## Similar Problems

### Valid Phone Numbers

Problem Statement:

You are given a list of tasks that need to be executed, each with a fixed start time and duration. Given a non-negative integer n that represents the cooldown period between two tasks of the same type, return the minimum number of CPU intervals needed to execute all the given tasks.

Example:

Input: tasks = [“A”,”A”,”A”,”B”,”B”,”B”], n = 2

Output: 8

Explanation:
A -> B -> idle -> A -> B -> idle -> A -> B

Approach:

To solve this problem, we can use a greedy approach. First, we need to count the frequency of each task. Then, we sort the tasks in decreasing order of their frequency. We also maintain a list of cooldowns for each task. Initially, all the cooldowns are 0.

We start processing the tasks in the order of their frequency. For each task, we find the task with the maximum cooldown. If the maximum cooldown is greater than 0, we wait for that many intervals before processing the current task. If the maximum cooldown is 0, we can process the current task immediately.

After processing the current task, we update the cooldown of all the tasks. The cooldown of the current task is set to n+1 (n is the given cooldown period). This is because we need to leave a gap of n intervals before processing the same task again. The cooldown of all the other tasks is reduced by 1.

We repeat this process until all the tasks are processed.

Code:

`class Solution { public: int leastInterval(vector<char>& tasks, int n) { unordered_map<char, int> freq; for (char t : tasks) { freq[t]++; } auto cmp = [&](char a, char b) { return freq[a] < freq[b]; }; priority_queue<char, vector<char>, decltype(cmp)> pq(cmp); for (auto p : freq) { pq.push(p.first); } vector<int> cooldown(26, 0); int intervals = 0; while (!pq.empty()) { vector<char> available; while (!pq.empty()) { char t = pq.top(); pq.pop(); if (cooldown[t-'A'] == 0) { available.push_back(t); break; } } if (available.empty()) { intervals++; } else { char t = available.front(); available.erase(available.begin()); freq[t]--; cooldown[t-'A'] = n+1; for (char c : available) { cooldown[c-'A']--; } if (freq[t] > 0) { pq.push(t); } intervals++; } for (int i = 0; i < 26; i++) { if (cooldown[i] > 0) { cooldown[i]--; } } } return intervals; } };`

Time Complexity:

The time complexity of this solution is O(nlogn), where n is the number of tasks. The sorting of tasks takes O(nlogn) time, and for each task, we can spend at most O(logn) time in the priority queue. Since we process each task exactly once, the total time complexity is O(nlogn).

## Step by Step Implementation For Task Scheduler

```class Solution {
public int leastInterval(char[] tasks, int n) {
int[] map = new int[26];
map[c - 'A']++;
Arrays.sort(map);
int max_val = map[25] - 1, idle_slots = max_val * n;
for (int i = 24; i >= 0 && map[i] > 0; i--) {
idle_slots -= Math.min(map[i], max_val);
}
}
}```
```There are a total of n tasks you have to pick, labeled from 0 to n-1. Some tasks may have dependencies, i.e. they need to be done before others.

Given the number of tasks and a list of dependencies, return a sequence of tasks you need to pick to finish all tasks.

You may assume that there will be no circular dependencies.

Example 1:

Input: n = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [0,2,1,3]
Explanation: There are a total of 4 tasks. You can do them in any order, but you should make sure to do task 0 first.

def findOrder(numCourses, prerequisites):

#Create a graph and add edges
graph = collections.defaultdict(list)
for u, v in prerequisites:
graph[u].append(v)

#Create a visited set to keep track of visited nodes
visited = set()

#Create a stack to keep track of the order of nodes
stack = []

#Create a function to do a depth first search
def dfs(node):
if node in visited:
return False
for neighbor in graph[node]:
if not dfs(neighbor):
return False
stack.append(node)
return True

#Do a depth first search for all nodes
for i in range(numCourses):
if not dfs(i):
return []
return stack[::-1]```
```var leastInterval = function(tasks, n) {
// create a hashmap to store the count of each task
let map = {};
map[t] = (map[t]||0) + 1;
}
// sort the tasks according to their count in descending order
let sortable = [];
for(let t in map){
sortable.push([t, map[t]]);
}
sortable.sort((a,b) => {
return b[1] - a[1];
});
// find the maximum count
let maxCount = sortable[0][1];
// find the number of tasks with maximum count
for(let i = 0; i < sortable.length; i++){
if(sortable[i][1] == maxCount){
}
}
// the minimum number of intervals needed is equal to the maximum count of any task plus the number of tasks with that maximum count minus 1, all multiplied by n plus 1
};```
```class Solution {
public:
int leastInterval(vector& tasks, int n) {
// Create a map of task counts
for (char c : tasks) {
}

// Sort the task counts in descending order
priority_queue pq;
for (auto it : taskCounts) {
pq.push(it.second);
}

// Perform the scheduling
int intervalCount = 0;
while (!pq.empty()) {
// Schedule up to n tasks at a time
for (int i = 0; i <= n; i++) {
if (!pq.empty()) {
pq.pop();
}
}

for (int count : scheduledTasks) {
if (--count > 0) {
pq.push(count);
}
}

// Add n+1 to the interval count (if there are more tasks to schedule)
intervalCount += pq.empty() ? scheduledTasks.size() : n + 1;
}

return intervalCount;
}
};```
```public class Solution {
public int LeastInterval(char[] tasks, int n) {
// Create a frequency map of all the tasks
{
{
}
else
{
}
}

// Sort the tasks by frequency (most frequent first)

// Initialize the result
var result = 0;

{
// Initialize the current task list

// Loop through the remaining tasks
for(int i = 0; i < sortedTasks.Count; i++)
{
// If the current task list is not full
{

// Decrement the index so that we don't skip a task
i--;
}
else
{
// Break out of the loop, we don't need to process any more tasks
break;
}
}

// If the sorted task list is empty, we can just return the result
{
}

// Otherwise, we need to update the frequencies of the tasks that we processed
{
}

// Sort the tasks by frequency (most frequent first)