#3031

Hard

LeetCode Problem Workspace

Minimum Time to Revert Word to Initial State II

The problem asks to calculate the minimum time required to revert a string to its initial state using specific operations each second.

string rolling hash string matching hash function

Problem Statement

You are given a string word and an integer k. In each second, you must perform two operations on word:

Remove the first k characters from the string. 2. Append the same k characters to the end of the string. The goal is to find the minimum number of seconds required to revert the string word to its initial state.

Examples

Example 1

Input: word = "abacaba", k = 3

Output: 2

At the 1st second, we remove characters "aba" from the prefix of word, and add characters "bac" to the end of word. Thus, word becomes equal to "cababac". At the 2nd second, we remove characters "cab" from the prefix of word, and add "aba" to the end of word. Thus, word becomes equal to "abacaba" and reverts to its initial state. It can be shown that 2 seconds is the minimum time greater than zero required for word to revert to its initial state.

Example 2

Input: word = "abacaba", k = 4

Output: 1

At the 1st second, we remove characters "abac" from the prefix of word, and add characters "caba" to the end of word. Thus, word becomes equal to "abacaba" and reverts to its initial state. It can be shown that 1 second is the minimum time greater than zero required for word to revert to its initial state.

Example 3

Input: word = "abcbabcd", k = 2

Output: 4

At every second, we will remove the first 2 characters of word, and add the same characters to the end of word. After 4 seconds, word becomes equal to "abcbabcd" and reverts to its initial state. It can be shown that 4 seconds is the minimum time greater than zero required for word to revert to its initial state.

Constraints

1 <= word.length <= 106
1 <= k <= word.length
word consists only of lowercase English letters.

Solution Approach

Identify the Longest Prefix-Suffix Pair

Find the longest suffix of the string that is also a prefix. This can be achieved by iterating through the string and checking for matching substrings in the prefix and suffix. The length of the longest matching substring determines the minimal number of operations.

Apply Rolling Hash for Efficiency

Use rolling hash to efficiently compute the matching substrings. By utilizing hash values, you can quickly check if a prefix matches a suffix without recalculating it from scratch every time, saving computational time especially for large strings.

Optimize to O(N) Time Complexity

The approach should be optimized to linear time O(N). This can be achieved by maintaining rolling hashes for prefixes and suffixes and then finding the largest length that is a multiple of k to avoid unnecessary recalculations and ensure that the solution is scalable.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final approach, particularly the implementation of rolling hash and prefix-suffix comparison. With rolling hash and the longest prefix-suffix matching, the solution can be optimized to O(N). The space complexity also depends on how hash values are stored and processed, typically O(N).

What Interviewers Usually Probe

Look for the candidate's ability to optimize the solution using hashing techniques.
Assess the candidate's understanding of prefix-suffix problems and their application to string manipulation.
Evaluate if the candidate can explain how they reduced the complexity of the problem to O(N).

Common Pitfalls or Variants

Common pitfalls

Candidates may attempt brute-force approaches, leading to inefficient O(N^2) time complexity.
Failing to use rolling hash may lead to suboptimal solutions for large inputs.
Not properly handling edge cases like when k is larger than the string length or when no operation is needed.

Follow-up variants

Instead of a fixed k, you could generalize this problem to work with variable values of k for different test cases.
Consider adding a constraint where the string may contain special characters or digits, complicating the prefix-suffix matching.
Instead of finding the smallest number of seconds, the problem could require finding the number of seconds needed to make the string reach any specific target state.

FAQ

What is the minimum time to revert a string to its initial state in the "Minimum Time to Revert Word to Initial State II" problem?

The minimum time is determined by finding the longest suffix which is also a prefix of the string, and whose length is a multiple of k. This allows the string to return to its initial state in the fewest operations.

How does the rolling hash optimize the "Minimum Time to Revert Word to Initial State II" problem?

The rolling hash helps efficiently compare prefixes and suffixes of the string without recalculating hashes from scratch each time, significantly reducing the computational cost.

What is the primary pattern to solve "Minimum Time to Revert Word to Initial State II"?

The primary pattern is to use string manipulation combined with rolling hash for efficient comparison of prefixes and suffixes.

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

The optimal time complexity is O(N), achieved by utilizing rolling hash and finding the longest matching prefix-suffix pair.

What are common pitfalls to avoid in "Minimum Time to Revert Word to Initial State II"?

Avoid brute-force solutions that result in O(N^2) time complexity, and be sure to handle edge cases like when k is larger than the string length.

terminal

Solution

Solution 1

#### Python3

class Hashing:
    __slots__ = ["mod", "h", "p"]

    def __init__(self, s: str, base: int, mod: int):
        self.mod = mod
        self.h = [0] * (len(s) + 1)
        self.p = [1] * (len(s) + 1)
        for i in range(1, len(s) + 1):
            self.h[i] = (self.h[i - 1] * base + ord(s[i - 1])) % mod
            self.p[i] = (self.p[i - 1] * base) % mod

    def query(self, l: int, r: int) -> int:
        return (self.h[r] - self.h[l - 1] * self.p[r - l + 1]) % self.mod


class Solution:
    def minimumTimeToInitialState(self, word: str, k: int) -> int:
        hashing = Hashing(word, 13331, 998244353)
        n = len(word)
        for i in range(k, n, k):
            if hashing.query(1, n - i) == hashing.query(i + 1, n):
                return i // k
        return (n + k - 1) // k

Continue Practicing

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