Microsoft Online Assessment (OA) - String Without 3 Identical Consecutive Letters
Given a string S having lowercase English letters, returns a string with no instances of three identical consecutive letters,
obtained from S by deleting the minimum possible number of letters.
Example 1:
Input: eedaaad
Output: eedaad
Explanation:
One occurrence of the letter a is deleted.
Example 2:
Input: xxxtxxx
Output: xxtxx
Explanation:
Note that the letter x can occur more than three times in the returned string if the occurrences are not consecutive.
Example 3:
Input: uuuuxaaaaxum
Output: uuxaaxum
Try it yourself
Implementation
1def filter_string(s: str) -> str:
2 news = s[0:2]
3 for i in range(2, len(s)):
4 # Do not append if the previous chars are the same
5 if s[i] != s[i - 1] or s[i] != s[i - 2]:
6 news += s[i]
7 return news
8
9if __name__ == '__main__':
10 s = input()
11 res = filter_string(s)
12 print(res)
131import java.util.Scanner;
2
3class Solution {
4 public static String filterString(String s) {
5 StringBuilder sb = new StringBuilder();
6 sb.append(s.charAt(0));
7 sb.append(s.charAt(1));
8 for (int i = 2; i < s.length(); ++i) {
9 if (s.charAt(i) != s.charAt(i - 1) || s.charAt(i) != s.charAt(i - 2)) {
10 sb.append(s.charAt(i));
11 }
12 }
13 return sb.toString();
14 }
15
16 public static void main(String[] args) {
17 Scanner scanner = new Scanner(System.in);
18 String s = scanner.nextLine();
19 scanner.close();
20 String res = filterString(s);
21 System.out.println(res);
22 }
23}
241function filterString(s) {
2 const res = [s.charAt(0), s.charAt(1)];
3 for (let i = 2; i < s.length; ++i) {
4 if (s.charAt(i) !== s.charAt(i - 1) || s.charAt(i) !== s.charAt(i - 2)) {
5 res.push(s[i]);
6 }
7 }
8 return res.join('');
9}
10
11function* main() {
12 const s = yield;
13 const res = filterString(s);
14 console.log(res);
15}
16
17class EOFError extends Error {}
18{
19 const gen = main();
20 const next = (line) => gen.next(line).done && process.exit();
21 let buf = '';
22 next();
23 process.stdin.setEncoding('utf8');
24 process.stdin.on('data', (data) => {
25 const lines = (buf + data).split('\n');
26 buf = lines.pop();
27 lines.forEach(next);
28 });
29 process.stdin.on('end', () => {
30 buf && next(buf);
31 gen.throw(new EOFError());
32 });
33}
34Got a question? Ask the Teaching Assistant anything you don't understand.
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