Problem Description

You have a string word of length n and an integer k where k divides n evenly.

You can perform operations to modify the string. In each operation:

• Pick any two positions i and j that are divisible by k (meaning positions 0, k, 2k, 3k, etc.)
• Replace the substring of length k starting at position i with the substring of length k starting at position j
• In other words, replace word[i..i+k-1] with word[j..j+k-1]

Your goal is to make the string k-periodic using the minimum number of operations.

A string is k-periodic if it consists of some pattern s of length k repeated multiple times. For example:

• "ababab" is 2-periodic because it's the pattern "ab" repeated 3 times
• "abcabcabc" is 3-periodic because it's the pattern "abc" repeated 3 times

The solution approach divides the string into chunks of length k and counts how many times each unique chunk appears. Since we want all chunks to be the same (to make it k-periodic), we keep the most frequent chunk and replace all others. The minimum operations needed equals the total number of chunks (n/k) minus the count of the most frequent chunk.

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

• The chunks are: "ab", "ab", "ca", "ba"
• "ab" appears 2 times (most frequent)
• Total chunks = 4
• Minimum operations = 4 - 2 = 2 (replace "ca" and "ba" with "ab")

Intuition

To make the string k-periodic, we need all segments of length k to be identical. Let's think about what happens when we divide the string into consecutive chunks of length k.

For example, if we have word = "abcdefabcxyz" and k = 3, we get chunks: ["abc", "def", "abc", "xyz"].

The key insight is that each operation allows us to copy one chunk over another. Since we can only replace chunks (not create new ones), we must work with the chunks we already have.

To minimize operations, we should:

1. Find which chunk appears most frequently (this will be our target pattern)
2. Keep all occurrences of this chunk unchanged
3. Replace all other chunks with this most frequent one

Why does this work? Consider that if we have m total chunks and the most frequent chunk appears f times, then:

• We need to change m - f chunks to match the most frequent one
• Each operation replaces exactly one chunk
• Therefore, we need exactly m - f operations

This greedy approach is optimal because:

• We must make all chunks identical (that's what k-periodic means)
• We minimize work by choosing the chunk that already appears most often
• Any other choice would require more replacements

The formula n/k - max(counter.values()) directly implements this logic:

• n/k gives us the total number of chunks
• max(counter.values()) gives us the frequency of the most common chunk
• Their difference is the minimum number of operations needed

Solution Approach

The implementation uses a counting approach with Python's Counter from the collections module.

Step-by-step breakdown:

1. Calculate the total length: Store n = len(word) for the string length.

2. Generate all k-length chunks: Use a list comprehension with range(0, n, k) to iterate through the string at intervals of k. For each starting position i, extract the substring word[i : i + k].

3. Count chunk frequencies: The Counter takes the list of all chunks and creates a frequency map. For example, if chunks are ["ab", "cd", "ab", "ef"], the Counter becomes {"ab": 2, "cd": 1, "ef": 1}.

4. Find the maximum frequency: Use .values() to get all frequency counts, then max() to find the highest frequency. This represents how many chunks we can keep unchanged.

5. Calculate minimum operations: The formula n // k - max(Counter(...).values()) computes:

   • n // k: Total number of chunks in the string
   • Subtract the maximum frequency to get chunks that need replacement

Example walkthrough: If word = "abcabdabc" and k = 3:

• Chunks extracted: ["abc", "abd", "abc"]
• Counter result: {"abc": 2, "abd": 1}
• Maximum frequency: 2
• Total chunks: 9 // 3 = 3
• Minimum operations: 3 - 2 = 1

The solution efficiently handles the problem in O(n) time complexity for string traversal and O(n/k) space complexity for storing the unique chunks in the Counter.

Example Walkthrough

Let's walk through the solution with word = "abcdabefabcd" and k = 4.

