How do you solve Minimum Time to Revert Word to Initial State II on LeetCode?

Minimum Time to Revert Word to Initial State II (LeetCode #3031) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the Z-function is crucial for optimizing string problems.

What is the brute force solution for Minimum Time to Revert Word to Initial State II?

The brute force approach for Minimum Time to Revert Word to Initial State II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves simulating the process of removing and adding characters to see how many seconds it takes to revert the word to its original state. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Time to Revert Word to Initial State II?

Minimum Time to Revert Word to Initial State II uses the following concepts: String, Rolling Hash, String Matching, Hash Function. The recommended patterns to study are: String Matching, Z-function, Prefix-Suffix Problems.

What are the interview tips for Minimum Time to Revert Word to Initial State II?

Always clarify the problem and constraints with the interviewer. Discuss your thought process and potential approaches before jumping into coding. Test your code with different inputs to ensure it handles all cases correctly.

What are common mistakes when solving Minimum Time to Revert Word to Initial State II?

Failing to consider edge cases, such as when k is greater than the length of the string. Not optimizing the solution when a pattern is recognized.

#3031

Minimum Time to Revert Word to Initial State II

Hard

StringRolling HashString MatchingHash FunctionString MatchingZ-functionPrefix-Suffix Problems

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses the Z-function to find the longest prefix which is also a suffix. This allows us to determine how many complete cycles of k can fit into the string, leading to a much faster solution.

⚙️

Algorithm

3 steps

1Step 1: Compute the Z-function for the string, which gives lengths of substrings starting from each position that match the prefix of the string.
2Step 2: Find the longest suffix that is also a prefix and whose length is a multiple of k.
3Step 3: Calculate the minimum time as the length of the string divided by k minus the length of the longest matching prefix-suffix divided by k.

solution.py22 lines

1# Full working Python code
2
3def z_function(s):
4    z = [0] * len(s)
5    l, r, n = 0, 0, len(s)
6    for i in range(1, n):
7        if i <= r:
8            z[i] = min(r - i + 1, z[i - l])
9        while i + z[i] < n and s[z[i]] == s[i + z[i]]:
10            z[i] += 1
11        if i + z[i] - 1 > r:
12            l, r = i, i + z[i] - 1
13    return z
14
15def min_time_to_revert(word, k):
16    n = len(word)
17    z = z_function(word)
18    max_suffix_prefix = 0
19    for i in range(n):
20        if z[i] % k == 0:
21            max_suffix_prefix = max(max_suffix_prefix, z[i])
22    return (n - max_suffix_prefix) // k

ℹ

Complexity note: The time complexity is O(n) because we compute the Z-function in linear time, and finding the maximum suffix-prefix is also linear. The space complexity is O(n) due to the storage of the Z-array.

1Understanding the Z-function is crucial for optimizing string problems.
2Identifying patterns in string manipulation can lead to more efficient algorithms.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.