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.In #830 Positions of Large Groups, the goal is to identify segments in a string where the same character appears consecutively at least three times. Each such segment is considered a large group, and you must return the starting and ending indices of these groups.
The most efficient way to solve this problem is by using a two-pointer or single pass traversal. Iterate through the string while tracking the start index of the current group. Whenever the character changes (or the end of the string is reached), calculate the length of the group. If the length is greater than or equal to 3, record the interval.
This approach works because consecutive characters naturally form groups during a linear scan. By maintaining the group start and detecting boundaries, you can collect all valid ranges without extra data structures. The method runs in O(n) time since each character is visited once, and it uses O(1) additional space apart from the output list.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Single Pass / Two-Pointer Group Tracking | O(n) | O(1) |
ThePrimeTime
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.
Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(n) for storing the result.
1
2def large_group_positions(s):
3 result = []
4 start = 0
5 for i in range(1, len(In this Python solution, we utilize list operations to determine the start and end of each group. If a group is determined to be large, its indices are appended to the result list.
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.
Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(n) for storing the result.
1using System;
using System.Collections.Generic;
public class Solution {
public IList<IList<int>> LargeGroupPositions(string s) {
List<IList<int>> result = new List<IList<int>>();
int slow = 0;
for (int fast = 0; fast < s.Length; fast++) {
if (fast == s.Length - 1 || s[fast] != s[fast + 1]) {
if (fast - slow + 1 >= 3) {
result.Add(new List<int> { slow, fast });
}
slow = fast + 1;
}
}
return result;
}
}
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While it may not be among the most frequently asked FAANG problems, it reflects common interview patterns involving string traversal and grouping. Understanding this problem helps build fundamentals used in many string-processing interview questions.
The two-pointer technique helps track the start and end of consecutive character groups efficiently. By advancing through the string and detecting when characters change, you can compute group lengths without repeatedly scanning the same characters.
No complex data structure is required for this problem. A simple list or array is used to store the resulting intervals, while the algorithm only tracks indices and characters during a single scan of the string.
The optimal approach is a single-pass traversal using two pointers or group tracking. You keep track of the start of each character group and record its indices when the group length reaches three or more. This avoids extra data structures and runs in linear time.
This C# implementation uses two pointers to achieve efficient operation, setting `slow` to mark group start, while `fast` iterates, appending results accordingly.