What is the time complexity of Minimum Time to Revert Word to Initial State II?

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

What data structures are used to solve Minimum Time to Revert Word to Initial State II?

This problem uses String, Rolling Hash, String Matching, Hash Function. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Time to Revert Word to Initial State II?

Insight 1: The core idea is to efficiently check if a shifted version of the word matches the original word. Insight 2: Z algorithm helps to efficiently compute the longest common prefix between the original string and its suffixes shifted by multiples of `k`.

Which companies ask LeetCode 3031. Minimum Time to Revert Word to Initial State II in interviews?

This problem is frequently asked at Sprinklr.

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Minimum Time to Revert Word to Initial State II - LeetCode 3031 Solution

Minimum Time to Revert Word to Initial State II - Complete Solution Guide

Minimum Time to Revert Word to Initial State II is LeetCode problem 3031, 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

You are given a 0-indexed string word and an integer k . At every second, you must perform the following operations: Remove the first k characters of word . Add any k characters to the end of word . Note that you do not necessarily need to add the same characters that you removed. However, you must perform both operations at every second. Return the minimum time greater than zero required for word to revert to its initial state . Example 1: Input: word = "abacaba", k = 3 Output: 2 Explanation: A

Detailed Explanation

The problem asks us to find the minimum number of 'seconds' it takes for a given string `word` to revert to its original state. In each second, we remove the first `k` characters of `word` and add any `k` characters to the end of `word`. The goal is to determine the smallest number of such operations that result in the original `word` being restored.

Solution Approach

The provided solution employs the Z algorithm to efficiently find the minimum time for the word to revert to its original state. The Z algorithm calculates a Z array, where Z[i] represents the length of the longest substring starting from index i which is also a prefix of the string. By using the Z array, the solution iterates through the string at intervals of `k`. If Z[i] is greater than or equal to `n - i`, it means that the suffix starting at index `i` is also a prefix of the original string. Therefore, after `i/k` operations, the word will revert to its initial state. If no such `i` is found, the minimum time is `(n + k - 1) / k`, which represents the minimum number of operations needed to remove all characters from the string and then add new characters to match the initial word.

Step-by-Step Algorithm

Step 1: Calculate the Z array using the Z algorithm. The Z array, `z`, stores, for each index `i`, the length of the longest common prefix between the entire string and the substring starting at index `i`.
Step 2: Iterate through the Z array at intervals of `k`, starting from `k`. For each index `i` (a multiple of `k`), check if `z[i] >= n - i`. This condition checks if the suffix of the string starting at index `i` matches a prefix of length `n - i` of the original string. If the condition is met, it means after `i / k` operations, the word will return to the initial state.
Step 3: If a match is found in Step 2, return `i / k` as the minimum time.
Step 4: If no match is found after iterating through all relevant indices, return `(n + k - 1) / k`. This represents the minimum time required to remove all characters of the word in chunks of size k, and then refill with `k` new characters in the last step to arrive to the original state.

Key Insights

Insight 1: The core idea is to efficiently check if a shifted version of the word matches the original word.
Insight 2: Z algorithm helps to efficiently compute the longest common prefix between the original string and its suffixes shifted by multiples of `k`.
Insight 3: Optimizing the search involves only considering shifts that are multiples of `k`, significantly reducing the search space.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Rolling Hash, String Matching, Hash Function.

Companies

Asked at: Sprinklr.

Minimum Time to Revert Word to Initial State II | LeetCode 3031

Hard

String Rolling Hash String Matching Hash Function

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

Solution Code

Available in:

class Solution {
    public int minimumTimeToInitialState(String word, int k) {
        int n = word.length();
        int[] z = new int[n];
        int l = 0, r = 0;
        for (int i = 1; i < n; i++) {
            if (i <= r) {
                z[i] = Math.min(r - i + 1, z[i - l]);
            }
            while (i + z[i] < n && word.charAt(z[i]) == word.charAt(i + z[i])) {
                z[i]++;
            }
            if (i + z[i] - 1 > r) {
                l = i;
                r = i + z[i] - 1;
            }

            
        }

        for (int i = k; i < n; i += k) {
            if (z[i] >= n - i) {
                return i / k;
            }
        }

        return (n + k - 1) / k;
    }
}

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Problem Info

Description

Problem 3031: Minimum Time to Revert Word to Initial State II - Detailed explanation and solution approaches.

