Leetcode #809: Expressive Words

In this guide, we solve Leetcode #809 Expressive Words in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

Sometimes people repeat letters to represent extra feeling. For example: "hello" -> "heeellooo" "hi" -> "hiiii" In these strings like "heeellooo", we have groups of adjacent letters that are all the same: "h", "eee", "ll", "ooo".

Quick Facts

Difficulty: Medium
Premium: No
Tags: Array, Two Pointers, String

Intuition

The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.

Moving one pointer at a time keeps the invariant intact and avoids nested loops.

Approach

Place pointers at the left and right ends and move them based on the comparison or target condition.

This yields a clean linear pass after any required sorting.

Steps:

Set left and right pointers.
Move a pointer based on the condition.
Update the best answer while scanning.

Example

Input: s = "heeellooo", words = ["hello", "hi", "helo"]
Output: 1
Explanation: 
We can extend "e" and "o" in the word "hello" to get "heeellooo".
We can't extend "helo" to get "heeellooo" because the group "ll" is not size 3 or more.

Python Solution

class Solution:
    def expressiveWords(self, s: str, words: List[str]) -> int:
        def check(s, t):
            m, n = len(s), len(t)
            if n > m:
                return False
            i = j = 0
            while i < m and j < n:
                if s[i] != t[j]:
                    return False
                k = i
                while k < m and s[k] == s[i]:
                    k += 1
                c1 = k - i
                i, k = k, j
                while k < n and t[k] == t[j]:
                    k += 1
                c2 = k - j
                j = k
                if c1 < c2 or (c1 < 3 and c1 != c2):
                    return False
            return i == m and j == n

        return sum(check(s, t) for t in words)

Complexity

The time complexity is $O(n \times m + \sum_{i=0}^{m-1} w_i)$ , where $n$ and $m$ are the lengths of the string $s$ and the array $\textit{words}$ , respectively, and $w_i$ is the length of the $i$ -th word in the array $\textit{words}$ . The space complexity is O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Python Solution

class Solution:
    def expressiveWords(self, s: str, words: List[str]) -> int:
        def check(s, t):
            m, n = len(s), len(t)
            if n > m:
                return False
            i = j = 0
            while i < m and j < n:
                if s[i] != t[j]:
                    return False
                k = i
                while k < m and s[k] == s[i]:
                    k += 1
                c1 = k - i
                i, k = k, j
                while k < n and t[k] == t[j]:
                    k += 1
                c2 = k - j
                j = k
                if c1 < c2 or (c1 < 3 and c1 != c2):
                    return False
            return i == m and j == n

        return sum(check(s, t) for t in words)

Complexity

The time complexity is

O(n \times m + \sum_{i=0}^{m-1} w_i)

, where

n

and

m

are the lengths of the string

s

and the array

\textit{words}

, respectively, and

w_i

is the length of the

i

-th word in the array

\textit{words}

. The space complexity is O(1).

Leetcode #809: Expressive Words

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #809: Expressive Words

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview