621. Task Scheduler

Description：
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.

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

方法1：數學推導

• Time complexity : $O\left ( n \right )$
• Space complexity : $O$
  ( n ) O\left ( n \right )
  思路：
  先找到出現次數最多的元素，然後試此元素的中間間隔為n；
  然後將其他元素插空；
  最後會發現有可能會有元素只能在最後，取個max即可。

class Solution {
public:
	int leastInterval(vector<char>& tasks, int n) {
		unordered_map<char, int> hash_map;
		int count = 0;
		for (auto t : tasks) {
			hash_map[t]++;
			count = max(hash_map[t], count);
		}
		int res = (count - 1) * (n + 1);
		for (auto t : hash_map)
			if (t.second == count)
				res++;
		return max((int)tasks.size(), res);
    }
};