944. Delete Columns to Make Sorted
- Hardness: \(\color{green}\textsf{Easy}\)
- Ralated Topics:
Array
、String
一、題目
You are given an array of n
string strs
, all of the same length.
The string s 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 aove 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 3.
Constraints:
n == strs.length
1 <= n <= 100
1 <= strs[i].length <= 1000
二、分析
- 根據題意,逐個檢查是否字元有隨著列增加而呈
lexicographically sorted
。 - 注意迴圈的走法,若是外圈為欄,內圈為列的話,在發現沒有排序時,可以提早跳出,加快速度。
三、解題
1. String
- Time complexity: \(O(m\times n)\)
- Space complexity: \(O(1)\)
int minDeletionSize(vector<string>& strs) {
int m = strs.size(), n = strs[0].size();
int cnt = 0;
for (int col = 0; col < n; ++col) {
for (int row = 1; row < m; ++row) {
if (strs[row-1][col] > strs[row][col]) {
cnt++;
break;
}
}
}
return cnt;
}