Step 1: Divide into chunks

• Start at positions 0, 4, 8 (multiples of k=4)
• Extract chunks of length 4:
  • Position 0: "abcd"
  • Position 4: "abef"
  • Position 8: "abcd"

Step 2: Count chunk frequencies

• Create frequency map:
  • "abcd" appears 2 times
  • "abef" appears 1 time

Step 3: Identify the most frequent chunk

• Maximum frequency = 2 (for chunk "abcd")
• This will be our target pattern

Step 4: Calculate minimum operations

• Total chunks = 12 ÷ 4 = 3
• Chunks to keep unchanged = 2 (the "abcd" chunks)
• Chunks to replace = 3 - 2 = 1
• Answer: 1 operation needed

Verification:

• We need to replace "abef" at position 4 with "abcd"
• After 1 operation: "abcdabcdabcd" (k-periodic with pattern "abcd")

This demonstrates why keeping the most frequent chunk minimizes operations - we only need to change 1 chunk instead of 2 if we had chosen "abef" as our target.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string word. The algorithm iterates through the string once with a step size of k using range(0, n, k), which creates n/k iterations. For each iteration, it extracts a substring of length k using word[i:i+k], which takes O(k) time. Therefore, the total time for creating all substrings is O(n/k * k) = O(n). The Counter construction processes these n/k substrings in O(n/k) time with O(k) time for hashing each substring, resulting in O(n) total time. Finding the maximum value in the counter takes O(n/k) time. Overall, the time complexity is O(n).

The space complexity is O(n). The Counter stores at most n/k unique substrings, each of length k. In the worst case where all substrings are unique, the total space required is O(n/k * k) = O(n). The generator expression word[i:i+k] for i in range(0, n, k) doesn't create all substrings at once but the Counter still needs to store them, resulting in O(n) space complexity.

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

Common Pitfalls

1. Empty Input or Edge Cases with Counter

The code assumes the Counter will always have values to take the maximum from. If the input string is empty or if there's an issue with chunk generation, max(substring_counter.values()) will fail with a ValueError because max() cannot operate on an empty sequence.

Solution: Add a guard clause to handle empty strings or validate that chunks were generated:

def minimumOperationsToMakeKPeriodic(self, word: str, k: int) -> int:
    n = len(word)
  
    # Handle edge case of empty string
    if n == 0:
        return 0
  
    k_length_substrings = [word[i:i + k] for i in range(0, n, k)]
  
    # Ensure we have substrings to work with
    if not k_length_substrings:
        return 0
  
    substring_counter = Counter(k_length_substrings)
    max_frequency = max(substring_counter.values())
  
    return n // k - max_frequency

2. Assuming k Always Divides n Evenly

While the problem states that k divides n evenly, in a real implementation you might encounter cases where this isn't true. The current code would create a final chunk of length less than k, leading to incorrect results.

Solution: Add validation to ensure k divides n evenly:

def minimumOperationsToMakeKPeriodic(self, word: str, k: int) -> int:
    n = len(word)
  
    # Validate that k divides n evenly
    if n % k != 0:
        raise ValueError(f"k={k} does not divide n={n} evenly")
  
    # Rest of the implementation...

3. Integer Division vs Float Division

Using n / k instead of n // k would return a float, which could cause issues in the subtraction operation or if the result needs to be strictly an integer.

Solution: Always use integer division (//) when calculating the number of blocks:

total_blocks = n // k  # Correct: integer division
# NOT: total_blocks = n / k  # Would return float

4. Memory Inefficiency with Large Strings

Creating a list of all substrings at once with list comprehension stores all chunks in memory simultaneously. For very large strings, this could cause memory issues.

Solution: Use a generator expression with Counter directly to process chunks on-the-fly:

def minimumOperationsToMakeKPeriodic(self, word: str, k: int) -> int:
    n = len(word)
  
    # Use generator expression instead of list comprehension
    substring_counter = Counter(word[i:i + k] for i in range(0, n, k))
  
    # Alternative: process incrementally
    # substring_counter = Counter()
    # for i in range(0, n, k):
    #     substring_counter[word[i:i + k]] += 1
  
    max_frequency = max(substring_counter.values())
    return n // k - max_frequency

5. Not Handling k = 0 or Negative k

If k is 0 or negative, the range function would either cause an infinite loop or behave unexpectedly.

Solution: Validate k at the beginning of the function:

def minimumOperationsToMakeKPeriodic(self, word: str, k: int) -> int:
    if k <= 0:
        raise ValueError(f"k must be positive, got k={k}")
  
    n = len(word)
    # Rest of the implementation...