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// 2365. Task Scheduler II
#include "leetcode.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.
*/
class Solution {
public:
long long taskSchedulerII(vector<int> &tasks, int space) {
unordered_map<int, long long> last;
long long ans = 0;
for (int t : tasks) {
if (last.count(t))
last[t] = ans = max(ans, last[t] + space) + 1;
else
last[t] = ++ans;
}
return ans;
}
};
int main() {
Solution obj;
vector<int> t1 = {1, 2, 1, 2, 3, 1}, t2 = {5, 8, 8, 5};
printf("%lld\n", obj.taskSchedulerII(t1, 3)); // expect: 9
printf("%lld\n", obj.taskSchedulerII(t2, 2)); // expect: 6
}
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