### Description

#### LeetCode 621

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 1:**

Input:tasks = ["A","A","A","B","B","B"], n = 2Output:8Explanation:A -> B -> idle -> A -> B -> idle -> A -> B.

**Note:**

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

### Analysis

The length of result should be the length of tasks plus the number of idle slots. So we could find the number of idle slots first. It is easy to know that the maximum idle slots should be the maximum time of repeated tasks times n. So we can initialize the number of idle slots as this and use it to minus the times of every tasks. If one task’s time is the maximum time, then the last interval does not need idle slots.

### Solution

class Solution { public int leastInterval(char[] tasks, int n) { int[] map=new int[26]; for(int i=0;i<tasks.length;i++) { map[tasks[i]-'A']++; } Arrays.sort(map); int maxval=map[25]-1; int idle_slots=maxval*n; for(int i=24;i>=0&&map[i]>0;i--) { idle_slots=idle_slots-Math.min(map[i],maxval); } return idle_slots>0?idle_slots+tasks.length:tasks.length; } }