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// 2273. Find Resultant Array After Removing Anagrams
#include "leetcode.h"
/*
* you are given a 0-indexed string array 'words' where 'words[i]' consists of
* lowercase english letters. in one operation, select any index 'i' such that
* '0 < i < words.length' and 'words[i - 1]' and 'words[i]' are anagrams, and
* delete 'words[i]' from 'words'. keep performing this operation as long as yu
* can select an index that satisfies the conditions. return 'words' after
* performining all operations. it can be shown that selecting the indices for
* each operation in any arbitrary order will lead to the same result. an
* anagram is a word or phrase formed by rearranging the letters of a different
* word or phrase using all the original letters exactly once.
*/
bool anagram(char *s, char *t) {
int hash[26] = {0};
if (strlen(s) != strlen(t))
return false;
for (int i = 0; i < strlen(s); i++)
hash[s[i] - 'a']++;
for (int i = 0; i < strlen(t); i++)
hash[t[i] - 'a']--;
for (int i = 0; i < 26; i++)
if (hash[i])
return false;
return true;
}
char **removeAnagrams(char **words, int wordsSize, int *returnSize) {
char **ans = (char **)calloc(wordsSize, sizeof(char *));
int idx = 0;
ans[idx++] = words[0];
for (int i = 1; i < wordsSize; i++)
if (!anagram(words[i - 1], words[i]))
ans[idx++] = words[i];
*returnSize = idx;
return ans;
}
int main() {
char *w1[] = {"abba", "baba", "bbaa", "cd", "cd"},
*w2[] = {"a", "b", "c", "d", "e"};
int rs1, rs2;
char **ra1 = removeAnagrams((char **)w1, ARRAY_SIZE(w1), &rs1);
char **ra2 = removeAnagrams((char **)w2, ARRAY_SIZE(w2), &rs2);
for (int i = 0; i < rs1; i++)
printf("%s ", ra1[i]);
printf("\n");
for (int i = 0; i < rs2; i++)
printf("%s ", ra2[i]);
printf("\n");
free(ra1);
free(ra2);
}
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