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// 2365. Task Scheduler II
#include <stdio.h>
#include <stdlib.h>
/*
* given a 0-indexed array of positive integers 'tasks', representing tasks that
* need to be completed in order, where 'tasks[i]' represents the type of the
* i'th task. you are also given a positve integer 'space' which represents the
* minimum number of days that must pass after the completion of a task before
* another task of the same type can be performed. each day until all tasks have
* been completed, you must either
* - complete the next task from 'tasks' or
* - take a break
* return the minimum number of days needed to complete all tasks.
*/
long long taskSchedulerII(int *tasks, int tasks_size, int space) {
long long ans = 0;
int n = tasks_size, m = 10 * tasks_size;
long long **hash = calloc(m, sizeof(long long *));
for (int i = 0; i < n; i++) {
int val = tasks[i], d = val;
while (1) {
if (!hash[d % m]) {
ans++;
hash[d % m] = malloc(2 * sizeof(long long));
hash[d % m][0] = tasks[i];
hash[d % m][1] = ans;
break;
} else if (hash[d % m][0] == tasks[i]) {
ans++;
if (ans >= (hash[d % m][1] + space + 1))
hash[d % m][1] = ans;
else {
hash[d % m][1] += (space + 1);
ans = hash[d % m][1];
}
break;
} else
d++;
}
}
for (int i = 0; i < m; i++)
if (hash[i])
free(hash[i]);
free(hash);
return ans;
}
int main() {
int t1[] = {1, 2, 1, 2, 3, 1}, t2[] = {5, 8, 8, 5};
printf("%lld\n", taskSchedulerII(t1, 6, 3)); // expect: 9
printf("%lld\n", taskSchedulerII(t2, 4, 2)); // expect: 6
}
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