What is the time complexity of Last Substring in Lexicographical Order?

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

What data structures are used to solve Last Substring in Lexicographical Order?

This problem uses Two Pointers, String. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Last Substring in Lexicographical Order?

Insight 1: We don't need to generate all substrings. Comparing substrings directly would be inefficient. Instead, we can efficiently identify the starting index of the last substring. Insight 2: The core idea is to maintain two pointers, 'i' and 'j', representing the starting indices of two potential candidates for the last substring. We compare the substrings starting at 'i' and 'j' character by character.

Which companies ask LeetCode 1163. Last Substring in Lexicographical Order in interviews?

This problem is frequently asked at Fastenal, MathWorks.

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Last Substring in Lexicographical Order - LeetCode 1163 Solution

Q: How to solve Last Substring in Lexicographical Order?

The provided solution employs a two-pointer approach to find the starting index of the lexicographically largest substring. It iteratively compares two substrings, indicated by indices 'i' and 'j', using a third index 'k' to track the matching prefix length. Based on the comparison, it updates 'i' and 'j' to converge towards the starting position of the last substring. This algorithm avoids generating and explicitly comparing all substrings, leading to an efficient solution.

Last Substring in Lexicographical Order - Complete Solution Guide

Last Substring in Lexicographical Order is LeetCode problem 1163, a Hard 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 , return the last substring of s in lexicographical order . Example 1: Input: s = "abab" Output: "bab" Explanation: The substrings are ["a", "ab", "aba", "abab", "b", "ba", "bab"]. The lexicographically maximum substring is "bab". Example 2: Input: s = "leetcode" Output: "tcode" Constraints: 1 <= s.length <= 4 * 10 5 s contains only lowercase English letters.

Detailed Explanation

The problem asks us to find the lexicographically largest substring within a given string 's'. Lexicographical order is essentially dictionary order. For example, 'abc' comes before 'abd', and 'baa' comes before 'bab'. The task is to return the substring of 's' that would appear last in a dictionary if all substrings were listed.

Solution Approach

The provided solution employs a two-pointer approach to find the starting index of the lexicographically largest substring. It iteratively compares two substrings, indicated by indices 'i' and 'j', using a third index 'k' to track the matching prefix length. Based on the comparison, it updates 'i' and 'j' to converge towards the starting position of the last substring. This algorithm avoids generating and explicitly comparing all substrings, leading to an efficient solution.

Step-by-Step Algorithm

Step 1: Initialize three pointers: 'i' to 0, 'j' to 1, and 'k' to 0. 'i' and 'j' represent the starting indices of two potential last substrings, and 'k' tracks the length of the matching prefix between the substrings starting at 'i' and 'j'.
Step 2: Iterate while 'j + k' is within the bounds of the string length 'n'.
Step 3: Compare the characters at 's[i + k]' and 's[j + k]'.
Step 4: If 's[i + k]' equals 's[j + k]', increment 'k' to compare the next characters.
Step 5: If 's[i + k]' is less than 's[j + k]', it means the substring starting at 'j' is lexicographically greater than the substring starting at 'i'. Update 'i' to 'j', 'j' to 'i + 1', and reset 'k' to 0.
Step 6: If 's[i + k]' is greater than 's[j + k]', it means the substring starting at 'i' is lexicographically greater. Update 'j' to 'j + k + 1', and reset 'k' to 0. This effectively skips the portion of the string that has already been determined to be smaller.
Step 7: After the loop finishes, 'i' will hold the starting index of the lexicographically last substring. Return the substring of 's' starting from index 'i'.

Key Insights

Insight 1: We don't need to generate all substrings. Comparing substrings directly would be inefficient. Instead, we can efficiently identify the starting index of the last substring.
Insight 2: The core idea is to maintain two pointers, 'i' and 'j', representing the starting indices of two potential candidates for the last substring. We compare the substrings starting at 'i' and 'j' character by character.
Insight 3: If s[i+k] == s[j+k], we increment k to compare the next characters. If s[i+k] < s[j+k], then the substring starting at 'j' is lexicographically larger, so we update i = j. If s[i+k] > s[j+k], the substring at 'i' is larger, so we advance 'j' beyond the matched portion.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Two Pointers, String.

Companies

Asked at: Fastenal, MathWorks.

Last Substring in Lexicographical Order | LeetCode 1163

Hard

Two Pointers String

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

Solution Code

Available in:

class Solution {
    public String lastSubstring(String s) {
        int n = s.length();
        int i = 0, j = 1, k = 0;
        while (j + k < n) {
            if (s.charAt(i + k) == s.charAt(j + k)) {
                k++;
            } else if (s.charAt(i + k) < s.charAt(j + k)) {
                i = j;
                j = i + 1;
                k = 0;
            } else {
                j = j + k + 1;
                k = 0;
            }
        }
        return s.substring(i);
    }
}

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Problem Info

Description

Problem 1163: Last Substring in Lexicographical Order - Detailed explanation and solution approaches.

