LeetCode: 944-Delete Columns to Make Sorted 解題紀錄

Last Updated on 2023-01-03 by Clay

題目

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

The strings can be arranged such that there is one on each line, making a grid. For example, strs = ["abc", "bce", "cae"] can be arranged as:

abc
bce
cae

You want to delete the columns that are not sorted lexicographically. In the above example (0-indexed), columns 0 ('a', 'b', 'c') and 2 ('c', 'e', 'e') are sorted while column 1 ('b', 'c', 'a') is not, so you would delete column 1.

Return the number of columns that you will delete.

Example 1:

Input: strs = ["cba","daf","ghi"]
Output: 1
Explanation: The grid looks as follows:
  cba
  daf
  ghi
Columns 0 and 2 are sorted, but column 1 is not, so you only need to delete 1 column.

Example 2:

Input: strs = ["a","b"]
Output: 0
Explanation: The grid looks as follows:
  a
  b
Column 0 is the only column and is sorted, so you will not delete any columns.

Example 3:

Input: strs = ["zyx","wvu","tsr"]
Output: 3
Explanation: The grid looks as follows:
  zyx
  wvu
  tsr
All 3 columns are not sorted, so you will delete all 3.

Constraints:

• n == strs.length
• 1 <= n <= 100
• 1 <= strs[i].length <= 1000
• strs[i] consists of lowercase English letters.

題目給定一個裝滿字串的陣列，我們要從縱軸的方向（columns）來判斷是否符合字典序（lexicographically），也就是是否符合 abc...xyz 這樣的順序。

如果不符合，我們要記錄下必須刪除的數量，最後回傳。

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解題思路

Brute Force

這題在 LeetCode 得到了強烈的倒讚，數量好像是讚的 4 倍左右。我想這是因為這題的解法大概就是暴力解，直接兩個 for-loop 完事，幾乎沒看到其他解法。

Brute Force 複雜度

Time Complexity	O(m*n)
Space Complexity	O(1)

C++ 範例程式碼

class Solution {
public:
    int minDeletionSize(vector<string>& strs) {
        // Init
        int n = 0;

        // Check
        for (int j=0; j<strs[0].size(); ++j) {
            for (int i=0; i<strs.size()-1; ++i) {
                if (strs[i][j] > strs[i+1][j]) {
                    ++n;
                    break;
                }
            }
        }

        return n;
    }
};

Python 範例程式碼

class Solution:
    def minDeletionSize(self, strs: List[str]) -> int:
        # Init
        n = 0

        # Checking
        for j in range(len(strs[0])):
            for i in range(len(strs)-1):
                if strs[i][j] > strs[i+1][j]:
                    n += 1
                    break

        return n

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References

• 944. Delete Columns to Make Sorted

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Read More

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