Introduction

In this blog post, we will explore how to solve the LeetCode problem "3137" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a string word of size n, and an integer k such that k divides n. In one operation, you can pick any two indices i and j, that are divisible by k, then replace the substring of length k starting at i with the substring of length k starting at j. That is, replace the substring word[i..i + k - 1] with the substring word[j..j + k - 1]. Return the minimum number of operations required to make word k-periodic. We say that word is k-periodic if there is some string s of length k such that word can be obtained by concatenating s an arbitrary number of times. For example, if word == “ababab”, then word is 2-periodic for s = "ab". Example 1: Input: word = "leetcodeleet", k = 4 Output: 1 Explanation: We can obtain a 4-periodic string by picking i = 4 and j = 0. After this operation, word becomes equal to "leetleetleet". Example 2: Input: word = "leetcoleet", k = 2 Output: 3 Explanation: We can obtain a 2-periodic string by applying the operations in the table below. i j word 0 2 etetcoleet 4 0 etetetleet 6 0 etetetetet Constraints: 1 <= n == word.length <= 105 1 <= k <= word.length k divides word.length. word consists only of lowercase English letters.

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Divide the input string into blocks of size k.
  • For each block, determine the most frequent block among all the blocks.
  • The minimum number of operations is the sum of the differences between the total number of blocks and the frequency of the most frequent block.

• Complexity:

  • Runtime: O(n), where n is the length of the input string.
  • Storage: O(n), for storing the blocks and their frequencies.

Code

    def min_operations_to_k_periodic(word: str, k: int) -> int:
    """
    Calculates the minimum number of operations required to make the input string k-periodic.

    Args:
        word: The input string.
        k: The length of the repeating substring.

    Returns:
        The minimum number of operations.
    """

    n = len(word)
    num_blocks = n // k
    blocks = []

    # Divide the string into blocks of size k
    for i in range(0, n, k):
        blocks.append(word[i:i + k])

    # Count the frequency of each block
    block_counts = {}
    for block in blocks:
        block_counts[block] = block_counts.get(block, 0) + 1

    # Find the most frequent block
    max_frequency = 0
    for block in block_counts:
        max_frequency = max(max_frequency, block_counts[block])

    # The minimum number of operations is the number of blocks minus the frequency of the most frequent block
    return num_blocks - max_frequency