Positions of Large Groups

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ArraysStringsTwo Pointers

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 <= 1000
s contains lowercase English letters only.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What defines a "large group"? Is the size a fixed value, or will it be provided as input?
What are the possible characters within the input string? Are we limited to lowercase English letters, or should I expect uppercase letters, numbers, or special characters?
If there are no large groups present, what should be the return value: an empty list, or null?
In the output, should the groups be sorted based on their start indices? Is there any specific order required?
Is the input string guaranteed to be non-null and non-empty? What should happen if the input is null or empty?

Brute Force Solution

Approach

The brute force method for finding large groups in a string is like carefully looking at every section of the string. We examine each character and its neighbors to identify sequences of identical characters. We then check if these sequences meet the minimum size requirement for a large group and record their positions.

Here's how the algorithm would work step-by-step:

Start at the very beginning of the string.
Look at the first character, and see how many consecutive characters after it are the same.
If the number of same characters in a row is big enough (meaning it forms a large group), record the starting and ending positions of this group.
Move to the next character in the string.
Repeat the process of checking for consecutive identical characters and recording large groups.
Continue until you have examined every character in the string.

Code Implementation

def find_large_groups_brute_force(input_string):
    large_groups = []
    string_length = len(input_string)

    for index in range(string_length):
        group_length = 1

        # Check consecutive chars.
        while index + group_length < string_length and \
                input_string[index] == input_string[index + group_length]:

            group_length += 1

        # Record start/end if large enough.
        if group_length >= 3:
            large_groups.append([index, index + group_length - 1])

        # Move the index forward to avoid re-counting.
        index += group_length - 1

    return large_groups

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string of length n. For each character, it potentially iterates through the consecutive identical characters to determine the group size. However, the inner loop's iterations are bounded by the string length and contribute to a single pass through the string. Therefore, the overall time complexity is O(n), as we visit each character a constant number of times.

Space Complexity

O(1) – The provided algorithm iterates through the string and identifies large groups. It stores the start and end positions of these groups, which are added to a list of results. The algorithm only utilizes a fixed number of variables to store the start and end indices of the current group, and a list to store the final output. The space required for storing the final output (the list of groups) depends on the number of large groups found in the input string but the question asks for the space complexity of provided solution and the given solution only needs to store two index variables and therefore the space complexity is constant, i.e., O(1).

Optimal Solution

Approach

The problem asks us to identify consecutive groups of the same character in a string that are of a certain size. Instead of comparing each character to every other character, we can walk through the string once, keeping track of the current group. This allows us to efficiently find all qualifying groups without redundant comparisons.

Here's how the algorithm would work step-by-step:

Start at the beginning of the string and consider the first character as the start of a potential group.
Move along the string, counting how many consecutive characters are the same as the starting character.
If the number of consecutive same characters is greater than or equal to the required size, record the starting and ending positions of this group.
If the character changes or we reach the end of the string, stop counting and check if the group meets the size requirement.
If the group is large enough, store its start and end positions.
Start looking for the next group from the character after the end of the previous group, and repeat the process until the end of the string.
The list of start and end positions of the large groups is your final answer.

Code Implementation

def positions_of_large_groups(s):    result = []
    start_position = 0
    index = 0
    string_length = len(s)

    while index < string_length:
        current_character = s[index]
        group_length = 0

        # Count consecutive chars.
        while index < string_length and s[index] == current_character:
            group_length += 1
            index += 1

        # Ensure the group is large enough.
        if group_length >= 3:
            result.append([start_position, index - 1])

        # Start the new group
        start_position = index

    return result

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string of length n only once. During this single pass, it counts consecutive characters to identify groups. The number of operations performed is directly proportional to the number of characters in the input string. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The dominant space usage comes from storing the list of large group start and end indices. In the worst-case scenario, where the string alternates between two characters (e.g., 'ababab'), each group will have a size of 1. Since the problem specifies that we only store groups of size at least 3, even in the worst case we would store O(N/3) which simplifies to O(N) index pairs. However, if we consider N as the length of the string s, we store at most a constant number of variables (start and list to store the results). Therefore the space used remains constant regardless of the input size. It's essentially a fixed number of integer variables.

Edge Cases

Case	How to Handle
Empty input string	Return an empty list immediately as there are no groups to find.
Input string with length 1	Return an empty list as a large group requires at least 3 consecutive characters.
Input string with length 2	Return an empty list, as a large group requires at least 3 consecutive identical characters.
String with all identical characters (e.g., 'aaaa')	The entire string should be identified as a single large group.
String with mixed large and small groups (e.g., 'aaabbbaaa')	The solution must correctly identify and report all large groups present.
Large group at the beginning of the string (e.g., 'aaabb')	The solution should correctly identify the start and end index of the group.
Large group at the end of the string (e.g., 'bbaaa')	The solution should correctly identify the start and end index of the group.
String containing non-ASCII characters	Ensure the solution properly handles UTF-8 or other encodings correctly; the character comparison should work regardless of the specific character set.

Empty input string

How to Handle:

Return an empty list immediately as there are no groups to find.

Input string with length 1

How to Handle:

Return an empty list as a large group requires at least 3 consecutive characters.

Input string with length 2

How to Handle:

Return an empty list, as a large group requires at least 3 consecutive identical characters.

String with all identical characters (e.g., 'aaaa')

How to Handle:

The entire string should be identified as a single large group.

String with mixed large and small groups (e.g., 'aaabbbaaa')

How to Handle:

The solution must correctly identify and report all large groups present.

Large group at the beginning of the string (e.g., 'aaabb')

How to Handle:

The solution should correctly identify the start and end index of the group.

Large group at the end of the string (e.g., 'bbaaa')

How to Handle:

The solution should correctly identify the start and end index of the group.

String containing non-ASCII characters

How to Handle:

Ensure the solution properly handles UTF-8 or other encodings correctly; the character comparison should work regardless of the specific character set.