What is the time complexity of Permutation in String?

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

What data structures are used to solve Permutation in String?

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

What are the key insights for solving Permutation in String?

Insight 1: The core idea is to recognize that we don't need to explicitly generate all permutations of `s1`. Instead, we can use a sliding window approach on `s2` and compare the character frequencies within the window to the character frequencies in `s1`. Insight 2: A hash table (or a fixed-size array in this case, given the limited character set) is an efficient way to store and compare character frequencies.

Which companies ask LeetCode 567. Permutation in String in interviews?

This problem is frequently asked at Agoda, Cisco, Expedia, Goldman Sachs, Revolut and 1 more companies.

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Permutation in String - LeetCode 567 Solution

Q: How to solve Permutation in String?

The provided solution employs a sliding window technique combined with character frequency counting. First, it checks if the length of `s1` exceeds that of `s2`. If it does, no permutation of `s1` can exist in `s2`, and the function returns `false`. Otherwise, it creates two frequency arrays (or hash maps) `s1_map` and `s2_map` to store the character counts for `s1` and the initial substring of `s2` with length equal to `s1`, respectively. It then compares these two arrays. If they are identical, it means that the initial substring of `s2` is a permutation of `s1`, and the function returns `true`. If not, it slides the window over `s2` by one character at a time. In each step, it updates `s2_map` by adding the frequency of the new character entering the window and subtracting the frequency of the character leaving the window. After each update, it compares `s1_map` and `s2_map` again. If they match at any point, the function returns `true`. If the loop completes without finding a match, it means that no permutation of `s1` exists in `s2`, and the function returns `false`.

Permutation in String - Complete Solution Guide

Permutation in String is LeetCode problem 567, 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 s1 and s2 , return true if s2 contains a permutation of s1 , or false otherwise. In other words, return true if one of s1 's permutations is the substring of s2 . Example 1: Input: s1 = "ab", s2 = "eidbaooo" Output: true Explanation: s2 contains one permutation of s1 ("ba"). Example 2: Input: s1 = "ab", s2 = "eidboaoo" Output: false Constraints: 1 <= s1.length, s2.length <= 10 4 s1 and s2 consist of lowercase English letters.

Detailed Explanation

The problem asks us to determine if a given string `s2` contains any permutation of another string `s1` as a substring. A permutation of `s1` is any arrangement of its characters. So, we need to check if any rearrangement of the characters in `s1` exists consecutively within `s2`. The input consists of two strings, `s1` and `s2`, containing lowercase English letters. The output is a boolean value: `true` if a permutation of `s1` is found within `s2`, and `false` otherwise. The constraints specify that the lengths of `s1` and `s2` are between 1 and 10,000, inclusive.

Solution Approach

The provided solution employs a sliding window technique combined with character frequency counting. First, it checks if the length of `s1` exceeds that of `s2`. If it does, no permutation of `s1` can exist in `s2`, and the function returns `false`. Otherwise, it creates two frequency arrays (or hash maps) `s1_map` and `s2_map` to store the character counts for `s1` and the initial substring of `s2` with length equal to `s1`, respectively. It then compares these two arrays. If they are identical, it means that the initial substring of `s2` is a permutation of `s1`, and the function returns `true`. If not, it slides the window over `s2` by one character at a time. In each step, it updates `s2_map` by adding the frequency of the new character entering the window and subtracting the frequency of the character leaving the window. After each update, it compares `s1_map` and `s2_map` again. If they match at any point, the function returns `true`. If the loop completes without finding a match, it means that no permutation of `s1` exists in `s2`, and the function returns `false`.

Step-by-Step Algorithm

Step 1: Check if `len(s1)` > `len(s2)`. If true, return `false` because a permutation of a longer string cannot be a substring of a shorter string.
Step 2: Initialize two frequency arrays, `s1_map` and `s2_map`, of size 26 (for the 26 lowercase English letters).
Step 3: Populate `s1_map` with the character frequencies of `s1`.
Step 4: Populate `s2_map` with the character frequencies of the first `len(s1)` characters of `s2`.
Step 5: Compare `s1_map` and `s2_map`. If they are equal, return `true` because the initial substring of `s2` is a permutation of `s1`.
Step 6: Iterate from `i = len(s1)` to `len(s2) - 1`. In each iteration:
Step 7: Increment the frequency of `s2[i]` in `s2_map` and decrement the frequency of `s2[i - len(s1)]` in `s2_map`. This is the sliding window update.
Step 8: Compare `s1_map` and `s2_map`. If they are equal, return `true`.
Step 9: If the loop completes without finding a match, return `false`.

Key Insights

Insight 1: The core idea is to recognize that we don't need to explicitly generate all permutations of `s1`. Instead, we can use a sliding window approach on `s2` and compare the character frequencies within the window to the character frequencies in `s1`.
Insight 2: A hash table (or a fixed-size array in this case, given the limited character set) is an efficient way to store and compare character frequencies.
Insight 3: If the length of `s1` is greater than `s2`, there can't be any permutation of `s1` inside `s2`. This is an important initial check for optimization.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Agoda, Cisco, Expedia, Goldman Sachs, Revolut, Yandex.

