What is the time complexity of Maximum Number of Words You Can Type?

The optimal solution has a time complexity of O(n*m).

What data structures are used to solve Maximum Number of Words You Can Type?

This problem uses Hash Table, String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Maximum Number of Words You Can Type?

Insight 1: The problem can be solved by iterating through each word in the `text` string and checking if any character in the word is present in the `brokenLetters` string. Insight 2: A simple boolean flag (`canType` in the provided solutions) can efficiently track whether a word contains a broken letter.

Which companies ask LeetCode 1935. Maximum Number of Words You Can Type in interviews?

This problem is frequently asked at Quora.

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Maximum Number of Words You Can Type - LeetCode 1935 Solution

Maximum Number of Words You Can Type - Complete Solution Guide

Maximum Number of Words You Can Type is LeetCode problem 1935, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly. Given a string text of words separated by a single space (no leading or trailing spaces) and a string brokenLetters of all distinct letter keys that are broken, return the number of words in text you can fully type using this keyboard . Example 1: Input: text = "hello world", brokenLetters = "ad" Output: 1 Explanation: We cannot type "world" because the 'd' key is broken. Example

Detailed Explanation

The problem asks you to count the number of words from a given string that can be typed using a malfunctioning keyboard. The input consists of two strings: `text`, containing words separated by spaces, and `brokenLetters`, containing the characters of broken keys. The output is the number of words from `text` that can be typed without using any broken keys. The problem assumes all letters are lowercase and that `brokenLetters` contains only distinct characters.

Solution Approach

The provided solutions all follow a similar approach. They first split the input `text` string into individual words. Then, they iterate through each word and check if it contains any character from the `brokenLetters` string. If a broken character is found, the word is considered untypeable. Otherwise, a counter is incremented. Finally, the counter (representing the number of typeable words) is returned.

Step-by-Step Algorithm

Step 1: Split the input string `text` into an array of words (using `text.split()` in Python, `text.split(" ")` in Java, `stringstream` in C++, or manual iteration in C).
Step 2: Iterate through each word in the array.
Step 3: For each word, iterate through the characters of `brokenLetters`.
Step 4: If a character from `brokenLetters` is found in the current word, set a flag (`canType`) to `false` and break the inner loop.
Step 5: If the flag (`canType`) is still `true` after checking all characters in `brokenLetters`, increment the `count` of typeable words.
Step 6: After processing all words, return the final `count`.

Key Insights

Insight 1: The problem can be solved by iterating through each word in the `text` string and checking if any character in the word is present in the `brokenLetters` string.
Insight 2: A simple boolean flag (`canType` in the provided solutions) can efficiently track whether a word contains a broken letter.
Insight 3: String splitting is a crucial preprocessing step to efficiently handle each word individually.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String.

Companies

Asked at: Quora.

Maximum Number of Words You Can Type | LeetCode 1935

Easy

Hash Table String

Time: O(n*m)Space: O(n)

Solution Code

Available in:

class Solution {
    public int canBeTypedWords(String text, String brokenLetters) {
        String[] words = text.split(" ");
        int count = 0;
        for (String word : words) {
            boolean canType = true;
            for (int i = 0; i < brokenLetters.length(); i++) {
                if (word.indexOf(brokenLetters.charAt(i)) != -1) {
                    canType = false;
                    break;
                }
            }
            if (canType) {
                count++;
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 1935: Maximum Number of Words You Can Type - Detailed explanation and solution approaches.

Topics

Hash Table String

Companies

Quora

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1935

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

6 steps

1
Step 1: Split the input string `text` into an array of words (using `text.split()` in Python, `text.split(" ")` in Java, `stringstream` in C++, or manual iteration in C).
2
Step 2: Iterate through each word in the array.
3
Step 3: For each word, iterate through the characters of `brokenLetters`.
4
Step 4: If a character from `brokenLetters` is found in the current word, set a flag (`canType`) to `false` and break the inner loop.
5
Step 5: If the flag (`canType`) is still `true` after checking all characters in `brokenLetters`, increment the `count` of typeable words.
6
Step 6: After processing all words, return the final `count`.

💡

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)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The outer loop iterates through each word in `text` (maximum of `m` iterations). The inner loop iterates through each character in `brokenLetters` (maximum of `n` iterations). The space used is proportional to the number of words generated by splitting the input string `text`.

📖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

Using Sets for faster lookup

Convert `brokenLetters` into a set for O(1) lookup time when checking if a character is broken. This slightly improves the time complexity.

Time: O(m + n) , where 'm' is the number of characters in `text` and 'n' is the number of characters in `brokenLetters`Space: O(n), to store the set of broken letters

Trade-offs: This approach uses slightly more space but improves time complexity. It is preferable when dealing with significantly larger inputs for `brokenLetters`.

✨

Key Insights

1Insight 1: The problem can be solved by iterating through each word in the `text` string and checking if any character in the word is present in the `brokenLetters` string.
2Insight 2: A simple boolean flag (`canType` in the provided solutions) can efficiently track whether a word contains a broken letter.
3Insight 3: String splitting is a crucial preprocessing step to efficiently handle each word individually.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

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

Edge Cases

Edge case 1: Empty `text` string. The solution correctly handles this by returning 0.
Edge case 2: Empty `brokenLetters` string. All words are typeable, so the solution correctly returns the number of words in `text`.
Edge case 3: A word containing multiple broken letters. The solution correctly identifies such words as untypeable because it exits the inner loop as soon as the first broken letter is found.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case of an empty `text` or `brokenLetters` string.

✗Common mistake 2: Not efficiently breaking the inner loop after finding a broken letter in a word (resulting in unnecessary iterations).

✅

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: Simulating a text input system with faulty keys, where you need to determine which words are still typeable.

🏦Real-world scenario 2: Analyzing user input to identify potential typing errors based on known faulty keys on a specific device.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

👨‍💻 From Experience

This is one of the first problems I solved when learning Hash Table. It's a great foundation for more complex problems.

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: October 14, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Maximum Number of Words You Can Type"?

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

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