What is the time complexity of Find All Anagrams in a String?

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

What data structures are used to solve Find All Anagrams in a String?

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

What are the key insights for solving Find All Anagrams in a String?

Insight 1: Using a sliding window approach allows us to efficiently scan through the string `s` without recomputing character frequencies for each substring. Insight 2: Representing character frequencies using an array of fixed size (26 for lowercase English letters) enables constant-time comparisons to check if two strings are anagrams.

Which companies ask LeetCode 438. Find All Anagrams in a String in interviews?

This problem is frequently asked at Bolt, Databricks, Intuit, Revolut, Snowflake and 3 more companies.

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Find All Anagrams in a String - LeetCode 438 Solution

Find All Anagrams in a String - Complete Solution Guide

Find All Anagrams in a String is LeetCode problem 438, 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

Given two strings s and p , return an array of all the start indices of p 's anagrams in s . You may return the answer in any order . Example 1: Input: s = "cbaebabacd", p = "abc" Output: [0,6] Explanation: The substring with start index = 0 is "cba", which is an anagram of "abc". The substring with start index = 6 is "bac", which is an anagram of "abc". Example 2: Input: s = "abab", p = "ab" Output: [0,1,2] Explanation: The substring with start index = 0 is "ab", which is an anagram of "ab". Th

Detailed Explanation

The problem requires you to find all starting indices of substrings within a given string `s` that are anagrams of another given string `p`. An anagram is a rearrangement of the letters in a word or phrase to produce a new word or phrase, using all the original letters exactly once. The input consists of two strings, `s` and `p`, and the output is a list of integer indices indicating the starting positions of the anagrams of `p` found in `s`. The length of both strings is between 1 and 3 * 10^4, and they contain only lowercase English letters.

Solution Approach

The provided solution employs a sliding window technique coupled with frequency counting to solve this problem. It initializes frequency maps for both the pattern `p` and the first window (substring of length `p`) in the string `s`. Then, it slides the window one character at a time across `s`, updating the frequency map for the current window by adding the new character and removing the leaving character. In each step, it checks if the current window's frequency map is identical to that of `p`. If they match, it means an anagram is found, and the starting index of the window is added to the result.

Step-by-Step Algorithm

Step 1: Check if the length of `s` is less than the length of `p`. If it is, return an empty list since `p` cannot be an anagram of any substring of `s`.
Step 2: Initialize two frequency arrays (or hash maps) `p_count` and `s_count` of size 26 (one for each lowercase English letter).
Step 3: Populate `p_count` with the frequencies of characters in `p`.
Step 4: Populate `s_count` with the frequencies of the first `len(p)` characters in `s`. This represents the initial sliding window.
Step 5: Compare `p_count` and `s_count`. If they are equal (meaning they represent the same character frequencies), add 0 (the starting index of the first window) to the result list.
Step 6: Iterate from index `len(p)` to the end of `s`. In each iteration:
Step 7: - Update `s_count` by incrementing the frequency of the character entering the window (the current character in `s`) and decrementing the frequency of the character leaving the window (the character at index `i - len(p)`).
Step 8: - Compare `p_count` and `s_count`. If they are equal, add `i - len(p) + 1` (the starting index of the current window) to the result list.
Step 9: Return the result list.

Key Insights

Insight 1: Using a sliding window approach allows us to efficiently scan through the string `s` without recomputing character frequencies for each substring.
Insight 2: Representing character frequencies using an array of fixed size (26 for lowercase English letters) enables constant-time comparisons to check if two strings are anagrams.
Insight 3: Pre-calculating the character frequency map for the pattern `p` allows us to compare it efficiently with each window in `s`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Sliding Window.

Companies

Asked at: Bolt, Databricks, Intuit, Revolut, Snowflake, Splunk, Turing, Yandex.

Find All Anagrams in a String | LeetCode 438

Medium

Hash Table String Sliding Window

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

Solution Code

Available in:

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

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        int ns = s.length();
        int np = p.length();
        if (ns < np) {
            return new ArrayList<>();
        }

        int[] p_count = new int[26];
        int[] s_count = new int[26];
        
        // Build frequency map for p and the first window in s
        for (int i = 0; i < np; i++) {
            p_count[p.charAt(i) - 'a']++;
            s_count[s.charAt(i) - 'a']++;
        }

        List<Integer> result = new ArrayList<>();
        // Check if the first window is an anagram
        if (Arrays.equals(p_count, s_count)) {
            result.add(0);
        }

        // Slide the window through the rest of the string
        for (int i = np; i < ns; i++) {
            // Add the new character entering the window from the right
            s_count[s.charAt(i) - 'a']++;
            // Remove the character leaving the window from the left
            s_count[s.charAt(i - np) - 'a']--;

            // Check if the current window's frequency map matches p's
            if (Arrays.equals(p_count, s_count)) {
                // The start index of the current window is (i - np + 1)
                result.add(i - np + 1);
            }
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 438: Find All Anagrams in a String - Detailed explanation and solution approaches.

