What is the time complexity of Find Common Characters?

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

What data structures are used to solve Find Common Characters?

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

What are the key insights for solving Find Common Characters?

The 'bottleneck' frequency: For any given character, its common count across all strings is dictated by its *lowest* frequency in any single string. If 'x' appears 5 times in word A, 2 times in word B, and 8 times in word C, then 'x' is common only 2 times. We must find this minimum across all words for each character. First word as the candidate filter: By initially building a frequency map only for `words[0]`, we automatically filter out any characters that cannot possibly be common to *all* strings. If a character isn't even in the first word, it certainly can't be in every word, saving unnecessary checks.

Which companies ask LeetCode 1002. Find Common Characters in interviews?

This problem is frequently asked at Tripadvisor.

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Find Common Characters - LeetCode 1002 Solution

Q: How to solve Find Common Characters?

The provided solution leverages the idea that if a character is to be common across *all* strings, it *must* at least be present in the very first string. This is our starting point. We first build a frequency map for the characters in `words[0]`. This map then serves as our candidate pool for common characters. For each character identified in this first word, we then iterate through *all* subsequent words in the `words` array. In each subsequent word, we count how many times that specific character appears. We continuously track the *minimum* count found for that character across all words. Once we've checked every word, this `min_count` represents the maximum number of times that character can be considered truly 'common'. Finally, we append this character to our result list `min_count` times, effectively incorporating the duplicate requirement. This process repeats for every unique character initially found in `words[0]`.

Find Common Characters - Complete Solution Guide

Find Common Characters is LeetCode problem 1002, 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

Given a string array words , return an array of all characters that show up in all strings within the words (including duplicates) . You may return the answer in any order . Example 1: Input: words = ["bella","label","roller"] Output: ["e","l","l"] Example 2: Input: words = ["cool","lock","cook"] Output: ["c","o"] Constraints: 1 <= words.length <= 100 1 <= words[i].length <= 100 words[i] consists of lowercase English letters.

Detailed Explanation

This problem asks us to identify all characters that are present in *every single string* within a given input array `words`. The critical detail here, which often trips up newcomers, is the 'including duplicates' clause. It's not just about whether a character exists in all strings, but rather, how many times it can *simultaneously* exist in all of them. For instance, if 'l' appears twice in "bella", twice in "label", and twice in "roller", then 'l' is common twice. But if 'o' appears twice in "cool", once in "lock", and twice in "cook", then 'o' is only common once, limited by its single appearance in "lock". We need to collect these characters, respecting their minimal common frequency, and return them as a list.

Solution Approach

The provided solution leverages the idea that if a character is to be common across *all* strings, it *must* at least be present in the very first string. This is our starting point. We first build a frequency map for the characters in `words[0]`. This map then serves as our candidate pool for common characters. For each character identified in this first word, we then iterate through *all* subsequent words in the `words` array. In each subsequent word, we count how many times that specific character appears. We continuously track the *minimum* count found for that character across all words. Once we've checked every word, this `min_count` represents the maximum number of times that character can be considered truly 'common'. Finally, we append this character to our result list `min_count` times, effectively incorporating the duplicate requirement. This process repeats for every unique character initially found in `words[0]`.

Step-by-Step Algorithm

Step 1: Initialize a frequency array (or hash map) to store the minimum frequency of each character (a-z). Initialize all frequencies to a high value (e.g., Integer.MAX_VALUE in Java, INT_MAX in C++).
Step 2: Iterate through each string in the input array.
Step 3: For each string, create a temporary frequency array (or hash map) to count character occurrences within that string.
Step 4: Update the minimum frequency array by comparing the current character frequencies with the minimum frequencies found so far. Use `min()` function to find the minimum frequency for each character across all strings processed so far.
Step 5: After processing all strings, iterate through the minimum frequency array. For each character, append it to the result array as many times as its minimum frequency indicates.
Step 6: Return the result array.

Key Insights

The 'bottleneck' frequency: For any given character, its common count across all strings is dictated by its *lowest* frequency in any single string. If 'x' appears 5 times in word A, 2 times in word B, and 8 times in word C, then 'x' is common only 2 times. We must find this minimum across all words for each character.
First word as the candidate filter: By initially building a frequency map only for `words[0]`, we automatically filter out any characters that cannot possibly be common to *all* strings. If a character isn't even in the first word, it certainly can't be in every word, saving unnecessary checks.
Iterative frequency intersection: Instead of generating full frequency maps for every single word and then merging them, this solution iteratively refines the count for each candidate character. It picks a character from the first word, then successively reduces its `min_count` as it checks its presence in subsequent words. This direct, character-by-character approach elegantly finds the shared minimum frequencies.

