How do you solve Consecutive Characters on LeetCode?

Consecutive Characters (LeetCode #1446) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The problem can be solved efficiently by recognizing patterns in consecutive characters.

What is the time complexity of Consecutive Characters?

The optimal solution for Consecutive Characters runs in O(n) time and uses O(1) space. We only traverse the string once, leading to O(n) time complexity, and we use a constant amount of space.

What is the brute force solution for Consecutive Characters?

The brute force approach for Consecutive Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible substring of the string and count the length of consecutive characters. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Consecutive Characters?

Consecutive Characters uses the following concepts: String. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Consecutive Characters?

Always think about the simplest solution first before optimizing. Discuss your thought process with the interviewer to show your understanding. Practice explaining your code clearly and concisely.

What are common mistakes when solving Consecutive Characters?

Not considering edge cases like strings with all identical characters. Forgetting to reset the count when characters change.

#1446

Consecutive Characters

Easy

StringTwo PointersSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of checking all substrings, we can traverse the string once and count consecutive characters, which reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize 'max_power' to 1 and 'current_power' to 1.
2Step 2: Loop through the string from the second character to the end.
3Step 3: If the current character is the same as the previous one, increment 'current_power'. Otherwise, reset 'current_power' to 1.
4Step 4: Update 'max_power' with the maximum of itself and 'current_power'.

solution.py11 lines

1# Full working Python code
2s = 'abbcccddddeeeeedcba'
3max_power = 1
4current_power = 1
5for i in range(1, len(s)):
6    if s[i] == s[i - 1]:
7        current_power += 1
8    else:
9        current_power = 1
10    max_power = max(max_power, current_power)
11print(max_power)

ℹ

Complexity note: We only traverse the string once, leading to O(n) time complexity, and we use a constant amount of space.

1The problem can be solved efficiently by recognizing patterns in consecutive characters.
2Using a single pass reduces time complexity significantly.

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