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

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 I?

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

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

Insight 1: The problem is essentially about finding the smallest integer `t` such that after `t` operations (each removing `k` characters), the remaining substring of the original `word` is a prefix of the original `word`. Insight 2: We can iterate through possible values of `t` (starting from 1) and check if the substring starting at index `t * k` is a prefix of the original word. If it is, then `t` is our answer.

Which companies ask LeetCode 3029. Minimum Time to Revert Word to Initial State I in interviews?

This problem is frequently asked at Sprinklr.

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Minimum Time to Revert Word to Initial State I - LeetCode 3029 Solution

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

Minimum Time to Revert Word to Initial State I is LeetCode problem 3029, 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

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' required for a given string `word` to revert to its original state. Each second involves removing the first `k` characters and appending any `k` characters to the end of the string. We need to find the smallest number of such operations that will result in the string being the same as the initial `word` string.

Solution Approach

The solution iterates through possible values of `t` (number of seconds) and checks if removing `t * k` characters from the beginning and appending some other `k` characters at each step would result in the original word. This is done by comparing the substring of `word` starting at index `t * k` with the original `word`. If they are the same (i.e., `word` starts with that substring), it means after `t` seconds, the string would revert to the initial state, so we return `t`. If the loop finishes without finding such a `t`, it means it takes `ceil(n/k)` operations to revert the string to original state.

Step-by-Step Algorithm

Step 1: Initialize `n` as the length of the input string `word` and `t` as 1 (number of seconds).
Step 2: Start a loop that continues as long as `t * k < n`. This condition ensures we still have remaining characters from original word after t operations.
Step 3: Inside the loop, check if the original `word` starts with the substring of `word` starting at index `t * k`. This can be done using string comparison methods like `startsWith` in Java and Python or `strncmp` in C.
Step 4: If the condition in Step 3 is true, it means after `t` operations the word has reverted to its original state. Return `t`.
Step 5: If the condition in Step 3 is false, increment `t` and continue the loop.
Step 6: If the loop finishes without returning, it means the minimum time required is when we add exactly some other characters, and return `t`, which is equivalent to `ceil(n/k)`.

Key Insights

Insight 1: The problem is essentially about finding the smallest integer `t` such that after `t` operations (each removing `k` characters), the remaining substring of the original `word` is a prefix of the original `word`.
Insight 2: We can iterate through possible values of `t` (starting from 1) and check if the substring starting at index `t * k` is a prefix of the original word. If it is, then `t` is our answer.
Insight 3: The loop condition `t * k < n` is crucial. If `t * k >= n`, after t operations, nothing from original word is left. It is then guaranteed that by adding some other characters, we can revert the word at `t = ceil(n/k)`, and we need to return `t` directly as the final answer.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Sprinklr.

Minimum Time to Revert Word to Initial State I | LeetCode 3029

Medium

String Rolling Hash String Matching Hash Function

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

Solution Code

Available in:

class Solution {
    public int minimumTimeToInitialState(String word, int k) {
        int n = word.length();
        int t = 1;
        while (t * k < n) {
            if (word.startsWith(word.substring(t * k))) {
                return t;
            }
            t += 1;
        }
        return t;
    }
}

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Problem Info

Description

Problem 3029: Minimum Time to Revert Word to Initial State I - Detailed explanation and solution approaches.

Topics

String Rolling Hash String Matching Hash Function

Companies

Sprinklr

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3029

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

6 steps

1
Step 1: Initialize `n` as the length of the input string `word` and `t` as 1 (number of seconds).
2
Step 2: Start a loop that continues as long as `t * k < n`. This condition ensures we still have remaining characters from original word after t operations.
3
Step 3: Inside the loop, check if the original `word` starts with the substring of `word` starting at index `t * k`. This can be done using string comparison methods like `startsWith` in Java and Python or `strncmp` in C.
4
Step 4: If the condition in Step 3 is true, it means after `t` operations the word has reverted to its original state. Return `t`.
5
Step 5: If the condition in Step 3 is false, increment `t` and continue the loop.
6
Step 6: If the loop finishes without returning, it means the minimum time required is when we add exactly some other characters, and return `t`, which is equivalent to `ceil(n/k)`.

💡

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 time complexity is determined by the while loop and the string comparison within it. The loop iterates a maximum of `n/k` times. The string comparison involves comparing at most `n` characters. The space complexity is constant because we only use a few 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

Rolling Hash

Use rolling hash to calculate hash values of substrings and compare hash values instead of direct string comparison. This can potentially improve average case time complexity.

Time: O(n) in average case (with careful hash function choice to avoid collisions), O(n^2) in worst case.Space: O(1)

Trade-offs: Rolling hash might improve average case performance, but increases code complexity and introduces the risk of hash collisions, which can lead to incorrect results. The provided solution is simpler and more reliable if performance is acceptable.

✨

Key Insights

1Insight 1: The problem is essentially about finding the smallest integer `t` such that after `t` operations (each removing `k` characters), the remaining substring of the original `word` is a prefix of the original `word`.
2Insight 2: We can iterate through possible values of `t` (starting from 1) and check if the substring starting at index `t * k` is a prefix of the original word. If it is, then `t` is our answer.
3Insight 3: The loop condition `t * k < n` is crucial. If `t * k >= n`, after t operations, nothing from original word is left. It is then guaranteed that by adding some other characters, we can revert the word at `t = ceil(n/k)`, and we need to return `t` directly as the final answer.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: When k is equal to the length of the word. In this case, the word reverts to its initial state after 1 operation. The loop will not be executed, and the function will return 1.
Edge case 2: When word is very short. Then the loop might not even execute once. In this case we also return 1.
Edge case 3: When the prefix substring of size (n - t*k) is always a prefix of original word before the operations reaches the point where t*k > n. In this case, the first match would cause the solution to return prematurely, leading to the correct solution.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect loop termination condition. It's important to use `t * k < n` to ensure that the substring being compared is a valid substring within the bounds of the original `word`.

✗Common mistake 2: Not considering the case where the word reverts to its initial state only after reaching the very end of string by repetitive operations. In this case return the final value of `t`.

✗Common mistake 3: Off-by-one errors when calculating the starting index of the substring to compare.

✅

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. This pattern of searching for repeating prefixes is useful in identifying redundancies within data, which can be exploited for compression.

🏦Real-world scenario 2: DNA sequencing. Similar pattern matching algorithms are used to identify repeating sequences in DNA strands, which can be indicators of genetic traits or mutations.

🎮Real-world scenario 3: Security/Cryptography. Checking if a rotated or shifted version of a key is identical to the original key is a related problem.

💡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: October 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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.

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

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 rolling hash techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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