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.

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].

Solutions

Credit: concise Java Solution O(N) time O(26) space

Idea:

Consider the example AAABBBCCDDE with n = 3

1. Construct a frame using the most frequent types of tasks. The frame consists of many blocks. Each block is a sequence of most frequent types of tasks. ABxxABxxAB for the example.

2. For each remaining type of tasks, add the tasks between the blocks. For the example,

   1. Add C: ABCxABCxAB

   2. Add D: ABCDABCDAB

   3. Add E: ABCDEABCDAB

class Solution {
public:
    int leastInterval(vector<char>& tasks, int n) {
        if(tasks.empty()) return 0;
        vector<int> freqs(26);
        for(auto task : tasks)
            ++freqs[task - 'A'];
        sort(freqs.begin(), freqs.end(), greater<int>());
        int f = freqs[0], i = 1;
        while(i < 26 && freqs[i] == f) ++i;
        int frame_size = max(i, n + 1) * (f - 1) + i;
        int num = i * f;
        while(i < 26 && freqs[i] > 0){
            num += freqs[i];
            ++i;
        }
        return max(num, frame_size);
    }
};

The above solution can be simplified to:

class Solution {
public:
    int leastInterval(vector<char>& tasks, int n) {
        if(tasks.empty()) return 0;
        vector<int> freqs(26);
        for(auto task : tasks)
            ++freqs[task - 'A'];
        sort(freqs.begin(), freqs.end(), greater<int>());
        int f = freqs[0], i = 1;
        while(i < 26 && freqs[i] == f) ++i;
        int frame_size = max(i, n + 1) * (f - 1) + i;
        int num = tasks.size();
        return max(num, frame_size);
    }
};

Because at the end num == nums.size().