Permutation in String | LeetCode 567

Medium

Hash Table Two Pointers String Sliding Window

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

Solution Code

Available in:

class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int len1 = s1.length();
        int len2 = s2.length();
        if (len1 > len2) {
            return false;
        }

        int[] s1_map = new int[26];
        int[] s2_map = new int[26];

        for (int i = 0; i < len1; i++) {
            s1_map[s1.charAt(i) - 'a']++;
            s2_map[s2.charAt(i) - 'a']++;
        }

        if (java.util.Arrays.equals(s1_map, s2_map)) {
            return true;
        }

        for (int i = len1; i < len2; i++) {
            s2_map[s2.charAt(i) - 'a']++;
            s2_map[s2.charAt(i - len1) - 'a']--;

            if (java.util.Arrays.equals(s1_map, s2_map)) {
                return true;
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 567: Permutation in String - Detailed explanation and solution approaches.

Topics

Hash Table Two Pointers String Sliding Window

Companies

AgodaCiscoExpediaGoldman SachsRevolutYandex

Stats

Difficulty:Medium

Languages:4

Companies:6

Problem #:567

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 `len(s1)` > `len(s2)`. If true, return `false` because a permutation of a longer string cannot be a substring of a shorter string.
2
Step 2: Initialize two frequency arrays, `s1_map` and `s2_map`, of size 26 (for the 26 lowercase English letters).
3
Step 3: Populate `s1_map` with the character frequencies of `s1`.
4
Step 4: Populate `s2_map` with the character frequencies of the first `len(s1)` characters of `s2`.
5
Step 5: Compare `s1_map` and `s2_map`. If they are equal, return `true` because the initial substring of `s2` is a permutation of `s1`.
6
Step 6: Iterate from `i = len(s1)` to `len(s2) - 1`. In each iteration:
7
Step 7: Increment the frequency of `s2[i]` in `s2_map` and decrement the frequency of `s2[i - len(s1)]` in `s2_map`. This is the sliding window update.
8
Step 8: Compare `s1_map` and `s2_map`. If they are equal, return `true`.
9
Step 9: If the loop completes without finding a match, return `false`.

💡

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 dominant operation is iterating through `s2` using the sliding window. The frequency array operations (incrementing, decrementing, and comparing) take constant time because the size of the alphabet is fixed. Hence, the time complexity is linear with respect to the length of `s2`.

📖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

Sort both `s1` and substrings of `s2` of length `len(s1)`. Then, compare the sorted strings. This checks if the substring is a permutation.

Time: O(n * m log m), where n is the length of `s2` and m is the length of `s1`. The outer loop is O(n), and sorting the substring is O(m log m).Space: O(m) for the sorted substring.

Trade-offs: Sorting is simpler to understand but less efficient in terms of time complexity compared to the frequency map approach, especially when the lengths of s1 and s2 are large. The sorting approach involves repeated sorting operations. Use this approach when the code simplicity is prioritised, but when time complexity is a major concern, prefer the main solution approach.

✨

Key Insights

1Insight 1: The core idea is to recognize that we don't need to explicitly generate all permutations of `s1`. Instead, we can use a sliding window approach on `s2` and compare the character frequencies within the window to the character frequencies in `s1`.
2Insight 2: A hash table (or a fixed-size array in this case, given the limited character set) is an efficient way to store and compare character frequencies.
3Insight 3: If the length of `s1` is greater than `s2`, there can't be any permutation of `s1` inside `s2`. This is an important initial check for optimization.

📚

Key Terms Explained

For Students

Hash Table

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

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: `s1` is an empty string. The code correctly handles this because the initial length check `if len1 > len2` will return `false`. Thus, when s1 is empty, if s2 is also empty, it will return `false`. However if s2 is non-empty, the code will incorrectly calculate the frequencies for the initial s2_map based on s1, resulting in wrong outputs. To make the code robust, we need to add a specific check if s1 is empty return True immediately if s2 is non-empty and return false if s2 is empty.
Edge case 2: `s2` is an empty string and `s1` is not. The code correctly handles this because the initial length check will return `false`.
Edge case 3: `s1` and `s2` are equal. The initial comparison of `s1_map` and `s2_map` will find a match, and the code will return `true` efficiently.
Edge case 4: `s1` consists of the same repeating character, and `s2` contains a substring with the same repeating character. For example, `s1 = 'aaa'` and `s2 = 'baaaabc'`. The algorithm correctly finds the permutation.
Edge case 5: `s1` is longer than `s2`, resulting in no possible permutation. The code handles this with the initial length check.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly updating the frequency array `s2_map` during the sliding window process. It's important to add the new character and subtract the old character correctly.

✗Common mistake 2: Forgetting the initial length check `len(s1) > len(s2)`, which is crucial for optimization and avoids unnecessary computations.

✗Common mistake 3: Not handling the case where either `s1` or `s2` is empty. This can lead to errors when accessing characters beyond the string boundaries.

✅

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 searching. This algorithm can be adapted to search for approximate matches of a pattern in a large text, where the pattern may have slight variations.

🏦Real-world scenario 2: Anagram detection. Determining if one string is an anagram of another is a specific case of permutation finding. This can be used in word games or plagiarism detection.

🎮Real-world scenario 3: DNA sequence analysis. Identifying specific patterns or sequences within a DNA strand, where variations (permutations) of a core sequence might be of interest.

💡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: June 28, 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.

@InterviewProJun 12

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

@CodeNinjaJun 5

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

@TechLeadMay 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.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Permutation in 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 two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?