Minimum Number of Pushes to Type Word II

lcpy gen -n 3016   # by problem number
lcpy gen -s minimum_number_of_pushes_to_type_word_ii   # by problem name

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.

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.

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.

from collections import Counter


class Solution:
    # Time: O(n)
    # Space: O(1)
    def minimum_pushes(self, word: str) -> int:
        counts = sorted(Counter(word).values(), reverse=True)
        return sum(count * (index // 8 + 1) for index, count in enumerate(counts))