Problem:
Score : 400 points

Problem Statement

Given a string t, we will call it unbalanced if and only if the length of t is at least 2, and more than half of the letters in t are the same. For example, both voodoo and melee are unbalanced, while neither noon nor a is.

You are given a string s consisting of lowercase letters. Determine if there exists a (contiguous) substring of s that is unbalanced. If the answer is positive, show a position where such a substring occurs in s.

Constraints

2 ≦ |s| ≦ 10^5

s consists of lowercase letters.

Partial Score

200 points will be awarded for passing the test set satisfying 2 ≦ N ≦ 100.

Input

The input is given from Standard Input in the following format:

s

Output

If there exists no unbalanced substring of s, print -1 -1.

If there exists an unbalanced substring of s, let one such substring be s_a s_{a+1} ... s_{b} (1 ≦ a < b ≦ |s|), and print a b. If there exists more than one such substring, any of them will be accepted.

Sample Input 1

needed

Sample Output 1

2 5

The string s_2 s_3 s_4 s_5 = eede is unbalanced. There are also other unbalanced substrings. For example, the output 2 6 will also be accepted.

Sample Input 2

atcoder

Sample Output 2

-1 -1

The string atcoder contains no unbalanced substring.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
	private static final int MOD = (int)Math.pow(10, 9) + 7;
	public static void main(String[] args) {
		FastReader sc = new FastReader();
      	String s = sc.next();
      	// 對每個字母計數
      	int[] charCnt = new int[26];
      	// 每個字母最遠到哪裡，例如finalLoc[0] = 15,代表a最後一次出現的位置
      	int[] finalLoc = new int[26];
      	int len = s.length();
      	
      	for (int i = 0; i < len; i++) {
      		char c = s.charAt(i);
      		charCnt[c - 'a']++;
      		finalLoc[c - 'a'] = i;
      	}
      	
      	for (int i = 0; i < len; i++) {
      		char c = s.charAt(i);
      		
      		if (charCnt[c - 'a'] > 0) {
      			int dist = finalLoc[c - 'a'] - i + 1;
      			if (dist >= 2) {
      				if (charCnt[c - 'a'] > (dist / 2));
      				System.out.println((i+1) + " " + (finalLoc[c - 'a'] + 1));
      				System.exit(0);
      			}
      			charCnt[c - 'a']--;
      		}
      	}
      	
      	System.out.println("-1 -1");
	}
	
    static class FastReader {
	    BufferedReader br;
	    StringTokenizer st;
	
	    public FastReader() {
	      br = new BufferedReader(new InputStreamReader(System.in));
	    }
	  
	    String next() { 
	      while (st == null || !st.hasMoreElements()) {
	        try {
	          st = new StringTokenizer(br.readLine());
	        } catch (IOException  e) {
	          e.printStackTrace();
	        }
	      }
	      return st.nextToken();
	    }
	
	    int nextInt() {
	      return Integer.parseInt(next());
	    }
	
	    long nextLong() {
	      return Long.parseLong(next());
	    }
	
	    double nextDouble() { 
	      return Double.parseDouble(next());
	    }
	
	    String nextLine() {
	      String str = "";
	        try{
	            str = br.readLine();
	        } catch (IOException e) {
	            e.printStackTrace();
	        }
	        return str;
	      }
	  }
}