What is the time complexity of Expressive Words?

The optimal solution has a time complexity of O(n*m*max(len(s), len(word))).

What data structures are used to solve Expressive Words?

This problem uses Array, Two Pointers, String. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Expressive Words?

Insight 1: The core idea is to compare groups of consecutive identical characters in the given string `s` and each word in `words` individually. We only care about the characters and their counts, not the specific positions. Insight 2: If a character group in `word` is longer than the corresponding group in `s`, the word cannot be stretchy.

Which companies ask LeetCode 809. Expressive Words in interviews?

This problem is frequently asked at Cisco.

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Expressive Words - LeetCode 809 Solution

Expressive Words - Complete Solution Guide

Expressive Words is LeetCode problem 809, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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" . You are given a string s and an array of query strings words . A query word is stretchy if it can be made to be equal to s by any number of applications of the following extension operation: choose a group consisting of characters c , and add some number of charact

Detailed Explanation

The problem asks us to determine how many words in a given array of words (`words`) are 'stretchy' with respect to a given string `s`. A word is considered 'stretchy' if it can be transformed into `s` by extending groups of identical characters in the word. Specifically, we can add characters to a group of identical characters if the resulting group in `s` has a length of 3 or more. The goal is to count how many words from `words` satisfy this condition.

Solution Approach

The solution iterates through each word in the `words` array and checks if it's stretchy with respect to the input string `s`. The `is_stretchy` function compares character groups in `s` and the current word. It checks if the character matches, if the group length in `word` is greater than `s`, and if the group length in `s` is less than 3 but not equal to the group length in `word`.

Step-by-Step Algorithm

Step 1: Iterate through each `word` in the `words` array.
Step 2: For each `word`, call the `is_stretchy` function to determine if it's stretchy with respect to `s`.
Step 3: Inside `is_stretchy`, use two pointers, `i` for `s` and `j` for `word`, to traverse the strings.
Step 4: Compare the characters at `s[i]` and `word[j]`. If they don't match, the word is not stretchy; return `false`.
Step 5: If the characters match, determine the length of the current character group in both `s` and `word`.
Step 6: Compare the lengths of the character groups. If the length in `word` is greater than the length in `s`, the word is not stretchy; return `false`.
Step 7: If the length in `s` is less than 3, the lengths must be equal for the word to be stretchy. If they are not equal, the word is not stretchy; return `false`.
Step 8: Advance the pointers `i` and `j` to the next character group.
Step 9: After processing all characters, check if both pointers have reached the end of their respective strings. If so, the word is stretchy; return `true`. Otherwise, return `false`.

Key Insights

Insight 1: The core idea is to compare groups of consecutive identical characters in the given string `s` and each word in `words` individually. We only care about the characters and their counts, not the specific positions.
Insight 2: If a character group in `word` is longer than the corresponding group in `s`, the word cannot be stretchy.
Insight 3: If a character group in `s` has a length less than 3, the corresponding group in `word` must have the same length as the group in `s` to be stretchy.

Complexity Analysis

Time Complexity: O(n*m*max(len(s), len(word)))

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, String.

Companies

Asked at: Cisco.

Expressive Words | LeetCode 809

Medium

Array Two Pointers String

Time: O(n*m*max(len(s), len(word)))Space: O(1)

Solution Code

Available in:

class Solution {
    public int expressiveWords(String s, String[] words) {
        
        int stretchy_count = 0;
        for (String word : words) {
            if (is_stretchy(s, word)) {
                stretchy_count++;
            }
        }
        
        return stretchy_count;
    }
    
    private boolean is_stretchy(String s, String word) {
        int n = s.length();
        int m = word.length();
        int i = 0;
        int j = 0;

        while (i < n && j < m) {
            if (s.charAt(i) != word.charAt(j)) {
                return false;
            }

            char char_group = s.charAt(i);

            int start_i = i;
            while (i < n && s.charAt(i) == char_group) {
                i++;
            }
            int len_s = i - start_i;

            int start_j = j;
            while (j < m && word.charAt(j) == char_group) {
                j++;
            }
            int len_w = j - start_j;

            if (len_w > len_s) {
                return false;
            }

            if (len_s < 3 && len_w != len_s) {
                return false;
            }
        }

        return i == n && j == m;
    }
}

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Problem Info

Description

Problem 809: Expressive Words - Detailed explanation and solution approaches.

