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Formatted question description: https://leetcode.ca/all/830.html

830. Positions of Large Groups (Easy)

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

Companies:
Adobe

Related Topics:
Array

Solution 1.

• Java
• C++
• Python
• Go

• class Solution {
      public List<List<Integer>> largeGroupPositions(String S) {
          List<List<Integer>> largeGroupPositions = new ArrayList<List<Integer>>();
          if (S.length() < 3)
              return largeGroupPositions;
          int begin = 0;
          char prev = S.charAt(0);
          int length = S.length();
          for (int i = 1; i < length; i++) {
              char c = S.charAt(i);
              if (c != prev) {
                  if (i - begin >= 3) {
                      int end = i - 1;
                      List<Integer> largeGroupPosition = new ArrayList<Integer>();
                      largeGroupPosition.add(begin);
                      largeGroupPosition.add(end);
                      largeGroupPositions.add(largeGroupPosition);
                  }
                  begin = i;
              }
              prev = c;
          }
          if (length - begin >= 3) {
              int end = length - 1;
              List<Integer> largeGroupPosition = new ArrayList<Integer>();
              largeGroupPosition.add(begin);
              largeGroupPosition.add(end);
              largeGroupPositions.add(largeGroupPosition);
          }
          return largeGroupPositions;
      }
  }
  
  ############
  
  class Solution {
      public List<List<Integer>> largeGroupPositions(String s) {
          int n = s.length();
          int i = 0;
          List<List<Integer>> ans = new ArrayList<>();
          while (i < n) {
              int j = i;
              while (j < n && s.charAt(j) == s.charAt(i)) {
                  ++j;
              }
              if (j - i >= 3) {
                  ans.add(Arrays.asList(i, j - 1));
              }
              i = j;
          }
          return ans;
      }
  }
  

• // OJ: https://leetcode.com/problems/positions-of-large-groups/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      vector<vector<int>> largeGroupPositions(string S) {
          vector<vector<int>> ans;
          for (int N = S.size(), i = 0; i < N;) {
              int j = i + 1;
              while (j < N && S[j] == S[i]) ++j;
              if (j - i >= 3) ans.push_back({ i, j - 1 });
              i = j;
          }
          return ans;
      }
  };
  

• class Solution:
      def largeGroupPositions(self, s: str) -> List[List[int]]:
          i, n = 0, len(s)
          ans = []
          while i < n:
              j = i
              while j < n and s[j] == s[i]:
                  j += 1
              if j - i >= 3:
                  ans.append([i, j - 1])
              i = j
          return ans
  
  ############
  
  class Solution:
      def largeGroupPositions(self, S):
          """
          :type S: str
          :rtype: List[List[int]]
          """
          groups = []
          before_index, before_char = 0, S[0]
          for i, s in enumerate(S):
              if s != before_char:
                  if i - before_index >= 3:
                      groups.append([before_index, i - 1])
                  before_index = i
                  before_char = s
          if i - before_index >= 2:
              groups.append([before_index, i])
          return groups
  

• func largeGroupPositions(s string) [][]int {
  	i, n := 0, len(s)
  	ans := [][]int{}
  	for i < n {
  		j := i
  		for j < n && s[j] == s[i] {
  			j++
  		}
  		if j-i >= 3 {
  			ans = append(ans, []int{i, j - 1})
  		}
  		i = j
  	}
  	return ans
  }
  

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