621. Task Scheduler

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].

FB Followup:

1. Output the order as original task

Thoughts:

1. count most frequent task, say k.
2. create k chunks, then fill less frequent chars into the gaps into each chunk
3. compare the length of the task vs (number of chunks) * (least length of each chunk) + last chunks size = ( k -1 ) * (n + 1) + (# of most frequent chars)

Code:

class Solution {
public:
    int leastInterval(vector<char>& tasks, int n) {
        int count [26];  fill_n(count, 26, 0); // need to fill each time since OJ seems to re-use the same 
        // allocated space for running multiple tests
        for(char task : tasks){
            count[task - 'A']++;
        }
        sort(begin(count),end(count));
        // count how many most frequent chars
        int i = 25;
        while(i >= 0 && count[i] == count[25])
             i--;
        int tasks_len = tasks.size();
        return max(tasks_len, (count[25] - 1)*(n + 1) + 25 - i);
    }
};

Code: C+

class Solution {
public:
    int leastInterval(vector<char>& tasks, int n) {
        vector<int> counter (26);
        int max_val = 0;
        int maxCount = 0;
        for (auto c : tasks){
            counter[c - 'A']++;
            if(max_val == counter[c - 'A']) maxCount ++;
            else if(max_val < counter[c - 'A']){
                max_val = counter[c- 'A'];
                maxCount = 1;
            } 
        }
        int partCount = max_val - 1;
        int partLen = n - (maxCount - 1); // could be negative
        int emptySlots = partCount * partLen;
        int residualTasks = tasks.size() - max_val * maxCount;
        int idles = max(0, emptySlots - residualTasks);

        return tasks.size() + idles; 

    }
};