diff options
-rw-r--r-- | leetcode.h | 1 | ||||
-rw-r--r-- | min_num_push_word2.c | 45 | ||||
-rw-r--r-- | min_num_push_word2.py | 33 |
3 files changed, 79 insertions, 0 deletions
@@ -1,4 +1,5 @@ #include "lib/uthash/src/uthash.h" +#include <assert.h> #include <ctype.h> #include <limits.h> #include <math.h> diff --git a/min_num_push_word2.c b/min_num_push_word2.c new file mode 100644 index 0000000..1b6d057 --- /dev/null +++ b/min_num_push_word2.c @@ -0,0 +1,45 @@ +// 3016. Minimum Number of Pushes to Type Word II +#include "leetcode.h" + +/* + * 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. + */ + +int cmp(const int *const a, const int *const b) { return -(*a - *b); } + +int minimumPushes(char *word) { + int *lower_freq = (int *)calloc(26, sizeof(int)); + for (int i = 0; word[i] != '\0'; i++) { + assert(word[i] >= 'a' && word[i] <= 'z'); + lower_freq[word[i] - 'a']++; + } + qsort(lower_freq, 26, sizeof(int), &cmp); + int ans = 0; + for (int i = 1; 1; i++) { + const int base_idx = 8 * (i - 1); + for (int j = 0; j < 8; j++) { + const int curr_idx = base_idx + j; + if (!(curr_idx < 26 && lower_freq[curr_idx])) + return ans; + ans += i * lower_freq[curr_idx]; + } + } + free(lower_freq); + return ans; +} + +int main() { + char *w1 = "abcde", *w2 = "xyzxyzxyzxyz", *w3 = "aabbccddeeffgghhiiiiii"; + printf("%d\n", minimumPushes(w1)); // expect: 5 + printf("%d\n", minimumPushes(w2)); // expect: 12 + printf("%d\n", minimumPushes(w3)); // expect: 24 +} 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")) |