621. Task Scheduler.cpp

Given a characters array tasks, representing the tasks a CPU needs to do, 
where each letter represents a different task. Tasks could be done in any order. 
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.

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:

1 <= task.length <= 104
tasks[i] is upper-case English letter.
The integer n is in the range [0, 100].












class Solution {
public:
    int leastInterval(vector<char>& tasks, int n) {
        vector <int> freq(26, 0);
        
        for (auto &x: tasks)  freq[x - 'A']++;
        
        int maxi = *max_element(freq.begin(), freq.end());
        
        int maxiCnt = 0;
        
        for (auto &x: freq) {
            if (maxi == x) maxiCnt++;
        }
        
        int partCnt = maxi - 1;
        
        int partLength = n - (maxiCnt - 1);
        
        int availableTasks = tasks.size() - (maxi * maxiCnt);
        
        int emptySlots = partCnt * partLength;
        
        int idle = max(0, emptySlots -availableTasks);
        
        return tasks.size() + idle;
    }
};