] A) {
        int row = A.length;
        char[][] chs = new char[row][];
        for(int i = 0; i < row; i++)
            chs[i] = A[i].toCharArray();
        int col = A[0].length();
        int res = 0;
        for(int j = 0; j < col; j++){
            int i = 1;
            for(; i < row; i++){
                if(chs[i][j] < chs[i - 1][j])
                    break;
            }
            if(i < row)
                res++;
        }
        return res;
    }
}