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

Minimum Time to Revert Word to Initial State I (LeetCode #3029) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Finding the longest prefix-suffix helps minimize operations.

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

The optimal solution for Minimum Time to Revert Word to Initial State I runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only make a single pass through the string to find the longest prefix-suffix.

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

The brute force approach for Minimum Time to Revert Word to Initial State I 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 repeatedly until the string returns to its original state. It’s straightforward but inefficient as it checks each configuration. 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 I?

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

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

Clarify the problem requirements before jumping into coding. Discuss your thought process and any assumptions with the interviewer. Test edge cases, such as empty strings or very small k values.

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

Not considering cases where k is larger than the string length. Failing to check if the string can revert in fewer operations.

#3029

Minimum Time to Revert Word to Initial State I

Medium

StringRolling HashString MatchingHash FunctionString MatchingPrefix-SuffixRolling Hash

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach leverages the concept of finding the longest prefix that is also a suffix. This allows us to determine how many operations are needed to revert the string back to its original state efficiently.

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Algorithm

5 steps

1Step 1: Find the length of the string, n.
2Step 2: Initialize a variable to store the longest prefix-suffix length.
3Step 3: Iterate through the string to find the longest suffix which is also a prefix and is a multiple of k.
4Step 4: Calculate the number of operations needed as n divided by k minus the longest prefix-suffix length divided by k.
5Step 5: Return the result.

solution.py7 lines

1def minTimeToRevert(word, k):
2    n = len(word)
3    lps = 0
4    for i in range(1, n):
5        if word[:i] == word[n-i:n]:
6            lps = i
7    return (n - lps) // k

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Complexity note: The time complexity is O(n) because we only make a single pass through the string to find the longest prefix-suffix.

1Finding the longest prefix-suffix helps minimize operations.
2Understanding string manipulation is crucial for efficiency.

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