621. Task Scheduler

/**

 * 621. Task Scheduler

 * 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.

 *

 * problem asks to find max time it will take to finish the tasks given a cool down period between the slots.

*/

The problem is interesting in that it has a reusable pattern that can be applied to other problems.

See my solution to the problem here.

https://github.com/zcoderz/leetcode/blob/main/src/main/java/face_book/medium/TaskScheduler.java

Essentially the problem asks to calculate number of cpu cycles needed to finish the work while recognizing cool down of N cycles between same task.

The problem can be solved as:

1. Sort the tasks by frequency
2. Find the highest frequency.
3. Subtract 1 from the highest frequency and multiply it by cool down period. This gets us the number of cycles we would need to wait between repeating that task. Note that we can do other tasks during this time. Call the number calculated here: open_slots.
4. Work your way from next highest frequency of task to least highest frequency of task while reducing open_slots (variable calculated in 3) with that frequency.
5. Total cpu cycles = total tasks + open_slots

Here is the code:

public int leastInterval(char[] tasks, int n) {
    int num = 'Z' - 'A' + 1;
    int[] taskFreq = new int[num];
    for (char task : tasks) {
        taskFreq[task - 'A'] = taskFreq[task - 'A'] + 1;
    }
    //sort the tasks by their frequency
    Arrays.sort(taskFreq);
    int maxFreq = taskFreq[num - 1];
    //open slots is a very intelligent concept.
    //it indicates the max wait that is bound to occur if there were no
    //other tasks that can get done during the wait
    int openSlots = (maxFreq - 1) * n;
    //remove the open slots until all slots have been used
    for (int j = num - 2; j >= 0 && openSlots > 0 && taskFreq[j] > 0; j--) {
        //taking a min here so that if two tasks have same highest frequency , you can only reduce at max
        //(maxFrequency -1)
        openSlots -= Math.min(taskFreq[j], maxFreq - 1);
    }
    //number slots needed are total work unit + remaining open slots
    return tasks.length + openSlots;
}