What is the time complexity of Find Words That Can Be Formed by Characters?

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

What data structures are used to solve Find Words That Can Be Formed by Characters?

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

What are the key insights for solving Find Words That Can Be Formed by Characters?

**Frequency Mapping for Availability:** The core of solving this problem lies in accurately tracking character counts. By first creating a frequency map of `chars`, we establish a clear 'budget' of available characters. This allows for quick, O(1) lookups to determine if a character exists and in what quantity. **Per-Word Character Usage:** A crucial understanding is that the character availability from `chars` 'resets' for each word. This means we aren't cumulatively depleting `chars` across the entire `words` array. Our comparison for each word should always be against the *original* `char_counts` derived from `chars`, effectively granting each word a fresh pool of characters.

Which companies ask LeetCode 1160. Find Words That Can Be Formed by Characters in interviews?

This problem is frequently asked at Atlassian, Datadog, Karat.

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Find Words That Can Be Formed by Characters - LeetCode 1160 Solution

Find Words That Can Be Formed by Characters - Complete Solution Guide

Find Words That Can Be Formed by Characters is LeetCode problem 1160, 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

You are given an array of strings words and a string chars . A string is good if it can be formed by characters from chars (each character can only be used once for each word in words ). Return the sum of lengths of all good strings in words . Example 1: Input: words = ["cat","bt","hat","tree"], chars = "atach" Output: 6 Explanation: The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6. Example 2: Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr" Outp

Detailed Explanation

This problem asks us to sift through a list of `words` and identify which ones can be 'formed' using the characters available in a separate string, `chars`. A word is considered 'good' if we can spell it out entirely using characters from `chars`. The critical detail here is that each character from `chars` can only be utilized once *per word*. This isn't a global depletion scenario; if `chars` has two 'a's, both 'cat' and 'hat' could potentially use an 'a' without 'cat' exhausting the 'a' supply for 'hat'. We ultimately need to sum up the lengths of all such 'good' words.

Solution Approach

The most intuitive and efficient way to handle character availability and usage is through frequency counting. The provided solution first builds a comprehensive frequency map, `char_counts`, for the `chars` string. This map acts as our 'budget' – it tells us exactly how many of each character we have at our disposal. Then, for each `word` in the `words` array, the solution performs a character-by-character check. It constructs a temporary frequency map, `word_counts`, for the *current* word. As it processes each character in the word, it immediately compares the character's count in `word_counts` against its available count in `char_counts`. If a character is either not present in `chars` at all, or the current word requires more of that character than `chars` provides, we know this word cannot be formed. In such cases, a flag `is_good` is set to `False`, and we efficiently break out of the inner loop, moving to the next word. If the word is fully processed and `is_good` remains `True`, it means we successfully 'budgeted' all its characters, and its length is added to our `total_length` sum.

Step-by-Step Algorithm

Create a frequency map (dictionary in Python, array in Java/C/C++) to store the counts of each character in the input string `chars`.
Iterate through each word in the `words` array.
For each word, create a frequency map of its characters.
Compare the word's character frequencies with the frequencies in `chars`. If any character in the word has a higher frequency than in `chars`, mark the word as 'bad' and proceed to the next word.
If a word is deemed 'good', add its length to the `total_length`.
Return the final `total_length`.

Key Insights

**Frequency Mapping for Availability:** The core of solving this problem lies in accurately tracking character counts. By first creating a frequency map of `chars`, we establish a clear 'budget' of available characters. This allows for quick, O(1) lookups to determine if a character exists and in what quantity.
**Per-Word Character Usage:** A crucial understanding is that the character availability from `chars` 'resets' for each word. This means we aren't cumulatively depleting `chars` across the entire `words` array. Our comparison for each word should always be against the *original* `char_counts` derived from `chars`, effectively granting each word a fresh pool of characters.
**Early Exit Optimization:** For any given `word`, as soon as we encounter a character that isn't available in `chars` or is needed in a quantity greater than what `chars` provides, we know immediately that the word cannot be formed. There's no need to continue checking the rest of that `word`'s characters. Breaking out of the inner loop early significantly improves efficiency, especially for longer words that are quickly found to be 'not good'.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, String, Counting.

