621. Task Scheduler Leet code Daily Challenge Using JavaScript





You are given an array of CPU tasks, each represented by letters A to Z, and a cooling time, n. Each cycle or interval allows the completion of one task. Tasks can be completed in any order, but there's a constraint: identical tasks must be separated by at least n intervals due to cooling time.

​Return the minimum number of intervals required to complete all tasks.

Example 1:

Input: tasks = ["A","A","A","B","B","B"], n = 2

Output: 8

Explanation: A possible sequence is: A -> B -> idle -> A -> B -> idle -> A -> B.

After completing task A, you must wait two cycles before doing A again. The same applies to task B. In the 3rd interval, neither A nor B can be done, so you idle. By the 4th cycle, you can do A again as 2 intervals have passed.

Example 2:

Input: tasks = ["A","C","A","B","D","B"], n = 1

Output: 6

Explanation: A possible sequence is: A -> B -> C -> D -> A -> B.

With a cooling interval of 1, you can repeat a task after just one other task.

Example 3:

Input: tasks = ["A","A","A", "B","B","B"], n = 3

Output: 10

Explanation: A possible sequence is: A -> B -> idle -> idle -> A -> B -> idle -> idle -> A -> B.

There are only two types of tasks, A and B, which need to be separated by 3 intervals. This leads to idling twice between repetitions of these tasks.

Constraints:

  • 1 <= tasks.length <= 104
  • tasks[i] is an uppercase English letter.
  • 0 <= n <= 100

Using JavaScript Solutions

/**
 * @param {character[]} tasks
 * @param {number} n
 * @return {number}
 */
var leastInterval = function(tasks, n) {
     // Count the frequency of each task
    let taskFreq = new Array(26).fill(0);
    for (let task of tasks) {
        taskFreq[task.charCodeAt(0) - 'A'.charCodeAt(0)]++;
    }

    // Sort the frequencies in descending order
    taskFreq.sort((a, b) => b - a);

    // Find the maximum frequency (most frequent task count)
    let maxFreq = taskFreq[0];

    // Count the tasks with the same count as the most frequent task
    let sameFreqCount = 1;
    for (let i = 1; i < taskFreq.length; i++) {
        if (taskFreq[i] === maxFreq) {
            sameFreqCount++;
        } else {
            break;
        }
    }

    // Calculate the total intervals
    let totalIntervals = (maxFreq - 1) * (n + 1) + sameFreqCount;

    // Return the maximum of total intervals and the total number of tasks
    return Math.max(totalIntervals, tasks.length);
 
};


Output

Case 1.

Input: tasks = ["A","A","A","B","B","B"], n = 2

Output: 8


Case 2.

Input: tasks = ["A","C","A","B","D","B"], n = 1

Output: 6


Case 3.

Input: tasks = ["A","A","A", "B","B","B"], n = 3

Output: 10