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// 1170. Compare Strings by Frequency of the Smallest Character
#include "leetcode.h"
/*
* let the function 'f(s)' be the frequency of the lexicographically smallest
* character in a non empty string 's'. for example, if 's = dcce' then 'f(s) =
* 2' because the lexicographically smallest character is 'c', which has a
* frequency of 2. you are given an array of strings 'words' and another array
* of query strings 'queries'. for each query 'queries[i]', count the number of
* words in 'words' such that 'f(queries[i]) < f(w)' for each 'w' in 'words'.
* return an integer array 'answer' where each 'answer[i]' is the answer to the
* i'th query
*/
int lex_freq(char *w) {
int *map = (int *)calloc('z' - 'a' + 1, sizeof(int)), res;
char smallest = 'z';
for (int i = 0; w[i]; i++) {
map[w[i] - 'a']++;
smallest = fmin(smallest, w[i]);
}
res = map[smallest - 'a'];
free(map);
return res;
}
int *numSmallerByFrequency(char **queries, int queriesSize, char **words,
int wordsSize, int *returnSize) {
int *word_freq = (int *)malloc(wordsSize * sizeof(int));
int *ans = (int *)calloc(queriesSize, sizeof(int)), freq;
*returnSize = queriesSize;
for (int i = 0; i < wordsSize; i++)
word_freq[i] = lex_freq(words[i]);
for (int i = 0; i < queriesSize; i++) {
freq = lex_freq(queries[i]);
for (int j = 0; j < wordsSize; j++)
if (freq < word_freq[j])
ans[i]++;
}
free(word_freq);
return ans;
}
int main() {
char *q1[] = {"cbd"}, *q2[] = {"bbb", "cc"};
char *w1[] = {"zaaaz"}, *w2[] = {"a", "aa", "aaa", "aaaa"};
int rs1, rs2;
int *nsbf1 = numSmallerByFrequency((char **)q1, ARRAY_SIZE(q1), (char **)w1,
ARRAY_SIZE(w1), &rs1);
int *nsbf2 = numSmallerByFrequency((char **)q2, ARRAY_SIZE(q2), (char **)w2,
ARRAY_SIZE(w2), &rs2);
for (int i = 0; i < rs1; i++)
printf("%d ", nsbf1[i]); // expect: 1
printf("\n");
for (int i = 0; i < rs2; i++)
printf("%d ", nsbf2[i]); // expect: 1 2
printf("\n");
free(nsbf1), free(nsbf2);
}
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