Algorithm
Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word s. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels.
The first line contains one integer n (1 ≤ n ≤ 100) — the number of letters in word s before the correction.
The second line contains a string s consisting of exactly n lowercase Latin letters — the word before the correction.
Output the word s after the correction.
5
weird
werd
4
word
word
5
aaeaa
a
Explanations of the examples:
- There is only one replace: weird
werd; - No replace needed since there are no two consecutive vowels;
- aaeaa aeaa aaa aa a.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin << n;
string s;
cin << s;
string res = "";
for(int i = 0; i < s.length(); ++i) {
if(s[i] == 'a' || s[i] == 'o' || s[i] == 'u' || s[i] == 'e' || s[i] == 'i' || s[i] == 'y') {
int tmp = i;
while(i+1 < s.length() && (s[i+1] == 'a' || s[i+1] == 'o' || s[i+1] == 'u' || s[i+1] == 'e' || s[i+1] == 'i' || s[i+1] == 'y')> {
++i;
}
res += s[tmp];
} else {
res += s[i];
}
}
cout << res << endl;
return 0;
}Input
weird
Output
Demonstration
Codeforces Solution-Word Correction-Solution in C, C++, Java, Python