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// 2182. Construct String With Repeat Limit
#include "leetcode.h"
/*
* you are given a string 's' and an integer 'repeatLimit'. construct a new
* string 'repeatLimitedString' using the characters of 's' such that no letter
* appears more than 'repeatLimit' times in a row. you do not have to use all
* characters from 's'. return the lexicographically largest
* 'repeatLimitedString' possible. a string 'a' is lexicographically larger than
* string 'b' if in the first position where 'a' and 'b' differ, string 'a' has
* a letter that appears later in the alphabet than the corresponding letter in
* 'b'. if the first 'min(a.len, b.len)' characters do not differ, then the
* longer string is the lexicographically larger one
*/
int *get_cnt(char *s) {
int n = strlen(s);
int *cnt = (int *)calloc(26, sizeof(int));
for (int i = 0; i < n; i++)
cnt[s[i] - 'a']++;
return cnt;
}
char *repeatLimitedString(char *s, int repeatLimit) {
int n = strlen(s), *cnt = get_cnt(s), idx = 0;
char *ans = (char *)calloc(n + 1, sizeof(char));
for (int i = 'z' - 'a'; i >= 0;) {
if (!cnt[i]) {
i--;
continue;
}
char c = i + 'a';
if (!idx || ans[idx - 1] != c) {
int copy = cnt[i] <= repeatLimit ? cnt[i] : repeatLimit;
memset(&ans[idx], c, copy);
idx += copy;
cnt[i] -= copy;
} else {
int next = i - 1;
while (next >= 0 && !cnt[next])
next--;
if (next < 0)
break;
else {
ans[idx++] = next + 'a';
cnt[next]--;
}
}
}
ans[idx] = '\0';
return ans;
}
int main() {
char *s1 = "cczazcc", *s2 = "aababab";
char *rls1 = repeatLimitedString(s1, 3);
char *rls2 = repeatLimitedString(s2, 2);
printf("%s\n", rls1); // expect: "zzcccac"
printf("%s\n", rls2); // expect: "bbabaa"
free(rls1), free(rls2);
}
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