Companies

Asked at: Atlassian, Datadog, Karat.

Find Words That Can Be Formed by Characters | LeetCode 1160

Easy

Array Hash Table String Counting

Time: O(n*k)Space: O(1)

Solution Code

Available in:

class Solution {
    public int countCharacters(String[] words, String chars) {
        int[] charCounts = new int[26];
        for (char c : chars.toCharArray()) {
            charCounts[c - 'a']++;
        }

        int totalLength = 0;
        for (String word : words) {
            int[] wordCounts = new int[26];
            boolean good = true;
            for (char c : word.toCharArray()) {
                int index = c - 'a';
                wordCounts[index]++;
                if (wordCounts[index] > charCounts[index]) {
                    good = false;
                    break;
                }
            }
            if (good) {
                totalLength += word.length();
            }
        }

        return totalLength;
    }
}

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Problem Info

Description

Problem 1160: Find Words That Can Be Formed by Characters - Detailed explanation and solution approaches.

Topics

Array Hash Table String Counting

Companies

AtlassianDatadogKarat

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:1160

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
Create a frequency map (dictionary in Python, array in Java/C/C++) to store the counts of each character in the input string `chars`.
2
Iterate through each word in the `words` array.
3
For each word, create a frequency map of its characters.
4
Compare the word's character frequencies with the frequencies in `chars`. If any character in the word has a higher frequency than in `chars`, mark the word as 'bad' and proceed to the next word.
5
If a word is deemed 'good', add its length to the `total_length`.
6
Return the final `total_length`.

💡

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*k)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The frequency maps are of fixed size (26 elements for lowercase letters). The iteration over `words` and characters within the words dominates the time complexity. The space used is independent of the lengths of `words` or `chars`, making it O(1).

📖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

Instead of frequency maps, we could use sets to check if the characters of a word are a subset of characters in `chars`. This would require additional processing to handle character duplication and wouldn't be as efficient as frequency counting.

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

Trade-offs: This approach might be simpler to understand but less efficient than frequency counting, especially for larger inputs. Frequency counting provides better performance for scenarios with many duplicate characters.

✨

Key Insights

1**Frequency Mapping for Availability:** The core of solving this problem lies in accurately tracking character counts. By first creating a frequency map of `chars`, we establish a clear 'budget' of available characters. This allows for quick, O(1) lookups to determine if a character exists and in what quantity.
2**Per-Word Character Usage:** A crucial understanding is that the character availability from `chars` 'resets' for each word. This means we aren't cumulatively depleting `chars` across the entire `words` array. Our comparison for each word should always be against the *original* `char_counts` derived from `chars`, effectively granting each word a fresh pool of characters.
3**Early Exit Optimization:** For any given `word`, as soon as we encounter a character that isn't available in `chars` or is needed in a quantity greater than what `chars` provides, we know immediately that the word cannot be formed. There's no need to continue checking the rest of that `word`'s characters. Breaking out of the inner loop early significantly improves efficiency, especially for longer words that are quickly found to be 'not good'.

📚

Key Terms Explained

For Students

Array

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

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

Empty `words` array: The solution correctly handles this edge case by returning 0.
Empty `chars` string: The solution correctly returns 0 because no words can be formed.

⚠️

Common Mistakes to Avoid

✗Forgetting to check if each character in the word exists within `chars` before incrementing the word count, potentially leading to incorrect results.

✗Not handling the edge case where `chars` is empty or `words` is empty. The solution needs to return 0 in these scenarios.

✅

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?

💼Text processing: Determining which words from a text can be formed using a limited set of available characters (e.g., checking spelling with a limited keyboard layout).

🏦Game development: In word games, it can be used to efficiently check if a player can form words from a given set of letters (Scrabble, Words with Friends).

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. 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: August 20, 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

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find Words That Can Be Formed by Characters"?

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

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