Complexity Analysis

Time Complexity: O(n*m*k)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, String.

Companies

Asked at: Tripadvisor.

Find Common Characters | LeetCode 1002

Easy

Array Hash Table String

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {
    public List<String> commonChars(String[] words) {
        int[] minFreq = new int[26];
        Arrays.fill(minFreq, Integer.MAX_VALUE);

        for (String word : words) {
            int[] charFreq = new int[26];
            for (char c : word.toCharArray()) {
                charFreq[c - 'a']++;
            }
            for (int i = 0; i < 26; i++) {
                minFreq[i] = Math.min(minFreq[i], charFreq[i]);
            }
        }

        List<String> commonChars = new ArrayList<>();
        for (int i = 0; i < 26; i++) {
            for (int j = 0; j < minFreq[i]; j++) {
                commonChars.add(String.valueOf((char) ('a' + i)));
            }
        }

        return commonChars;
    }
}

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Problem Info

Description

Problem 1002: Find Common Characters - Detailed explanation and solution approaches.

Topics

Array Hash Table String

Companies

Tripadvisor

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1002

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: Initialize a frequency array (or hash map) to store the minimum frequency of each character (a-z). Initialize all frequencies to a high value (e.g., Integer.MAX_VALUE in Java, INT_MAX in C++).
2
Step 2: Iterate through each string in the input array.
3
Step 3: For each string, create a temporary frequency array (or hash map) to count character occurrences within that string.
4
Step 4: Update the minimum frequency array by comparing the current character frequencies with the minimum frequencies found so far. Use `min()` function to find the minimum frequency for each character across all strings processed so far.
5
Step 5: After processing all strings, iterate through the minimum frequency array. For each character, append it to the result array as many times as its minimum frequency indicates.
6
Step 6: Return the result array.

💡

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*k)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant factor in time complexity is the nested loops: iterating through words (n), characters in each word (m), and then potentially comparing frequencies for each character (k). Space complexity is constant because we're using a fixed-size array to store character frequencies. The size of the output array is also bounded by the input size, but the space used to construct the result does not dominate the overall space complexity.

📖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 a hashmap for counting

Instead of an array, we can use a hashmap to store character frequencies. This makes the code slightly more readable but potentially less efficient due to hash map lookups. The Python solution uses this approach.

Time: O(m*n*k)Space: O(k), where k is the number of unique characters (at most 26)

Trade-offs: Hashmaps offer better readability but might have slightly higher constant time factors compared to arrays for this specific problem. The difference is negligible for small alphabets like English lowercase letters.

✨

Key Insights

1The 'bottleneck' frequency: For any given character, its common count across all strings is dictated by its *lowest* frequency in any single string. If 'x' appears 5 times in word A, 2 times in word B, and 8 times in word C, then 'x' is common only 2 times. We must find this minimum across all words for each character.
2First word as the candidate filter: By initially building a frequency map only for `words[0]`, we automatically filter out any characters that cannot possibly be common to *all* strings. If a character isn't even in the first word, it certainly can't be in every word, saving unnecessary checks.
3Iterative frequency intersection: Instead of generating full frequency maps for every single word and then merging them, this solution iteratively refines the count for each candidate character. It picks a character from the first word, then successively reduces its `min_count` as it checks its presence in subsequent words. This direct, character-by-character approach elegantly finds the shared minimum frequencies.

📚

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

Edge case 1: Empty input array. The solution handles this gracefully; it returns an empty array.
Edge case 2: Strings with no common characters. The solution correctly returns an empty array in this scenario.
Edge case 3: Strings with very long lengths (approaching 100). The solution will still function correctly, but the time taken will increase linearly with string length.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating minimum frequencies. Failing to properly update the minimum frequency for each character across all strings leads to incorrect results.

✗Common mistake 2: Not handling edge cases like empty input or strings with no common characters. Robust code should explicitly consider 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?

💼Real-world scenario 1: Analyzing text data to find common words or characters across multiple documents. This is helpful in natural language processing tasks like identifying keywords or analyzing writing style.

🏦Real-world scenario 2: In database systems, finding common attributes across multiple datasets for data integration or comparison purposes.

💡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

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

⚠️ 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: September 11, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 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 Common 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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