LeetCode #621 — MEDIUM

Task Scheduler

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array of CPU tasks, each labeled with a letter from A to Z, and a number n. Each CPU interval can be idle or allow the completion of one task. Tasks can be completed in any order, but there's a constraint: there has to be a gap of at least n intervals between two tasks with the same label.

Return the minimum number of CPU 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 intervals before doing A again. The same applies to task B. In the 3^rd interval, neither A nor B can be done, so you idle. By the 4^th interval, 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

Step 01

Brute Force Baseline

Problem summary: You are given an array of CPU tasks, each labeled with a letter from A to Z, and a number n. Each CPU interval can be idle or allow the completion of one task. Tasks can be completed in any order, but there's a constraint: there has to be a gap of at least n intervals between two tasks with the same label. Return the minimum number of CPU intervals required to complete all tasks.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Greedy

Example 1

["A","A","A","B","B","B"]
2

Example 2

["A","C","A","B","D","B"]
1

Example 3

["A","A","A", "B","B","B"]
3

Core Insight

What unlocks the optimal approach

There are many different solutions for this problem, including a greedy algorithm.
For every cycle, find the most frequent letter that can be placed in this cycle. After placing, decrease the frequency of that letter by one.
Use Priority Queue.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #621: Task Scheduler
class Solution {
    public int leastInterval(char[] tasks, int n) {
        int[] cnt = new int[26];
        int x = 0;
        for (char c : tasks) {
            c -= 'A';
            ++cnt[c];
            x = Math.max(x, cnt[c]);
        }
        int s = 0;
        for (int v : cnt) {
            if (v == x) {
                ++s;
            }
        }
        return Math.max(tasks.length, (x - 1) * (n + 1) + s);
    }
}

// Accepted solution for LeetCode #621: Task Scheduler
func leastInterval(tasks []byte, n int) int {
	cnt := make([]int, 26)
	x := 0
	for _, c := range tasks {
		c -= 'A'
		cnt[c]++
		x = max(x, cnt[c])
	}
	s := 0
	for _, v := range cnt {
		if v == x {
			s++
		}
	}
	return max(len(tasks), (x-1)*(n+1)+s)
}

# Accepted solution for LeetCode #621: Task Scheduler
class Solution:
    def leastInterval(self, tasks: List[str], n: int) -> int:
        cnt = Counter(tasks)
        x = max(cnt.values())
        s = sum(v == x for v in cnt.values())
        return max(len(tasks), (x - 1) * (n + 1) + s)

// Accepted solution for LeetCode #621: Task Scheduler
use std::collections::{BinaryHeap, VecDeque, HashMap};

impl Solution {
    pub fn least_interval(tasks: Vec<char>, n: i32) -> i32 {
        let mut count = HashMap::new();
        let mut max_heap = BinaryHeap::new();
        for task in tasks {
            let count = count.entry(task).or_insert(0);
            *count += 1;
        }
        for (key, val) in count.iter() {
            max_heap.push(*val);
        }
        let mut time = 0;
        let mut deque: VecDeque<(i32,i32)> = VecDeque::new();
        while deque.len() > 0 || max_heap.len() > 0 {
            time += 1;

            if max_heap.len() == 0 {
                time = deque[0].1;
            }
            else {
                let cnt = max_heap.pop().unwrap() - 1;
                if cnt != 0 {
                    deque.push_back((cnt, time + n));
                }
            }
            if deque.len() > 0 && deque[0].1 == time {
                max_heap.push(deque.pop_front().unwrap().0);
            }
        }
        time
    }
}

// Accepted solution for LeetCode #621: Task Scheduler
function leastInterval(tasks: string[], n: number): number {
    const charMap = new Map();
    let maxCharCount = 0;
    let maxChar = tasks[0];

    for (let char of tasks) {
        charMap.set(char, (charMap.get(char) || 0) + 1);
        if (charMap.get(char) > maxCharCount) {
            maxCharCount = charMap.get(char);
            maxChar = char;
        }
    }

    let idleCount = (maxCharCount - 1) * n;

    charMap.forEach((count, char) => {
        // 'return' inside forEach() serve as 'continue'
        if (char === maxChar) {
            return;
        }
        if (count === maxCharCount) {
            idleCount -= count - 1;
        } else {
            idleCount -= count;
        }
    });

    if (idleCount <= 0) {
        return tasks.length;
    }
    return tasks.length + idleCount;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.