Topics

Hash Table String Sliding Window

Companies

BoltDatabricksIntuitRevolutSnowflakeSplunkTuringYandex

Stats

Difficulty:Medium

Languages:4

Companies:8

Problem #:438

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: Check if the length of `s` is less than the length of `p`. If it is, return an empty list since `p` cannot be an anagram of any substring of `s`.
2
Step 2: Initialize two frequency arrays (or hash maps) `p_count` and `s_count` of size 26 (one for each lowercase English letter).
3
Step 3: Populate `p_count` with the frequencies of characters in `p`.
4
Step 4: Populate `s_count` with the frequencies of the first `len(p)` characters in `s`. This represents the initial sliding window.
5
Step 5: Compare `p_count` and `s_count`. If they are equal (meaning they represent the same character frequencies), add 0 (the starting index of the first window) to the result list.
6
Step 6: Iterate from index `len(p)` to the end of `s`. In each iteration:
7
Step 7: - Update `s_count` by incrementing the frequency of the character entering the window (the current character in `s`) and decrementing the frequency of the character leaving the window (the character at index `i - len(p)`).
8
Step 8: - Compare `p_count` and `s_count`. If they are equal, add `i - len(p) + 1` (the starting index of the current window) to the result list.
9
Step 9: Return the result list.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the single loop that iterates `n - len(p)` times where `n` is the length of `s`. Inside the loop, we perform constant-time operations (incrementing/decrementing counts and comparing arrays of size 26). The space complexity is constant because the size of the frequency arrays does not depend on the input size; it is fixed at 26.

📖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

Sorting

For each substring of `s` with length `len(p)`, sort the substring and `p` and compare them. If the sorted strings are equal, it's an anagram.

Time: O(n * k log k), where n is the length of `s` and k is the length of `p`. This comes from sorting each of the n-k+1 windows of size k.Space: O(k), for sorting and creating the substring.

Trade-offs: The sorting approach is less efficient than the sliding window approach because of the repeated sorting operations. However, it is potentially simpler to understand.

✨

Key Insights

1Insight 1: Using a sliding window approach allows us to efficiently scan through the string `s` without recomputing character frequencies for each substring.
2Insight 2: Representing character frequencies using an array of fixed size (26 for lowercase English letters) enables constant-time comparisons to check if two strings are anagrams.
3Insight 3: Pre-calculating the character frequency map for the pattern `p` allows us to compare it efficiently with each window in `s`.

📚

Key Terms Explained

For Students

Hash Table

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

Sliding Window

A moving "window" over data to process subsets efficiently.

String

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

Edge Cases

Edge case 1: `s` is shorter than `p`. The code correctly handles this by returning an empty list, as an anagram of `p` cannot exist within `s`.
Edge case 2: `s` and `p` are empty strings. Although the problem statement states that the string lengths will be at least 1, a robust solution should consider this. The provided code would likely have an issue if s and p were empty, but this falls outside the problem constraints.
Edge case 3: `s` and `p` are the same. The solution correctly identifies the index 0 as an anagram (the entire string is an anagram of itself).
Edge case 4: `p` contains repeated characters (e.g., "aab"). The frequency counting approach accurately handles this.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to update the frequency count when the window slides. It's crucial to both increment the count of the new character entering the window and decrement the count of the character leaving the window.

✗Common mistake 2: Incorrectly calculating the starting index of the anagram. It should be `i - len(p) + 1` where `i` is the current index in the loop.

✗Common mistake 3: Not handling the edge case where the length of `s` is less than the length of `p`.

✗Common mistake 4: Premature optimization. Beginners might try to manually compare the arrays to stop early. Comparing entire array is only 26 compares and already quite fast. Trying to optimize further may introduce bugs.

✅

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: Text analysis and information retrieval. Finding anagrams can be used to identify variations of words or phrases in a text corpus, which is helpful for search engines and document analysis.

🏦Real-world scenario 2: Cryptography. Anagram analysis can be used in some simple forms of code breaking, such as analyzing substitution ciphers.

🎮Real-world scenario 3: Bioinformatics. Analyzing DNA sequences can involve searching for specific patterns or motifs, and identifying anagrams of these motifs can be relevant in some biological contexts.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

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 18, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

❓

Frequently Asked Questions

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 do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 All Anagrams in a String"?

For this medium-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.

🤖 Need help with this problem?