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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 }