# 830. Positions of Large Groups

## Description

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".

A group is identified by an interval [start, end], where start and end denote the start and end indices (inclusive) of the group. In the above example, "xxxx" has the interval [3,6].

A group is considered large if it has 3 or more characters.

Return the intervals of every large group sorted in increasing order by start index.

Example 1:

Input: s = "abbxxxxzzy"
Output: [[3,6]]
Explanation: "xxxx" is the only large group with start index 3 and end index 6.


Example 2:

Input: s = "abc"
Output: []
Explanation: We have groups "a", "b", and "c", none of which are large groups.


Example 3:

Input: s = "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]
Explanation: The large groups are "ddd", "eeee", and "bbb".


Constraints:

• 1 <= s.length <= 1000
• s contains lowercase English letters only.

## Solutions

• 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;
}
}

• class Solution {
public:
vector<vector<int>> largeGroupPositions(string s) {
int n = s.size();
int i = 0;
vector<vector<int>> ans;
while (i < n) {
int j = i;
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


• 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
}