LMAN- Editorial

editorial
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PROBLEM LINK:

Code Battle: Holi Edition

Author: Naman Mittal
Tester: Aditya Rana
Editorialist: Deepanshu Jindal

DIFFICULTY:

CAKEWALK

PREREQUISITES:

Strings, Basic Programming

PROBLEM:

For a given string S, comprising of only lowercase English alphabets, eliminate the vowels from the string that occur between two consonants.

QUICK EXPLANATION:

Iterate over the string by checking adjacent characters. If a vowel is sandwiched between consonants, then don’t add it in the resultant string.

EXPLANATION:

Iterate over the string S. For any valid index i, if the character present at index i is a vowel, and characters present at index i-1 and i+1 are both consonants, then we will not consider vowel present at index i in our resultant string. Otherwise, we can unhesitatingly add them to our resultant string.

Complexities

Time Complexity: O(N) per test case, where N is the length of string.
Auxillary Space Complexity: O(N) per test case, where N is the length of string.

SOLUTIONS:

Setter's Solution 
#include <bits/stdc++.h>
using namespace std;
unordered_set<char> us({'a', 'e', 'i', 'o', 'u'});

bool isVowel(char c) {
  return us.find(c) != us.end();
}

string solve(string &s) {
  string ans = "";

  for(int i = 0; i < s.size(); ++i) {
    if(i == 0 || i == s.size() - 1) {
      ans += s[i];
      continue;
    }

    if(isVowel(s[i]) && !isVowel(s[i - 1]) && !isVowel(s[i + 1])) continue;
    ans += s[i];
  }

  return ans;
}

int main() {
  int t; cin >> t;
  while(t--) {
    string s; cin >> s;
    cout << solve(s) << endl;
  }
  return 0;
}
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