Problem Description

You start with a string word and an integer k. Every second, you must perform exactly two operations:

1. Remove the first k characters from word
2. Add any k characters to the end of word

The goal is to find the minimum number of seconds (greater than 0) needed to transform word back to its original state.

For example, if word = "abcdef" and k = 2:

• After 1 second: Remove "ab", leaving "cdef". You could add "ab" to get "cdefab"
• After 2 seconds: Remove "cd", leaving "efab". You could add "cd" to get "efabcd"
• After 3 seconds: Remove "ef", leaving "abcd". You could add "ef" to get "abcdef" (original string)

The solution uses rolling hash to efficiently check if a suffix of the original string matches a prefix. At each step i (where i is a multiple of k), it checks if word[i:] equals word[:n-i]. If they match, we know we can restore the original string in i/k seconds by:

• Removing the first i characters over i/k operations
• Adding back the appropriate characters to restore the original string

The algorithm iterates through positions k, 2k, 3k, ... and uses the Hashing class to compare substrings in O(1) time. The hash function uses polynomial rolling hash with base 13331 and modulo 998244353 to avoid collisions. If no early match is found, the answer is ceiling(n/k), which represents completely cycling through the entire string.

Intuition

The key insight is that after removing the first k characters and adding k characters to the end, we're essentially performing a circular shift of the string. After t seconds, we will have removed the first t*k characters from the original string.

For the string to return to its initial state, the remaining portion of the original string (after removing t*k characters) must match the beginning of the original string. In other words, if we've removed i = t*k characters, then word[i:] must equal word[:n-i] for us to be able to reconstruct the original string by adding the appropriate characters.

Why does this work? Consider what happens after t seconds:

• We've removed characters at positions [0, t*k) from the original string
• The current string starts with characters from position t*k of the original
• To get back to the original, we need the suffix word[t*k:] to match the prefix word[:n-t*k]

This is because if they match, we can add the removed characters word[:t*k] to the end, and the string becomes word[t*k:] + word[:t*k], which equals the original word when the suffix-prefix condition holds.

The rolling hash technique is used to efficiently compare these substrings in O(1) time rather than O(n) for direct string comparison. We check every possible position that's a multiple of k (since we can only remove k characters at a time), starting from k and going up to n. The first position where the suffix-prefix condition holds gives us our minimum time.

If no such position exists before reaching the end, it means we need to cycle through the entire string, which takes ceiling(n/k) seconds.

Solution Approach

The solution implements a rolling hash algorithm to efficiently check substring equality. Here's how the implementation works:

Rolling Hash Class (Hashing)

The Hashing class precomputes hash values for all prefixes of the string:

• h[i] stores the hash value of the prefix s[0:i]
• p[i] stores base^i mod mod for efficient hash computation
• The hash formula used is: hash(s) = (s[0]*base^(n-1) + s[1]*base^(n-2) + ... + s[n-1]) mod mod

The query(l, r) method returns the hash of substring s[l-1:r] in O(1) time using the formula:

hash(s[l-1:r]) = (h[r] - h[l-1] * p[r-l+1]) mod mod

Main Algorithm (minimumTimeToInitialState)

1. Initialize the hashing object with base 13331 and modulo 998244353 (large prime to minimize collisions)

2. Iterate through all positions that are multiples of k:

   for i in range(k, n, k):

   At each position i, we check if removing the first i characters would allow us to restore the original string.

3. For each position i, compare two substrings using hash values:

   • hashing.query(1, n - i): Hash of word[0:n-i] (the prefix we need to match)
   • hashing.query(i + 1, n): Hash of word[i:n] (the suffix that remains after removal)

   Note: The query uses 1-based indexing, so we add 1 to convert from 0-based.

4. If the hashes match, it means word[0:n-i] == word[i:n]. This indicates we can restore the original string in i // k seconds by:

   • Removing k characters i/k times
   • Adding back the appropriate characters to complete the original string

5. If no match is found in the loop, return (n + k - 1) // k, which is the ceiling division representing the time needed to completely cycle through the entire string.

The time complexity is O(n) for preprocessing the hash values and O(n/k) for checking positions, resulting in O(n) overall. The space complexity is O(n) for storing the hash arrays.

Example Walkthrough

Let's trace through the algorithm with word = "abacaba" and k = 3.

Step 1: Initialize Rolling Hash

• Create hash arrays for the string "abacaba" (length n = 7)
• The hash will help us compare substrings in O(1) time

Step 2: Check positions that are multiples of k

Position i = 3 (after 1 second):

• We check if removing the first 3 characters allows restoration
• Need to compare: word[0:4] ("abac") with word[3:7] ("caba")
• Using hash: query(1, 4) vs query(4, 7)
• "abac" ≠ "caba", so we can't restore in 1 second

Position i = 6 (after 2 seconds):

• We check if removing the first 6 characters allows restoration
• Need to compare: word[0:1] ("a") with word[6:7] ("a")
• Using hash: query(1, 1) vs query(7, 7)
• "a" = "a", so we CAN restore in 2 seconds!

