package com.fishercoder.solutions;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * 830. Positions of Large Groups
 *
 * In a string S of lowercase letters, these letters form consecutive groups of the same character.
 * For example, a string like S = "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z" and "yy".
 * Call a group large if it has 3 or more characters.  We would like the starting and ending positions of every large group.
 * The final answer should be in lexicographic order.
 *
 * Example 1:
 * Input: "abbxxxxzzy"
 * Output: [[3,6]]
 * Explanation: "xxxx" is the single large group with starting  3 and ending positions 6.
 *
 * Example 2:
 * Input: "abc"
 * Output: []
 * Explanation: We have "a","b" and "c" but no large group.
 *
 * Example 3:
 * Input: "abcdddeeeeaabbbcd"
 * Output: [[3,5],[6,9],[12,14]]
 *
 * Note:  1 <= S.length <= 1000
 */
public class _830 {
  public static class Solution1 {
    public List<List<Integer>> largeGroupPositions(String S) {
      List<List<Integer>> result = new ArrayList<>();
      char[] chars = S.toCharArray();
      for (int i = 0; i < chars.length; ) {
        char first = chars[i];
        int j = i + 1;
        while (j < chars.length && first == chars[j]) {
          j++;
        }
        if ((j - i) >= 3) {
          result.add(Arrays.asList(i, j - 1));
        }
        i = j;
      }
      return result;
    }
  }
}

package com.fishercoder.solutions;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class _830 {
    public static class Solution1 {
        public List<List<Integer>> largeGroupPositions(String S) {
            List<List<Integer>> result = new ArrayList<>();
            char[] chars = S.toCharArray();
            for (int i = 0; i < chars.length; ) {
                char first = chars[i];
                int j = i + 1;
                while (j < chars.length && first == chars[j]) {
                    j++;
                }
                if ((j - i) >= 3) {
                    result.add(Arrays.asList(i, j - 1));
                }
                i = j;
            }
            return result;
        }
    }
}

refactor 830