Introduction

In this blog post, we will explore how to solve the LeetCode problem "3016" 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 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 = "xyzxyzxyzxyz" Output: 12 Explanation: The remapped keypad given in the image provides the minimum cost. "x" -> one push on key 2 "y" -> one push on key 3 "z" -> one push on key 4 Total cost is 1 4 + 1 4 + 1 4 = 12 It can be shown that no other mapping can provide a lower cost. Note that the key 9 is not mapped to any letter: it is not necessary to map letters to every key, but to map all the letters. Example 3: Input: word = "aabbccddeeffgghhiiiiii" Output: 24 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 "f" -> one push on key 7 "g" -> one push on key 8 "h" -> two pushes on key 9 "i" -> one push on key 9 Total cost is 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 2 2 + 6 * 1 = 24. It can be shown that no other mapping can provide a lower cost. Constraints: 1 <= word.length <= 105 word consists of lowercase English letters.

Explanation

Here's a breakdown of the solution:

• Count Character Frequencies: Calculate the frequency of each character in the input string word.
• Prioritize by Frequency: Assign the most frequent characters to the smallest number of pushes (1, 2, 3, etc.) possible using the available keys (8 keys, max 3 pushes on each).

• Calculate Total Pushes: Sum the product of each character's frequency and its assigned number of pushes.

• Runtime Complexity: O(N), where N is the length of the word. This is due to iterating through the word once to count frequencies and then iterating at most 26 times to assign pushes.

• Storage Complexity: O(1), because we are using a constant amount of space for the character counts and the pushes list (both bounded by the size of the alphabet).

Code

    from collections import Counter

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

    Args:
        word: The string to type.

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

    counts = Counter(word)
    frequencies = sorted(counts.values(), reverse=True)
    pushes = [0] * len(frequencies)
    total_pushes = 0

    for i in range(len(frequencies)):
        pushes[i] = (i // 8) + 1
        total_pushes += frequencies[i] * pushes[i]

    return total_pushes