Problem statement

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 = 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].

Solution

This is also a pure math problem. Generally, it depends on the maximum number of tasks.

Basic idea is divided into three categories:

• Find the most appeared task, employ a hash table, indexed by the tasks.
• Count the number of most appeared task, calculate the return value.
• If the return value is less than the size of array(because the cooling time is too short). return the size of array.

Time complexity is O(n), space complexity is O(n).

class Solution {
public:
    int leastInterval(vector<char>& tasks, int n) {
        unordered_map<char, int> hash_table;
        int max_val = 0;
        for(auto c : tasks){
            hash_table[c]++;
            max_val = max(max_val, hash_table[c]);
        }
        int max_cnt = 0;
        for(auto it : hash_table){
            if(it.second == max_val){
                max_cnt++;
            }
        }
        int task_size = tasks.size();
        return max(task_size, (n + 1) * (max_val - 1) + max_cnt);
    }
};