Question

Formatted question description: https://leetcode.ca/all/944.html

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
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 = ["a","b"]
Output: 0
Explanation: D = {}
Example 3:

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

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


Algorithm

It is an operation to traverse a two-dimensional array by column. If the character at the current position is less than or equal to the character in the same column of the next line, skip and continue to compare the characters on the next line and the next line. Otherwise, it means that the column needs to be deleted, and the result res is incremented by 1, and the current column is broken. Refer to the code as follows:

Code

C++

class Solution {
public:
    int minDeletionSize(vector<string>& A) {
        int n = A.size(), len = A[0].size(), res = 0;
        for (int j = 0; j < len; ++j) {
            for (int i = 0; i < n - 1; ++i) {
                if (A[i][j] <= A[i + 1][j]) continue;
                ++res;
                break;
            }
        }
        return res;
    }
};

Java

  • class Solution {
        public int minDeletionSize(String[] A) {
            int rows = A.length, columns = A[0].length();
            int deletionSize = 0;
            for (int i = 0; i < columns; i++) {
                for (int j = 1; j < rows; j++) {
                    if (A[j].charAt(i) < A[j - 1].charAt(i)) {
                        deletionSize++;
                        break;
                    }
                }
            }
            return deletionSize;
        }
    }
    
  • // OJ: https://leetcode.com/problems/delete-columns-to-make-sorted
    // Time: O(MN)
    // Space: O(1)
    class Solution {
    public:
        int minDeletionSize(vector<string>& A) {
            int M = A.size(), N = A[0].size(), ans = 0;
            for (int i = 0; i < N; ++i) {
                int j = 1;
                for (; j < M && A[j][i] >= A[j - 1][i]; ++j);
                if (j < M) ans++;
            }
            return ans;
        }
    };
    
  • class Solution:
        def minDeletionSize(self, A):
            """
            :type A: List[str]
            :rtype: int
            """
            res = 0
            N = len(A[0])
            for i in range(N):
                col = [a[i] for a in A]
                if col != sorted(col):
                    res += 1
            return res
    

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