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.length1 <= n <= 1001 <= strs[i].length <= 1000strs[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。
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