#830

Easy

auto_awesomeString-driven solution strategy

LeetCode Problem Workspace

Positions of Large Groups

Identify and return intervals of consecutive, large character groups in a string of lowercase letters.

Problem Statement

You are given a string s consisting of lowercase English letters. A group of characters is defined as a consecutive sequence of the same letter. For example, in the string 'abbxxxxzyy', the groups are 'a', 'bb', 'xxxx', 'z', and 'yy'.

Your task is to return a list of intervals, each representing a large group. A group is considered large if it contains more than two characters. The result should consist of intervals of the form [start, end], where 'start' and 'end' are the indices of the first and last characters of a large group.

Examples

Example 1

Input: s = "abbxxxxzzy"

Output: [[3,6]]

"xxxx" is the only large group with start index 3 and end index 6.

Example 2

Input: s = "abc"

Output: []

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]]

The large groups are "ddd", "eeee", and "bbb".

Constraints

1 <= s.length <= 1000
s contains lowercase English letters only.

Solution Approach

Iterate Through the String

The first step is to iterate through the string and group consecutive characters. Maintain a pointer to track the start of a group and extend it until the character changes. Once a group is formed, check its length.

Check Group Size

For each group, verify if its size is greater than 2. If it is, store the interval [start, end] where the group starts and ends. This will be added to the final result.

Return the Result

After processing the string, return the list of intervals corresponding to large groups. Ensure that the intervals are correctly formatted and that the result only includes groups that are large enough.

Complexity Analysis

Metric	Value
Time	O(N)
Space	O(N)

The time complexity is O(N), where N is the length of the string, as we iterate through the string once. The space complexity is O(N), since we store intervals for the large groups found, which in the worst case could be as large as N/2 intervals.

What Interviewers Usually Probe

Candidates should demonstrate knowledge of efficient string iteration techniques.
Look for candidates to properly handle string boundary cases, such as transitions between characters.
Assess if candidates correctly identify the necessary conditions for a group to be considered large.

Common Pitfalls or Variants

Common pitfalls

Incorrectly including groups of size 2 as large groups.
Failing to handle boundary cases where groups start or end at the beginning or end of the string.
Overcomplicating the problem by using unnecessary data structures or multiple passes.

Follow-up variants

Consider handling uppercase letters or strings with mixed cases (though constraints specify lowercase).
What if the string contained digits instead of letters? Can the solution still work?
Optimize for a scenario where the string is very large (close to 1000 characters).

FAQ

What does 'Positions of Large Groups' refer to?

It refers to identifying groups of consecutive characters in a string that are larger than 2 characters and returning their intervals.

How do I identify large groups in the string?

A group is considered large if it contains more than two consecutive characters of the same type. The solution requires identifying such groups and returning their start and end indices.

What is the optimal time complexity for this problem?

The optimal time complexity is O(N), where N is the length of the string, since each character is checked once.

Can the solution handle strings up to the maximum length?

Yes, the solution is designed to handle strings up to 1000 characters, maintaining an O(N) time complexity.

What is a common mistake to avoid in this problem?

A common mistake is to incorrectly count groups that are not large enough, such as groups with only 2 characters.

terminal

Solution

Solution 1: Two Pointers

We use two pointers $i$ and $j$ to find the start and end positions of each group, then check if the group length is greater than or equal to $3$. If so, we add it to the result array.

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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

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