# 19.Task Scheduler

Given a characters array `tasks`, representing the tasks a CPU needs to do, where each letter represents a different task. Tasks could be done in any order. Each task is done in one unit of time. For each unit of time, the CPU could complete either one task or just be idle.

However, there is a non-negative integer `n` that represents the cooldown period between two **same tasks** (the same letter in the array), that is that there must be at least `n` units of time between any two same tasks.

```
Example 1:
Input: tasks = ["A","A","A","B","B","B"], n = 2
Output: 8
Explanation: 
A -> B -> idle -> A -> B -> idle -> A -> B
There is at least 2 units of time between any two same tasks.

Example 2:
Input: tasks = ["A","A","A","B","B","B"], n = 0
Output: 6
Explanation: On this case any permutation of size 6 would work since n = 0.
["A","A","A","B","B","B"]
["A","B","A","B","A","B"]
["B","B","B","A","A","A"]
...
And so on.

Example 3:
Input: tasks = ["A","A","A","A","A","A","B","C","D","E","F","G"], n = 2
Output: 16
Explanation: 
One possible solution is
A -> B -> C -> A -> D -> E -> A -> F -> G -> A -> idle -> idle -> A -> idle -> idle -> A
```

## Solution: (Priority Queue)

**Sorting according to the frequency**

```cpp
class Solution
{
public:
    int leastInterval(vector<char> &tasks, int n)
    {
        if (n == 0)
        {
            return tasks.size();
        }

        vector<int> arr(26, 0);
        for (int i = 0; i < tasks.size(); i++)
        {
            arr[tasks[i] - 65]++;
        }

        priority_queue<int, vector<int>> pq;

        for (int i = 0; i < 26; i++)
        {
            if (arr[i] > 0)
            {
                pq.push(arr[i]);
            }
        }

        vector<int> temp;
        int res = 0;

        while (!pq.empty())
        {

            int time = 1;

            temp.push_back(pq.top());
            pq.pop();

            for (int i = 0; i < n; i++)
            {
                if (!pq.empty())
                {
                    temp.push_back(pq.top());
                    pq.pop();
                    time++;
                }
            }

            for (int i = 0; i < temp.size(); i++)
            {
                if (temp[i] - 1 > 0)
                {
                    pq.push(temp[i] - 1);
                }
            }

            res += (pq.empty()) ? time : n + 1;
            temp.clear();
        }

        return res;
    }
};
```

**Time Complexity: O(n log n)**

## Solution :&#x20;

```cpp
class Solution
{
public:
    int leastInterval(vector<char> &tasks, int n)
    {
        if (n == 0)
        {
            return tasks.size();
        }

        vector<int> arr(26, 0);
        for (int i = 0; i < tasks.size(); i++)
        {
            arr[tasks[i] - 65]++;
        }

        sort(arr.begin(), arr.end(), greater<int>());

        int maxEle = arr[0] - 1; //getting the max frquency
        int idle = maxEle * n; //no of idle gaps 

        for (int i = 1; i < 26; i++)
        {
            idle -= min(maxEle, arr[i]); //filling the gaps
        }

        if (idle < 0)
        {
            return tasks.size();
        }

        return idle + tasks.size();
    }
};

```

**Time Complexity: O(n)**


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://soumyajit4419.gitbook.io/ds-algo/arrays/20.task-scheduler.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.