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Diffstat (limited to 'min_num_push_word2.py')
-rw-r--r-- | min_num_push_word2.py | 33 |
1 files changed, 33 insertions, 0 deletions
diff --git a/min_num_push_word2.py b/min_num_push_word2.py new file mode 100644 index 0000000..010c9a1 --- /dev/null +++ b/min_num_push_word2.py @@ -0,0 +1,33 @@ +# 3016. Minimum Number of Pushes to Type Word II +from collections import Counter + +""" +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. +""" + + +class Solution(object): + def minimumPushes(self, word): + """ + :type word: str + :rtype: int + """ + return sum( + (i // 8 + 1) * j for i, j in enumerate(sorted(Counter(word).values())[::-1]) + ) + + +if __name__ == "__main__": + obj = Solution() + print(obj.minimumPushes(word="abcde")) + print(obj.minimumPushes(word="xyzxyzxyzxyz")) + print(obj.minimumPushes(word="aabbccddeeffgghhiiiiii")) |