How do you solve Positions of Large Groups on LeetCode?

Positions of Large Groups (LeetCode #830) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Identifying groups efficiently can reduce time complexity.

What is the brute force solution for Positions of Large Groups?

The brute force approach for Positions of Large Groups is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each character in the string and counting consecutive characters. If a character repeats, we continue counting until we hit a different character. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Positions of Large Groups?

Positions of Large Groups uses the following concepts: String. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Positions of Large Groups?

Think about edge cases, such as strings with no large groups. Explain your thought process clearly as you code. Consider the trade-offs between time and space complexity.

What are common mistakes when solving Positions of Large Groups?

Not resetting the count after processing a group. Forgetting to check the last group after the loop ends.

#830

Positions of Large Groups

Easy

StringTwo PointersSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a single pass through the string to identify groups of characters. By keeping track of the current character and its count, we can efficiently determine if a group is large and record its indices.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty list to store the intervals of large groups.
2Step 2: Loop through the string while keeping track of the current character and its count.
3Step 3: When the character changes, check if the count of the previous character is 3 or more.
4Step 4: If it is, record the start and end indices of that group.
5Step 5: Continue until the end of the string and return the list of intervals.

solution.py16 lines

1# Full working Python code
2
3def largeGroupPositions(s):
4    intervals = []
5    n = len(s)
6    count = 1
7    for i in range(1, n):
8        if s[i] == s[i - 1]:
9            count += 1
10        else:
11            if count >= 3:
12                intervals.append([i - count, i - 1])
13            count = 1
14    if count >= 3:
15        intervals.append([n - count, n - 1])
16    return intervals

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the string, checking each character only once. The space complexity is O(n) in the worst case, where all characters are the same and form a large group.

1Identifying groups efficiently can reduce time complexity.
2Using a single pass to count characters minimizes overhead.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.