Task Scheduler — DSA Visualizer

Problem

You are given a list of CPU tasks labeled A–Z and an integer n representing a cooldown period. Between two tasks with the same label, the CPU must wait at least n intervals (idle counts). Return the minimum number of intervals to finish all tasks.

Example 1

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

Output: 8

Explanation: A→B→idle→A→B→idle→A→B = 8 intervals.

Example 2

Input: tasks = [A,A,A,B,B,B], n = 0

Output: 6

Explanation: No cooldown — run all 6 tasks back-to-back.

Example 3

Input: tasks = [A,A,A,A,B,B,B,C,D], n = 2

Output: 10

Explanation: A appears 4×; formula gives (4-1)×3 + 1 = 10 with 2 idle slots.

Constraints: 1 ≤ tasks.length ≤ 10⁴ | tasks[i] is uppercase A–Z | 0 ≤ n ≤ 100

Task Frequencies

Variables

maxFreq

—

maxCount

—

n (cooldown)

—

partCount

—

partLen

—

result

—

CPU Schedule Timeline

Most Frequent

Other Tasks

Idle

Tail

Formula Builder

maxFreq = highest task frequency

—

maxCount = tasks sharing maxFreq

—

partCount = maxFreq − 1

—

partLen = n + 1

—

emptySlots = partCount × partLen − (tasks − maxCount)

—

result = tasks.length + max(0, emptySlots)

—

Step Logic

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Algorithm

Count task frequencies; find maxFreq (highest frequency)

maxCount = number of tasks sharing maxFreq

partCount = maxFreq − 1 partitions; partLen = n + 1 slots each

emptySlots = partCount × partLen − (tasks.length − maxCount)

return tasks.length + max(0, emptySlots)

Time

O(n)

Space

O(1)

Why the Formula Works

The most frequent task (say A with frequency f) forces a minimum structure: f−1 partitions of width n+1 separate its appearances, with the final chunk appended at the end. Other tasks slot into the partitions; any leftover partition capacity becomes idle time. If there are enough tasks to fill everything, idle = 0 and the answer is simply tasks.length. Ties at maxFreq share the final tail.