What is the time complexity of Count the Number of Consistent Strings?

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

What data structures are used to solve Count the Number of Consistent Strings?

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

What are the key insights for solving Count the Number of Consistent Strings?

Using a set for `allowed` allows for O(1) lookup time to check if a character is allowed. Iterating through each word and checking each character against the `allowed` set is a straightforward approach.

Which companies ask LeetCode 1684. Count the Number of Consistent Strings in interviews?

This problem is frequently asked at Robinhood.

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Count the Number of Consistent Strings - LeetCode 1684 Solution

Count the Number of Consistent Strings - Complete Solution Guide

Count the Number of Consistent Strings is LeetCode problem 1684, 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 a string allowed consisting of distinct characters and an array of strings words . A string is consistent if all characters in the string appear in the string allowed . Return the number of consistent strings in the array words . Example 1: Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"] Output: 2 Explanation: Strings "aaab" and "baa" are consistent since they only contain characters 'a' and 'b'. Example 2: Input: allowed = "abc", words = ["a","b","c","ab","ac","bc"

Detailed Explanation

The problem asks you to count the number of strings in a given array `words` that are considered "consistent." A string is consistent if all its characters are present in another string, `allowed`. The `allowed` string contains only unique characters. The input consists of the `allowed` string and an array of strings `words`. The output is a single integer representing the count of consistent strings.

Solution Approach

The provided solutions all follow a similar approach: they iterate through each string in the `words` array. For each string, they iterate through its characters, checking if each character is present in the `allowed` string. If a character is not found in `allowed`, the string is marked as inconsistent, and the inner loop breaks. If the inner loop completes without finding any disallowed characters, the string is counted as consistent. Different implementations use different data structures (sets, boolean arrays) to optimize the character lookup process.

Step-by-Step Algorithm

Create a data structure (set or boolean array) to efficiently store and check the characters in `allowed`.
Iterate through each string in the `words` array.
For each string, iterate through its characters.
For each character, check if it's present in the `allowed` data structure.
If a character is not found in `allowed`, mark the string as inconsistent and break the inner loop.
If the inner loop completes without finding any inconsistent characters, increment the count of consistent strings.
Return the final count of consistent strings.

Key Insights

Using a set for `allowed` allows for O(1) lookup time to check if a character is allowed.
Iterating through each word and checking each character against the `allowed` set is a straightforward approach.
The problem can be efficiently solved using a boolean array to represent allowed characters instead of a set for potential memory optimizations, especially for larger alphabets.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Robinhood.

Count the Number of Consistent Strings | LeetCode 1684

Easy

Array Hash Table String Bit Manipulation Counting

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

Solution Code

Available in:

class Solution {
    public int countConsistentStrings(String allowed, String[] words) {
        int count = 0;
        boolean[] allowedChars = new boolean[26];
        for (char c : allowed.toCharArray()) {
            allowedChars[c - 'a'] = true;
        }

        for (String word : words) {
            boolean consistent = true;
            for (char c : word.toCharArray()) {
                if (!allowedChars[c - 'a']) {
                    consistent = false;
                    break;
                }
            }
            if (consistent) {
                count++;
            }
        }

        return count;
    }
}

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Problem Info

Description

Problem 1684: Count the Number of Consistent Strings - Detailed explanation and solution approaches.

Topics

Array Hash Table String Bit Manipulation Counting

Companies

Robinhood

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1684

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

7 steps

1
Create a data structure (set or boolean array) to efficiently store and check the characters in `allowed`.
2
Iterate through each string in the `words` array.
3
For each string, iterate through its characters.
4
For each character, check if it's present in the `allowed` data structure.
5
If a character is not found in `allowed`, mark the string as inconsistent and break the inner loop.
6
If the inner loop completes without finding any inconsistent characters, increment the count of consistent strings.
7
Return the final count of consistent strings.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The outer loop iterates `n` times (once for each word). The inner loop iterates at most `m` times (for the longest word). The set/boolean array has a constant space complexity because the size of the alphabet is fixed (26 lowercase English letters).

📖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

Bit Manipulation

Represent allowed characters using a single integer (32 bits would suffice for the 26 characters). Set bits corresponding to allowed characters. For each word, check if all characters' bits are set within the integer representation.

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

Trade-offs: Slightly more complex implementation, but can offer a marginal performance improvement due to bitwise operations.

✨

Key Insights

1Using a set for `allowed` allows for O(1) lookup time to check if a character is allowed.
2Iterating through each word and checking each character against the `allowed` set is a straightforward approach.
3The problem can be efficiently solved using a boolean array to represent allowed characters instead of a set for potential memory optimizations, especially for larger alphabets.

📚

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 `allowed` string: The solution correctly handles this case; no strings will be consistent.
Empty `words` array: The solution correctly returns 0 in this case.
Words containing characters outside the lowercase alphabet: The solutions provided assume only lowercase characters. Robust solutions should handle this scenario explicitly, or clearly state the assumption within the problem definition.

⚠️

Common Mistakes to Avoid

✗Forgetting to handle the case where a word is empty: Though the problem statement implies non-empty words, robust code should account for this possibility.

✗Inefficient character lookup: Using nested loops to check for character presence in `allowed` without using a set or other optimized data structure leads to O(m*n*k) time complexity, where k is the length of allowed, making the solution less efficient.

✅

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?

💼Data validation: Checking if input strings conform to a predefined set of allowed characters.

🏦Text processing: Filtering words based on specific allowed characters (e.g., removing non-alphanumeric 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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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 4, 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

❓

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 "Count the Number of Consistent Strings"?

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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