What is the time complexity of Unique Substrings in Wraparound String?

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

What data structures are used to solve Unique Substrings in Wraparound String?

This problem uses String, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Unique Substrings in Wraparound String?

Insight 1: We don't need to actually generate the infinite wraparound string. We only need to consider the consecutive subsequences (substrings) within the input string `s`. Insight 2: For each character in `s`, we only need to consider the longest consecutive wraparound sequence ending at that character. This reduces the number of substrings to check dramatically.

Which companies ask LeetCode 467. Unique Substrings in Wraparound String in interviews?

This problem is frequently asked at MAQ Software.

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Unique Substrings in Wraparound String - LeetCode 467 Solution

Unique Substrings in Wraparound String - Complete Solution Guide

Unique Substrings in Wraparound String is LeetCode problem 467, 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

We define the string base to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz" , so base will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd...." . Given a string s , return the number of unique non-empty substrings of s are present in base . Example 1: Input: s = "a" Output: 1 Explanation: Only the substring "a" of s is in base. Example 2: Input: s = "cac" Output: 2 Explanation: There are two substrings ("a", "c") of s in base. Example 3: Input: s

Detailed Explanation

The problem asks us to find the number of unique non-empty substrings of a given string `s` that are present in an infinite wraparound string `base` made up of the alphabet 'abcdefghijklmnopqrstuvwxyz' repeated infinitely. In essence, we want to find out how many distinct substrings from `s` exist within this infinite wraparound string. For instance, 'za' and 'ab' are valid consecutive substrings in the wraparound string, as is 'zab'. The input is a string `s` containing only lowercase English letters, and the output is an integer representing the count of unique substrings of `s` found in the wraparound string. The length of the input string `s` can be up to 10^5.

Solution Approach

The solution uses a dynamic programming approach to efficiently calculate the number of unique substrings. It iterates through the input string `s`, keeping track of the current length of a consecutive wraparound sequence. For each character, it updates an array `maxLength` where `maxLength[i]` stores the maximum length of a wraparound substring ending with the i-th character of the alphabet (where 'a' is the 0th character, 'b' is the 1st, etc.). This array helps prevent double-counting substrings. The final result is the sum of all values in the `maxLength` array, which represents the total number of unique substrings.

Step-by-Step Algorithm

Step 1: Initialize an array `maxLength` of size 26 with all elements set to 0. This array will store the maximum length of a wraparound string ending at each letter (a-z).
Step 2: Initialize a variable `currentLength` to 0. This variable tracks the length of the current consecutive wraparound string.
Step 3: Iterate through the input string `s` from left to right.
Step 4: In each iteration, check if the current character is a consecutive character in the wraparound string compared to the previous character. This is determined by checking if `(ord(s[i]) - ord(s[i-1])) % 26 == 1`. If it is, increment `currentLength`. Otherwise, reset `currentLength` to 1.
Step 5: Calculate the index `idx` corresponding to the current character `s[i]` in the `maxLength` array. This index is `ord(s[i]) - ord('a')`.
Step 6: Update `maxLength[idx]` to be the maximum of its current value and `currentLength`. This ensures that we store the longest wraparound string ending with this letter.
Step 7: After iterating through the entire string `s`, calculate the sum of all elements in the `maxLength` array. This sum represents the total number of unique substrings in the wraparound string that are also substrings of `s`.
Step 8: Return the calculated sum.

Key Insights

Insight 1: We don't need to actually generate the infinite wraparound string. We only need to consider the consecutive subsequences (substrings) within the input string `s`.
Insight 2: For each character in `s`, we only need to consider the longest consecutive wraparound sequence ending at that character. This reduces the number of substrings to check dramatically.
Insight 3: We can use dynamic programming to keep track of the maximum length of the consecutive wraparound sequences ending at each character of the alphabet. A hashmap/array of size 26 indexed by the alphabet can achieve this.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: MAQ Software.

