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.
strs = ["abc", "bce", "cae"] can be arranged as follows: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.class Solution {
public:
int minDeletionSize(vector<string>& strs) {
int count = 0;
int n = strs.size(), m = strs[0].size();
for(int i=0; i<m; i++){
for(int j=0; j<n-1; j++){
if(strs[j][i]>strs[j+1][i]){
count++;
break;
}
}
}
return count;
}
};class Solution {
public int minDeletionSize(String[] strs) {
int n = strs[0].length(), del = 0;
loop: for (int i = 0; i < n; i++) {
char prev = 'a';
for (String s : strs) {
char curr = s.charAt(i);
if (curr < prev) {
del++;
continue loop;
}
prev = curr;
}
}
return del;
}
}