0944 - Delete Columns to Make Sorted (Easy)
Problem Link
https://leetcode.com/problems/delete-columns-to-make-sorted/
Problem Statement
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.
Approach 1: Matrix Traversing
- C++
- Rust
class Solution {
public:
int minDeletionSize(vector<string>& s) {
int ans = 0;
// iterate cols
for (int col = 0; col < s[0].size(); col++) {
// iterate rows
for (int row = 1; row < s.size(); row++) {
// if the character in the previous row is greater than the character in the current row
// then we can delete this column
if (s[row - 1][col] > s[row][col]) {
ans += 1;
break;
}
}
}
return ans;
}
};
impl Solution {
pub fn min_deletion_size(strs: Vec<String>) -> i32 {
let mut ans = 0;
// iterate cols
for col in 0 .. strs[0].as_bytes().len() {
// iterate rows
for row in 1 .. strs.len() {
// if the character in the previous row is greater than the character in the current row
// then we can delete this column
if strs[row - 1].as_bytes()[col] > strs[row].as_bytes()[col] {
ans += 1;
break;
}
}
}
ans
}
}