# 621. Task Scheduler (Medium)

Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks. Tasks could be done without original order. Each task could be done in one interval. For each interval, CPU could finish one task or just be idle.

However, there is a non-negative cooling interval n that means between two same tasks, there must be at least n intervals that CPU are doing different tasks or just be idle.

You need to return the least number of intervals the CPU will take to finish 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.
```

Note:

1. The number of tasks is in the range [1, 10000].
2. The integer n is in the range [0, 100].

## Solutions

``````public class Solution {

public int leastInterval(char[] tasks, int n) {
if (tasks == null || tasks.length == 0) {
return 0;
}

if (n == 0) {
}

int[] count = new int[26];
for(char t : tasks){
count[t - 'A']++;
}

Arrays.sort(count);

int i = 25;
while(i >= 0 && count[i] == count[25]) {
i--;
}

// Assume the time as chunks
// initially, AXXX,AXXX,AXXX,A
// insert B, ABXX,ABXX,ABXX,AB
// ...

// (count[25]-1) means chunks - 1
// (n+1) means the chunk size
// 25 - i means the last chunk you don't have to count in the idle time
return Math.max(tasks.length, (count[25] - 1) * (n + 1) + 25 - i);
}
}
``````

### 2.

``````public class Solution {

// This problem is hard to figure out.

public int leastInterval(char[] tasks, int n) {
if (n == 0) {
}

Map<Character, Integer> taskFrequence = new HashMap<>();
for (char c : tasks) {
}

}

Queue<Integer> queue = new PriorityQueue<>(Comparator.reverseOrder());
for (char c : taskFrequence.keySet()) {
}

int currTime = 0;

Map<Integer, Integer> coolDown = new HashMap<>();

while (!queue.isEmpty() || !coolDown.isEmpty()) {

// get the remained execution times of certain task executed at currTime-n-1, it could be idle
// at that time if not exists in coolDown map
if (coolDown.containsKey(currTime - n - 1)) {

// put the remained execution times to the queue
queue.offer(coolDown.remove(currTime - n - 1));
}

if (!queue.isEmpty()) {
// pick the task of most quantity to execute
int remanent = queue.poll() - 1;

if (remanent != 0) {
coolDown.put(currTime, remanent);
}
}

currTime++;
}

return currTime;
}
}
``````

## Incorrect Solutions

### 1.

``````class Solution {

// Incorrect Solution

public int leastInterval(char[] tasks, int n) {
if (tasks.length == 0) {
return 0;
}

if (n == 0) {
}

int[] freq = new int[26];
for (char t : tasks) {
freq[t - 'A']++;
}

int[] occurrence = new int[26];
Arrays.fill(occurrence, -1);

int len = tasks.length;

int charCount = 0;
int totalCount = 0;

int idx = 0;
while (charCount < len) {
int i = idx % 26;

while (freq[i] == 0) {
idx++;
i = idx % 26;
}

idx++;
freq[i]--;
charCount++;
totalCount++;

if (occurrence[i] < 0) {
occurrence[i] = totalCount - 1;

continue;
}

int idle = n - (totalCount - 2 - occurrence[i]);

totalCount += Math.max(0, idle);

occurrence[i] = totalCount - 1;
}

}
}
``````

### 2.

``````class Solution {

// Incorrect Solution

public int leastInterval(char[] tasks, int n) {
int ans = 0;
if (tasks.length == 0) {
return ans;
}

if (n == 0) {
}

int[] freq = new int[26];
for (char t : tasks) {
freq[t - 'A']++;
}

int remain = tasks.length;

while (remain > 0) {
int count = 0;
for (int i = 0; i < 26; i++) {
if (freq[i] == 0) {
continue;
}

freq[i]--;
count++;

if (count > n) {
break;
}
}

remain -= count;
if (remain == 0) {
ans += count;
break;
}

int idle = 0;
if (count < n + 1) {
idle = n + 1 - count;
}

ans += idle + count;
}

return ans;
}
}
``````