What is the time complexity of Existence of a Substring in a String and Its Reverse?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Existence of a Substring in a String and Its Reverse?

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

What are the key insights for solving Existence of a Substring in a String and Its Reverse?

Insight 1: Reversing the string is a crucial step to compare substrings against their reversed counterparts. Insight 2: A nested loop (implicit in the `contains` or `find` operation) or a sliding window approach can efficiently check all possible substrings of length 2.

Which companies ask LeetCode 3083. Existence of a Substring in a String and Its Reverse in interviews?

This problem is frequently asked at Rubrik.

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Existence of a Substring in a String and Its Reverse - LeetCode 3083 Solution

Existence of a Substring in a String and Its Reverse - Complete Solution Guide

Existence of a Substring in a String and Its Reverse is LeetCode problem 3083, 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 s , find any substring of length 2 which is also present in the reverse of s . Return true if such a substring exists, and false otherwise. Example 1: Input: s = "leetcode" Output: true Explanation: Substring "ee" is of length 2 which is also present in reverse(s) == "edocteel" . Example 2: Input: s = "abcba" Output: true Explanation: All of the substrings of length 2 "ab" , "bc" , "cb" , "ba" are also present in reverse(s) == "abcba" . Example 3: Input: s = "abcd" Output: false E

Detailed Explanation

The problem asks to determine if a given string `s` contains a substring of length 2 that is also present in the reverse of `s`. The input is a string `s` containing only lowercase English letters. The output is a boolean value: `true` if such a substring exists, and `false` otherwise. The constraints specify that the length of `s` is between 1 and 100 characters.

Solution Approach

The provided solutions all follow a similar approach. First, they check if the input string's length is less than 2; if so, they return `false` because no substring of length 2 can exist. Then, they reverse the input string. Finally, they iterate through the original string, extracting substrings of length 2 and checking if each substring is present in the reversed string using built-in functions like `contains` (Java), `find` (C++), or `in` (Python) or manual string comparison (C). If any substring is found in the reversed string, the function returns `true`; otherwise, it returns `false` after checking all possible substrings.

Step-by-Step Algorithm

Step 1: Check if the length of the input string `s` is less than 2. If it is, return `false`.
Step 2: Reverse the input string `s` and store it in a new string `reversed_s`.
Step 3: Iterate through the original string `s` using a loop, extracting substrings of length 2.
Step 4: For each substring, check if it exists in `reversed_s`. If it does, return `true`.
Step 5: If the loop completes without finding any matching substring, return `false`.

Key Insights

Insight 1: Reversing the string is a crucial step to compare substrings against their reversed counterparts.
Insight 2: A nested loop (implicit in the `contains` or `find` operation) or a sliding window approach can efficiently check all possible substrings of length 2.
Insight 3: Handling edge cases such as strings with length less than 2 is essential to avoid index out-of-bounds errors.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String.

Companies

Asked at: Rubrik.

Existence of a Substring in a String and Its Reverse | LeetCode 3083

Easy

Hash Table String

Time: O(n^2)Space: O(n)

Solution Code

Available in:

class Solution {
    public boolean isSubstringPresent(String s) {
        if (s.length() < 2) {
            return false;
        }

        String reversedS = new StringBuilder(s).reverse().toString();

        for (int i = 0; i < s.length() - 1; i++) {
            String sub = s.substring(i, i + 2);
            if (reversedS.contains(sub)) {
                return true;
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 3083: Existence of a Substring in a String and Its Reverse - Detailed explanation and solution approaches.

Topics

Hash Table String

Companies

Rubrik

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3083

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

5 steps

1
Step 1: Check if the length of the input string `s` is less than 2. If it is, return `false`.
2
Step 2: Reverse the input string `s` and store it in a new string `reversed_s`.
3
Step 3: Iterate through the original string `s` using a loop, extracting substrings of length 2.
4
Step 4: For each substring, check if it exists in `reversed_s`. If it does, return `true`.
5
Step 5: If the loop completes without finding any matching substring, 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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n^2) because the outer loop iterates n-1 times (where n is the length of the string), and the `contains` or `find` operation within the loop has a time complexity of O(n) in the worst case (linear search). The space complexity is O(n) because we create a reversed string of length n. In Python, string slicing `[::-1]` creates a new string in O(n) space. In other languages, using `StringBuilder` (Java) can reduce the space to be amortized 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 a Hash Set

Create a hash set to store all substrings of length 2 from the original string. Then, iterate through the reversed string and check if any substring is present in the hash set. This approach improves the search time from O(n) to O(1) on average.

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

Trade-offs: This approach uses more space but reduces the time complexity. It's preferable for larger input strings.

✨

Key Insights

1Insight 1: Reversing the string is a crucial step to compare substrings against their reversed counterparts.
2Insight 2: A nested loop (implicit in the `contains` or `find` operation) or a sliding window approach can efficiently check all possible substrings of length 2.
3Insight 3: Handling edge cases such as strings with length less than 2 is essential to avoid index out-of-bounds errors.

📚

Key Terms Explained

For Students

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 string ("") or a string with length 1. The solution correctly handles this by returning `false`.
Edge case 2: String with all identical characters, e.g., "aaaa". The solution correctly identifies the repeating substring of length 2.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge case of strings with length less than 2, leading to index out-of-bounds errors.

✗Common mistake 2: Inefficiently checking for substring presence. Using a nested loop to compare substrings directly instead of using built-in `contains` or `find` functions leads to O(n^3) time complexity.

✅

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: Pattern recognition in DNA sequences. Finding short repeating sequences in reversed DNA strands might have implications for gene analysis.

🏦Real-world scenario 2: Detecting palindromic substrings in text processing. This problem is closely related to palindrome detection, which has applications in natural language processing and data compression.

💡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

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

⚠️ 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 24, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 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.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Existence of a Substring in a String and Its Reverse"?

For this easy-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?