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The problem “Positions of Large Groups” on LeetCode (#830) can be stated as follows:

You are given a string s of lowercase English letters that is sorted in non-decreasing order. A “large group” is a group of 3 or more consecutive characters that are the same.

Your task is to return a list of the starting and ending positions of every large group. The answer should be in the form of a list of lists.

Example:

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

Constraints:

Clarifying Questions

Before proceeding, let’s ensure all details are cleared:

  1. Q: Can the string s contain any characters other than lowercase English letters?
    • A: No, it only contains lowercase English letters.
  2. Q: Should overlapping groups be considered distinctly?
    • A: Overlapping is not an issue since each character is part of at most one group.

Strategy

  1. Initialization: Initialize an empty result list to store the positions of all large groups.
  2. Iteration: Use a pointer to traverse the string while keeping track of the start and end of the current group of characters.
  3. Check Group Size: For each group of consecutive same characters, if the length is 3 or more, add the start and end indices to the result list.
  4. Edge Cases: Make sure to handle the end of the string and any small lengths input.

Code

def largeGroupPositions(s: str):
    result = []
    i = 0
    n = len(s)
    
    while i < n:
        start = i
        while i < n and s[i] == s[start]:
            i += 1
        # Now s[start:i] is the large group we are checking
        if i - start >= 3:
            result.append([start, i - 1])
    return result

Time Complexity

The time complexity for this solution is O(n), where n is the length of the string s. This is because we only traverse the string once with a single loop. The space complexity is O(1) if we don’t account for the space used to store the result list, and O(k) where k is the number of large groups when including the output space.

By following this approach, we ensure that all edge cases are covered efficiently while maintaining optimal time complexity.

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