Unique Substrings in Wraparound String | LeetCode 467

Medium

String Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int findSubstringInWraproundString(String s) {
        int[] maxLength = new int[26];
        
        int currentLength = 0;
        
        for (int i = 0; i < s.length(); i++) {
            if (i > 0 && (s.charAt(i) - s.charAt(i-1) + 26) % 26 == 1) {
                currentLength += 1;
            } else {
                currentLength = 1;
            }
            
            int idx = s.charAt(i) - 'a';
            maxLength[idx] = Math.max(maxLength[idx], currentLength);
        }
        
        int sum = 0;
        for (int length : maxLength) {
            sum += length;
        }
        
        return sum;
    }
}

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Problem Info

Description

Problem 467: Unique Substrings in Wraparound String - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

MAQ Software

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:467

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

8 steps

1
Step 1: Initialize an array `maxLength` of size 26 with all elements set to 0. This array will store the maximum length of a wraparound string ending at each letter (a-z).
2
Step 2: Initialize a variable `currentLength` to 0. This variable tracks the length of the current consecutive wraparound string.
3
Step 3: Iterate through the input string `s` from left to right.
4
Step 4: In each iteration, check if the current character is a consecutive character in the wraparound string compared to the previous character. This is determined by checking if `(ord(s[i]) - ord(s[i-1])) % 26 == 1`. If it is, increment `currentLength`. Otherwise, reset `currentLength` to 1.
5
Step 5: Calculate the index `idx` corresponding to the current character `s[i]` in the `maxLength` array. This index is `ord(s[i]) - ord('a')`.
6
Step 6: Update `maxLength[idx]` to be the maximum of its current value and `currentLength`. This ensures that we store the longest wraparound string ending with this letter.
7
Step 7: After iterating through the entire string `s`, calculate the sum of all elements in the `maxLength` array. This sum represents the total number of unique substrings in the wraparound string that are also substrings of `s`.
8
Step 8: Return the calculated sum.

💡

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 algorithm iterates through the input string `s` once, performing constant-time operations in each iteration. Therefore, the time complexity is directly proportional to the length of `s`. The space complexity is constant because the size of the `maxLength` array is fixed at 26 and does not depend on the input size.

📖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

Brute Force Approach

Generate all possible substrings of `s` and check if each substring exists in the infinite wraparound string. Use a set to store unique substrings and return the size of the set.

Time: O(n^3), where n is the length of `s`. Generating all substrings takes O(n^2), and checking if a substring is present in the wraparound string takes O(n).Space: O(n^2) in the worst case, for storing all the unique substrings in the set.

Trade-offs: The brute force approach is simple to understand but highly inefficient for large input strings. The dynamic programming approach is much more efficient and suitable for larger inputs. Use Brute force only if string length is very small.

✨

Key Insights

1Insight 1: We don't need to actually generate the infinite wraparound string. We only need to consider the consecutive subsequences (substrings) within the input string `s`.
2Insight 2: For each character in `s`, we only need to consider the longest consecutive wraparound sequence ending at that character. This reduces the number of substrings to check dramatically.
3Insight 3: We can use dynamic programming to keep track of the maximum length of the consecutive wraparound sequences ending at each character of the alphabet. A hashmap/array of size 26 indexed by the alphabet can achieve this.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: Empty string. The code correctly handles an empty input string because the loop does not execute and returns 0.
Edge case 2: Single-character string. The code works correctly for single-character strings because the `currentLength` will be 1, and the corresponding entry in `maxLength` will be set to 1.
Edge case 3: String with repeated characters (e.g., 'aaaaa'). The algorithm correctly handles repeated characters by resetting the `currentLength` to 1 whenever the characters are not consecutive in the wraparound string, avoiding overcounting.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Overcounting substrings. A naive approach might count overlapping substrings multiple times. The `maxLength` array is used to prevent this overcounting by ensuring that only the longest wraparound string ending at each character is considered.

✗Common mistake 2: Incorrect wraparound check. The modulo operator (%) is crucial for handling the wraparound condition ('z' followed by 'a'). Failing to use the modulo operator, or using it incorrectly, will lead to incorrect results.

✗Common mistake 3: Forgetting to consider substrings of length one. Remember that the problem asks for the number of unique *non-empty* substrings. Each individual character in `s` is a valid substring if it exists within the wraparound string.

✅

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: DNA sequence analysis. The wraparound concept can be analogous to circular DNA or RNA sequences, where identifying unique substrings or patterns is important for bioinformatics research.

🏦Real-world scenario 2: Cyclic buffer manipulation. When dealing with cyclic buffers (e.g., in audio or video processing), this algorithm can be adapted to efficiently count distinct data segments within the buffer.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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 16, 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

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

@TechLeadJun 29

❓

Frequently Asked Questions

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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 "Unique Substrings in Wraparound String"?

For this medium-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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