Why does this work? After 2 seconds of operations:

• Second 1: Remove "aba" → "caba", add any 3 chars (say "xyz") → "cabaxyz"
• Second 2: Remove "cab" → "axyz", add "aba" → "axyzaba"
• But wait, since word[0:1] = word[6:7] = "a", we can be smarter:
• Second 1: Remove "aba" → "caba", add "aba" → "cabaaba"
• Second 2: Remove "cab" → "aaba", add "cab" → "aabacab"
• Actually, let me reconsider...
• Second 1: Remove "aba" → "caba", we need to add 3 chars that will help us get back
• Second 2: Remove "cab" → "a" + (our 3 added chars)
• Since word[6:7] = "a" matches word[0:1] = "a", we can add "bacaba" over the 2 operations to restore "abacaba"

The answer is 2 seconds.

If no match was found at any position, we would return ⌈7/3⌉ = 3 seconds.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The analysis breaks down as follows:

• The Hashing.__init__ method performs a single loop from 1 to len(s) + 1, which takes O(n) time where n is the length of the string.
• The Hashing.query method performs constant time operations (arithmetic and modulo operations), so it takes O(1) time.
• In the minimumTimeToInitialState method:
  • Creating the Hashing object takes O(n) time.
  • The for loop iterates from k to n with step k, resulting in at most n/k iterations.
  • Each iteration calls hashing.query twice, each taking O(1) time.
  • Therefore, the loop takes O(n/k) time.
• Since k ≥ 1, the loop complexity O(n/k) is bounded by O(n).
• The overall time complexity is O(n) + O(n/k) = O(n).

Space Complexity: O(n)

The space analysis:

• The Hashing class stores two arrays h and p, each of size len(s) + 1 = n + 1.
• These arrays take O(n) space.
• The Solution class uses only a constant amount of additional space for variables like i and n.
• Therefore, the total space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Hash Collision Leading to False Positives

The most critical pitfall in this rolling hash solution is assuming that equal hash values always mean equal strings. Hash collisions can occur even with a good hash function, leading to incorrect results.

Problem Example:

# Even with base=13331 and mod=998244353, collisions are possible
# The algorithm might incorrectly identify a match when hashes are equal
# but the actual substrings are different

Solution: Add a verification step when hash values match to ensure the substrings are actually equal:

class Solution:
    def minimumTimeToInitialState(self, word: str, k: int) -> int:
        hashing = Hashing(word, base=13331, mod=998244353)
        n = len(word)
      
        for i in range(k, n, k):
            # First check hash equality
            if hashing.query(1, n - i) == hashing.query(i + 1, n):
                # Verify actual string equality to avoid hash collision
                if word[:n-i] == word[i:]:
                    return i // k
      
        return (n + k - 1) // k

2. Integer Overflow in Hash Calculation

Without proper modulo operations, the hash values can overflow, especially for long strings.

Problem Example:

# Incorrect: Missing modulo in intermediate calculations
self.h[i] = self.h[i - 1] * base + ord(s[i - 1])  # Can overflow!

Solution: Always apply modulo operations at each step:

# Correct: Apply modulo at each operation
self.h[i] = (self.h[i - 1] * base + ord(s[i - 1])) % mod
self.p[i] = (self.p[i - 1] * base) % mod

3. Off-by-One Errors in Index Conversion

The Hashing class uses 1-based indexing while Python uses 0-based indexing, making it easy to make indexing errors.

Problem Example:

# Incorrect indexing - forgetting to adjust for 1-based indexing
if hashing.query(0, n - i - 1) == hashing.query(i, n - 1):  # Wrong!

Solution: Document the indexing convention clearly and be consistent:

# Correct: Convert 0-based to 1-based indexing
# To check word[0:n-i] == word[i:n]
if hashing.query(1, n - i) == hashing.query(i + 1, n):
    return i // k

4. Using Double Hashing for Collision Resistance

For production code where correctness is critical, use double hashing with different bases and moduli:

class Solution:
    def minimumTimeToInitialState(self, word: str, k: int) -> int:
        # Use two different hash functions
        hash1 = Hashing(word, base=13331, mod=998244353)
        hash2 = Hashing(word, base=31, mod=10**9 + 7)
        n = len(word)
      
        for i in range(k, n, k):
            # Check both hashes match
            if (hash1.query(1, n - i) == hash1.query(i + 1, n) and
                hash2.query(1, n - i) == hash2.query(i + 1, n)):
                return i // k
      
        return (n + k - 1) // k

This dramatically reduces the probability of collision from approximately 1/mod to 1/(mod1 * mod2).