2365. Task Scheduler II

Problem Statement

You are 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 ith task.

You are also given a positive 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.

Example 1:

Input: tasks = [1,2,1,2,3,1], space = 3
Output: 9
Explanation:
One way to complete all tasks in 9 days is as follows:
Day 1: Complete the 0th task.
Day 2: Complete the 1st task.
Day 3: Take a break.
Day 4: Take a break.
Day 5: Complete the 2nd task.
Day 6: Complete the 3rd task.
Day 7: Take a break.
Day 8: Complete the 4th task.
Day 9: Complete the 5th task.
It can be shown that the tasks cannot be completed in less than 9 days.

Example 2:

Input: tasks = [5,8,8,5], space = 2
Output: 6
Explanation:
One way to complete all tasks in 6 days is as follows:
Day 1: Complete the 0th task.
Day 2: Complete the 1st task.
Day 3: Take a break.
Day 4: Take a break.
Day 5: Complete the 2nd task.
Day 6: Complete the 3rd task.
It can be shown that the tasks cannot be completed in less than 6 days.

Constraints:

• 1 <= tasks.length <= 105

• 1 <= tasks[i] <= 109

• 1 <= space <= tasks.length

Intuition

Approach:

Maintain a map to keep track of the next time, When it can be execute

Links

https://leetcode.com/problems/task-scheduler-ii/description/

Video Links

Approach 1:

C++

class Solution {
public:
    long long taskSchedulerII(vector<int>& tasks, int space) {
        unordered_map<int, long long> mp;
        long long cur_time=0;

        for(auto &it: tasks){
            if(mp.find(it) == mp.end()){
                mp[it] = cur_time+space+1;
                cur_time++;
            }
            else{
                // For taking the idle time, current time is less so have to wait
                if(cur_time < mp[it]){
                    cur_time = mp[it];
                }

                // No ideal time required, Directly shift the time ahead
                mp[it] = cur_time+space+1;
                cur_time++;
            }
        }

        return cur_time;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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