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 <= 1000s contains lowercase English letters only.This approach involves iterating through the string to detect groups of consecutive characters. We maintain a start index for the current group and check whenever a character changes. If the length of a group is 3 or more, we record the start and end indices.
The function iterates through the string, marking the start of a group. Whenever a character change is detected, it checks the size of the group. Groups of size 3 or more are considered 'large', and their start and end indices are added to the results.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(n) for storing the result.
The two-pointer technique is another efficient way to solve this problem. The idea is to have a slow pointer marking the start of the group and a fast pointer iterating through the string. When the character at the fast pointer changes or reaches the end, check the length of the group.
This C solution uses two pointers, `slow` marking the start of a group and `fast` iterating through the string. When a change is detected, the interval is evaluated for its 'large' status.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(n) for storing the result.
| Approach | Complexity |
|---|---|
| Iterative Group Detection | Time Complexity: O(n), where n is the length of the string. |
| Two Pointer Technique | Time Complexity: O(n), where n is the length of the string. |
Leetcode Interviews • ThePrimeTime • 1,196,166 views views
Watch 9 more video solutions →Practice Positions of Large Groups with our built-in code editor and test cases.
Practice on FleetCode