Topics

Two Pointers String

Companies

FastenalMathWorks

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1163

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

7 steps

1
Step 1: Initialize three pointers: 'i' to 0, 'j' to 1, and 'k' to 0. 'i' and 'j' represent the starting indices of two potential last substrings, and 'k' tracks the length of the matching prefix between the substrings starting at 'i' and 'j'.
2
Step 2: Iterate while 'j + k' is within the bounds of the string length 'n'.
3
Step 3: Compare the characters at 's[i + k]' and 's[j + k]'.
4
Step 4: If 's[i + k]' equals 's[j + k]', increment 'k' to compare the next characters.
5
Step 5: If 's[i + k]' is less than 's[j + k]', it means the substring starting at 'j' is lexicographically greater than the substring starting at 'i'. Update 'i' to 'j', 'j' to 'i + 1', and reset 'k' to 0.
6
Step 6: If 's[i + k]' is greater than 's[j + k]', it means the substring starting at 'i' is lexicographically greater. Update 'j' to 'j + k + 1', and reset 'k' to 0. This effectively skips the portion of the string that has already been determined to be smaller.
7
Step 7: After the loop finishes, 'i' will hold the starting index of the lexicographically last substring. Return the substring of 's' starting from index 'i'.

💡

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 outer `while` loop iterates at most n times. The variable `k` might increment, but in each iteration of the `while` loop, either `j` gets incremented or `i` gets updated, ensuring that each character of `s` is visited a constant number of times. Hence, linear time complexity. The space complexity is constant because we only use a fixed number of integer variables.

📖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

Suffix Array

Construct a suffix array, which stores all suffixes of the string in lexicographical order. The last element of the suffix array would be the lexicographically largest substring.

Time: O(n log n) or O(n) depending on the suffix array construction algorithmSpace: O(n)

Trade-offs: While theoretically interesting, suffix array construction is more complex to implement and has a higher time complexity (O(n log n) in general or sometimes optimized to O(n) using sophisticated algorithms) compared to the two-pointer approach. The two-pointer approach offers a simpler and more efficient solution for this specific problem.

Brute Force

Generate all substrings of the string and then find the lexicographically largest one.

Time: O(n^2) to generate all substrings, and O(n) to compare strings, and O(n^3) overall.Space: O(n^2) in the worst case (e.g., storing all substrings)

Trade-offs: This approach is simple to understand but highly inefficient, especially for large input strings. It generates every possible substring, which the problem does not require. It is clearly less preferable due to the higher time and space complexity.

✨

Key Insights

1Insight 1: We don't need to generate all substrings. Comparing substrings directly would be inefficient. Instead, we can efficiently identify the starting index of the last substring.
2Insight 2: The core idea is to maintain two pointers, 'i' and 'j', representing the starting indices of two potential candidates for the last substring. We compare the substrings starting at 'i' and 'j' character by character.
3Insight 3: If s[i+k] == s[j+k], we increment k to compare the next characters. If s[i+k] < s[j+k], then the substring starting at 'j' is lexicographically larger, so we update i = j. If s[i+k] > s[j+k], the substring at 'i' is larger, so we advance 'j' beyond the matched portion.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

String

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

Edge Cases

Edge case 1: String with only one character (e.g., s = "a"). The algorithm correctly returns "a" because the while loop condition (j + k < n) is false from the start, and s[i:] resolves to s[0:], which is "a".
Edge case 2: String with all identical characters (e.g., s = "aaaa"). The algorithm correctly returns "aaaa". 'i' remains at 0, 'j' keeps incrementing, and 'k' keeps incrementing as long as j+k < n. Because all characters are the same, i never changes.
Edge case 3: Empty string. According to the problem constraints, the string length will always be greater or equal to 1. However, if the string was empty, the code would have an indexing error at s[i+k], so a check would have to be added to handle this corner case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case when characters are equal properly. Failing to increment 'k' to compare subsequent characters can lead to incorrect results.

✗Common mistake 2: Incrementing 'j' too aggressively. Incrementing 'j' by 1 after each comparison, regardless of whether there was a matching prefix, can lead to missing the optimal substring. The algorithm needs to jump ahead correctly based on matched prefix length.

✗Common mistake 3: Assuming that the first character being the largest in the alphabet will automatically give us the correct answer. 'z' may appear early in the string, but a later substring starting with 'y' and having a larger continuation might be the correct answer.

✅

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 editors and IDEs often use similar algorithms for tasks like code completion and finding the best match for a user's input. While this specific problem is not directly used, the core string comparison principles apply.

🏦Real-world scenario 2: Bioinformatics uses lexicographical ordering for comparing DNA sequences. While simplified, this problem shares similarities with tasks involving finding the 'best' or most significant sequence within a larger genetic dataset.

💡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

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

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 23, 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 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

Frequently Asked Questions

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.

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Last Substring in Lexicographical Order"?

For this hard-level Two Pointers 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?