Task Scheduler

You are given a char array representing tasks CPU need to do. It contains capital letters A to Z where each letter represents a different task. Tasks could be done without the original order of the array. 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.

You need to return the least number of units of times that the CPU will take to finish all the given 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

Constraints:

• The number of tasks is in the range [1, 10000].

• The integer n is in the range [0, 100].

/**
 * The key is to find out how many idles do we need. Let's first look at how to
 * arrange them. it's not hard to figure out that we can do a "greedy
 * arrangement": always arrange task with most frequency first. 
 * E.g. we have following tasks : 3 A, 2 B, 1 C. and we have n = 2. According to what we have
 * above, we should first arrange A, and then B and C. Imagine there are "slots"
 * and we need to arrange tasks by putting them into "slots". Then A should be
 * put into slot 0, 3, 6 since we need to have at least n = 2 other tasks
 * between two A. After A put into slots, it looks like this:
 * 
 * A ? ? A ? ? A "?" is "empty" slots.
 * 
 * Now we can use the same way to arrange B and C. The finished schedule should
 * look like this:
 * 
 * A B C A B # A 
 * "#" is idle
 */
public class Solution {
    public int leastInterval(char[] tasks, int n) {
        int[] counter = new int[26];
        int max = 0; // Max frequency
        int maxCount = 0; // Number of characters at max frequency
        for (char task : tasks) {
            counter[task - 'A']++;
            if (max == counter[task - 'A'])
                maxCount++;
            else if (max < counter[task - 'A']) {
                max = counter[task - 'A'];
                maxCount = 1;
            }
        }

        int partCount = max - 1;
        int partLength = n - (maxCount - 1);
        int emptySlots = partCount * partLength;
        int availableTasks = tasks.length - max * maxCount;
        int idles = Math.max(0, emptySlots - availableTasks);

        return tasks.length + idles;
    }
}