Watch 10 video solutions for Positions of Large Groups, a easy level problem involving String. This walkthrough by CodeJulian has 1,294 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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.Problem Overview: Given a string s, you need to return the start and end indices of every group of identical characters that appears three or more times consecutively. These are called large groups. The result should list each group's starting and ending index in ascending order.
Approach 1: Iterative Group Detection (O(n) time, O(1) space)
This approach scans the string once while tracking the start index of the current character group. Iterate through the string and compare each character with the previous one. When the character changes, the current group ends. Check the group length using i - start; if it is at least 3, record [start, i-1]. Then update start to the current index and continue scanning. After the loop, run the same check for the final group since it might reach the end of the string. The algorithm performs a single pass over the string, giving O(n) time complexity and constant O(1) extra space (excluding the output). This method works well for problems involving consecutive patterns in a string.
Approach 2: Two Pointer Technique (O(n) time, O(1) space)
The two pointers pattern provides a clean way to track character groups. Maintain two indices: left marks the start of a group and right expands forward while characters remain the same. While s[right] == s[left], keep moving right. Once a different character appears or the string ends, compute the group size using right - left. If the size is at least 3, append [left, right-1] to the result. Then move left to right to begin scanning the next group. Each character is visited at most twice (once by each pointer), so the time complexity remains O(n) with O(1) additional space. This version clearly separates group detection and expansion logic, which many engineers find easier to reason about.
Recommended for interviews: Interviewers typically expect the linear scan or two-pointer solution because both run in O(n) time and require only constant extra space. Starting with the iterative group detection approach shows you understand how to track contiguous segments in a string. Implementing the two-pointer variation demonstrates familiarity with a widely used pattern for segment scanning problems.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Iterative Group Detection | O(n) | O(1) | Simple implementation when scanning consecutive characters in a string |
| Two Pointer Technique | O(n) | O(1) | Preferred when solving segment or range problems using expanding windows |