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-rw-r--r--task_scheduler.cpp36
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diff --git a/task_scheduler.cpp b/task_scheduler.cpp
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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
-}