#1446

Easy

auto_awesomeString-driven solution strategy

LeetCode Problem Workspace

Consecutive Characters

Find the power of a string by determining the length of its longest substring with a unique character.

Problem Statement

Given a string s, you are tasked with finding the power of the string. The power of a string is defined as the length of the longest substring that contains only one unique character.

For example, in the string leetcode, the longest consecutive substring is 'ee', which has a length of 2. Your goal is to determine and return this length for any given string s.

Examples

Example 1

Input: s = "leetcode"

Output: 2

The substring "ee" is of length 2 with the character 'e' only.

Example 2

Input: s = "abbcccddddeeeeedcba"

Output: 5

The substring "eeeee" is of length 5 with the character 'e' only.

Constraints

1 <= s.length <= 500
s consists of only lowercase English letters.

Solution Approach

Single Pass Solution

Iterate over the string, tracking the length of consecutive characters that are the same. When a character differs, reset the counter and update the maximum length if necessary.

Efficient Tracking

While iterating, maintain a running count of consecutive characters. When encountering a new character, compare and update the maximum power found so far.

Space Optimization

Only track the current power and the maximum power found. This ensures an optimal solution in terms of both time and space complexity, keeping space usage to O(1).

Complexity Analysis

Metric	Value
Time	O(N)
Space	O(1)

The time complexity of this approach is O(N) where N is the length of the string s, as we are only making a single pass through the string. The space complexity is O(1) since we are only using a few variables to store intermediate values, not requiring extra space proportional to the input size.

What Interviewers Usually Probe

Ability to identify patterns in consecutive strings.
Efficient use of space while solving string problems.
Correct understanding of linear time complexity for string manipulation.

Common Pitfalls or Variants

Common pitfalls

Not updating the maximum length correctly after encountering different characters.
Mismanaging the substring length when the character count resets.
Overcomplicating the problem with unnecessary data structures.

Follow-up variants

What if the string contains no repeated characters? This would still work, as each character would be a substring of length 1.
What if the string is made up of the same character throughout? The maximum power would be the length of the entire string.
How would the solution change if the string was very long? The current solution remains optimal with O(N) time complexity.

FAQ

What is the power of a string in the 'Consecutive Characters' problem?

The power of a string is the length of its longest substring containing only one unique character.

How do I solve the 'Consecutive Characters' problem efficiently?

You can solve it with a linear scan of the string, tracking consecutive characters and updating the maximum length as needed.

Can the 'Consecutive Characters' problem be solved in constant space?

Yes, the problem can be solved in O(1) space by maintaining just two variables: one for the current length of consecutive characters and one for the maximum length found.

What is the time complexity of the 'Consecutive Characters' problem?

The time complexity is O(N), where N is the length of the string, because we traverse the string once.

Are there any edge cases I should be aware of for this problem?

Yes, make sure to handle cases where the string has no repeated characters or consists of just one repeated character throughout.

terminal

Solution

Solution 1: Traversal and Counting

We define a variable $\textit{t}$ to represent the length of the current consecutive characters, initially $\textit{t}=1$.

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class Solution:
    def maxPower(self, s: str) -> int:
        ans = t = 1
        for a, b in pairwise(s):
            if a == b:
                t += 1
                ans = max(ans, t)
            else:
                t = 1
        return ans

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