Introduction

In this blog post, we will explore how to solve the LeetCode problem "3014" 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 containing distinct lowercase English letters. Telephone keypads have keys mapped with distinct collections of lowercase English letters, which can be used to form words by pushing them. For example, the key 2 is mapped with ["a","b","c"], we need to push the key one time to type "a", two times to type "b", and three times to type "c" . It is allowed to remap the keys numbered 2 to 9 to distinct collections of letters. The keys can be remapped to any amount of letters, but each letter must be mapped to exactly one key. You need to find the minimum number of times the keys will be pushed to type the string word. Return the minimum number of pushes needed to type word after remapping the keys. An example mapping of letters to keys on a telephone keypad is given below. Note that 1, *, #, and 0 do not map to any letters. Example 1: Input: word = "abcde" Output: 5 Explanation: The remapped keypad given in the image provides the minimum cost. "a" -> one push on key 2 "b" -> one push on key 3 "c" -> one push on key 4 "d" -> one push on key 5 "e" -> one push on key 6 Total cost is 1 + 1 + 1 + 1 + 1 = 5. It can be shown that no other mapping can provide a lower cost. Example 2: Input: word = "xycdefghij" Output: 12 Explanation: The remapped keypad given in the image provides the minimum cost. "x" -> one push on key 2 "y" -> two pushes on key 2 "c" -> one push on key 3 "d" -> two pushes on key 3 "e" -> one push on key 4 "f" -> one push on key 5 "g" -> one push on key 6 "h" -> one push on key 7 "i" -> one push on key 8 "j" -> one push on key 9 Total cost is 1 + 2 + 1 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 12. It can be shown that no other mapping can provide a lower cost. Constraints: 1 <= word.length <= 26 word consists of lowercase English letters. All letters in word are distinct.

Explanation

Here's a solution to the problem, along with an explanation of the approach:

• Frequency Analysis: Calculate the frequency of each character in the input word. This is crucial because more frequent characters should ideally be mapped to keys that require fewer pushes.

• Greedy Assignment: Sort the character frequencies in descending order. Assign the most frequent characters to keys that require the fewest pushes (i.e., one push), then the next most frequent to keys that require two pushes, and so on. The keys available are 2-9, total 8 keys.

• Cost Calculation: Calculate the total cost based on the greedy assignment. The cost is the sum of frequency * pushes for each character.

• Runtime Complexity: O(n log n), where n is the length of the word (due to sorting), because in the worst case, all characters can be distinct. The frequency calculation is O(n), and the cost calculation is O(n), but sorting dominates. Storage Complexity: O(1), since we only store a fixed size frequency map (26 letters) and a few constant size variables.

Code

    def minimum_pushes(word: str) -> int:
    """
    Calculates the minimum number of pushes needed to type the string after remapping the keys.

    Args:
        word: The input string containing distinct lowercase English letters.

    Returns:
        The minimum number of pushes needed to type the word.
    """

    frequencies = {}
    for char in word:
        frequencies[char] = frequencies.get(char, 0) + 1

    sorted_frequencies = sorted(frequencies.values(), reverse=True)

    total_pushes = 0
    push_count = 1
    keys_available = 8

    for i, frequency in enumerate(sorted_frequencies):
        total_pushes += frequency * push_count
        if (i + 1) % keys_available == 0:
            push_count += 1

    return total_pushes