Topics

Array Two Pointers String

Companies

Cisco

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:809

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Iterate through each `word` in the `words` array.
2
Step 2: For each `word`, call the `is_stretchy` function to determine if it's stretchy with respect to `s`.
3
Step 3: Inside `is_stretchy`, use two pointers, `i` for `s` and `j` for `word`, to traverse the strings.
4
Step 4: Compare the characters at `s[i]` and `word[j]`. If they don't match, the word is not stretchy; return `false`.
5
Step 5: If the characters match, determine the length of the current character group in both `s` and `word`.
6
Step 6: Compare the lengths of the character groups. If the length in `word` is greater than the length in `s`, the word is not stretchy; return `false`.
7
Step 7: If the length in `s` is less than 3, the lengths must be equal for the word to be stretchy. If they are not equal, the word is not stretchy; return `false`.
8
Step 8: Advance the pointers `i` and `j` to the next character group.
9
Step 9: After processing all characters, check if both pointers have reached the end of their respective strings. If so, the word is stretchy; return `true`. Otherwise, return `false`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*m*max(len(s), len(word)))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the nested loops: the outer loop iterating through the words and the inner `is_stretchy` function comparing the character counts. The space complexity is constant as no extra space is used that scales with the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Run Length Encoding

Encode both the string `s` and each word in `words` using run-length encoding (RLE). Then compare the RLE representations. This pre-processing step could improve efficiency slightly.

Time: O(n*m), where 'n' is the number of words, and 'm' is the maximum length of s or any word. RLE encoding takes O(len(s)) and O(len(word)), while comparison after encoding takes O(min(number of encoded groups))Space: O(len(s) + len(word)), where each run-length encoded sequence can take up this much space in the worst case (all distinct characters).

Trade-offs: The RLE approach might be slightly faster if the string `s` and the words in `words` contain many repeated characters. However, the overhead of encoding each string might outweigh the benefits if the strings are relatively short or have few repeated characters. Adds extra space complexity.

✨

Key Insights

1Insight 1: The core idea is to compare groups of consecutive identical characters in the given string `s` and each word in `words` individually. We only care about the characters and their counts, not the specific positions.
2Insight 2: If a character group in `word` is longer than the corresponding group in `s`, the word cannot be stretchy.
3Insight 3: If a character group in `s` has a length less than 3, the corresponding group in `word` must have the same length as the group in `s` to be stretchy.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty String s or word: The code handles empty strings correctly because the while loop condition checks for `i < n` and `j < m`, preventing out-of-bounds access.
Edge case 2: word length is significantly larger than s: If `word` contains a character not in `s`, the function will return `false` immediately. If `word` contains the same sequence of characters as `s`, but is longer, the check `if (len_w > len_s)` handles this.
Edge case 3: `s` contains a sequence of length 1 or 2, while `word` contains a sequence of a different length for the same character: If `s` has a group of length 1 or 2, the corresponding group in `word` must have the same length. The `if (len_s < 3 && len_w != len_s)` check ensures this.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case when the length of a group in `s` is less than 3. The code must enforce that if the group length in `s` is less than 3, the corresponding group length in `word` must be equal to it.

✗Common mistake 2: Incorrectly incrementing the pointers `i` and `j` after processing a character group. The loop condition should ensure that the entire string is traversed.

✗Common mistake 3: Forgetting to compare all the characters in both S and the word. The check `i == n && j == m` at the end is crucial for determining that all characters were indeed processed and thus it is 'stretchy'.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression techniques: Similar concepts are used in run-length encoding (RLE), which is a simple form of data compression where runs of data are stored as a single data value and count. The 'stretchy' concept can be seen as a generalization of RLE.

🏦Real-world scenario 2: Text processing and pattern matching: In text editors or search engines, a 'stretchy' search could allow users to find matches even if the text contains slight variations in the repetition of characters.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 29, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Expressive Words"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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