// 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 &tasks, int space) { unordered_map 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 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 }