Topics

String Rolling Hash String Matching Hash Function

Companies

Sprinklr

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3031

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

4 steps

1
Step 1: Calculate the Z array using the Z algorithm. The Z array, `z`, stores, for each index `i`, the length of the longest common prefix between the entire string and the substring starting at index `i`.
2
Step 2: Iterate through the Z array at intervals of `k`, starting from `k`. For each index `i` (a multiple of `k`), check if `z[i] >= n - i`. This condition checks if the suffix of the string starting at index `i` matches a prefix of length `n - i` of the original string. If the condition is met, it means after `i / k` operations, the word will return to the initial state.
3
Step 3: If a match is found in Step 2, return `i / k` as the minimum time.
4
Step 4: If no match is found after iterating through all relevant indices, return `(n + k - 1) / k`. This represents the minimum time required to remove all characters of the word in chunks of size k, and then refill with `k` new characters in the last step to arrive to the original state.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The Z algorithm's linear time complexity arises from amortized analysis; each character is visited and compared a limited number of times. The space complexity is directly proportional to the string's length due to the Z array storage.

📖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

String comparison with sliding window

Iteratively shift the original string by multiples of `k` characters and compare if it matches the original string. This would involve creating shifted strings and comparing them.

Time: O(n^2) in the worst case, where n is the length of the word.Space: O(n), to store the shifted strings.

Trade-offs: This approach is simpler to understand but less efficient than the Z algorithm due to the string comparisons. The Z algorithm provides a more optimized approach for pattern matching.

Rolling Hash

Compute the hash of the original string and then iteratively shift the string by `k` and compute its hash. Compare the hashes to see if the strings are equal.

Time: O(n) on average, if hash collisions are rare. O(n^2) in the worst case with many collisions.Space: O(1), assuming a fixed-size hash.

Trade-offs: Can be faster than naive string comparison if collisions are managed well. Implementing a robust rolling hash function can be complex, and handling collisions adds overhead. Z algorithm is often preferred for guaranteed linear time complexity.

✨

Key Insights

1Insight 1: The core idea is to efficiently check if a shifted version of the word matches the original word.
2Insight 2: Z algorithm helps to efficiently compute the longest common prefix between the original string and its suffixes shifted by multiples of `k`.
3Insight 3: Optimizing the search involves only considering shifts that are multiples of `k`, significantly reducing the search space.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: `k` is equal to the length of the word `n`. In this case, only one operation is needed to revert the string (remove all characters and add the original characters again). The algorithm correctly handles this because, in the loop it will check `z[k]` which is `z[n]`. Since it will not enter the loop the final calculation will be (n + n - 1) / n which is always 1.
Edge case 2: The string `word` is already in its initial state after a few operations less than `(n + k - 1) // k`. The loop checks for this condition and returns the minimum operations required.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge case where `k` is a multiple of the string length correctly, resulting in incorrect minimum time calculation.

✗Common mistake 2: Implementing the Z algorithm incorrectly, leading to incorrect Z array values and, consequently, a wrong final answer.

✗Common mistake 3: Forgetting to consider the case where it requires one complete rotation (removing all characters and putting them back) when the string doesn't revert before the last possible operation.

✅

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: Data compression algorithms often use pattern matching techniques similar to this to identify repeating patterns in data streams.

🏦Real-world scenario 2: In network security, intrusion detection systems utilize pattern matching to detect known attack signatures in network traffic. This problem's logic can be used to efficiently search for such signatures.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: October 30, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

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.

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 "Minimum Time to Revert Word to Initial State II"?

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

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