1. Count tasks
Tally how often each task appears.
counts[task] = (counts[task] ?? 0) + 1Typorithm · medium
Task Scheduler
▷ Interactive algorithm labtask-schedulergreedyfrequencyidle-slotsgreedy-choice
Saved on this device
Full lesson · 8 learning steps Open any exercise for focused practice
10%
Step 1 of 8 · intro card
Learn Task Scheduler
Greedy
Frame the schedule around the most frequent task's cooldown gaps, then fill or overflow those gaps.
Identical tasks need n cooldown slots between runs.
The most frequent task creates (max - 1) gaps of size n + 1, plus the final batch of max-frequency tasks.
If there are enough other tasks to fill every idle slot, the answer is simply the total number of tasks.
Preparing algorithm lab…
01
Count frequencies first.
02
Skeleton: (max - 1) * (n + 1) + ties.
03
Answer is max(skeleton, total tasks).
function leastInterval(tasks, n) {
const counts = {};
for (const task of tasks) {
counts[task] = (counts[task] ?? 0) + 1;
}
const values = Object.values(counts);
const max = Math.max(...values);
const maxCount = values.filter((c) => c === max).length;
const slots = (max - 1) * (n + 1) + maxCount;
return Math.max(slots, tasks.length);
}Read the implementation
Connect each part of the code to the concept before you type.
2. Busiest task
The most frequent task forces the schedule's skeleton.
const max = Math.max(...values)3. Ties
Count how many tasks share the maximum frequency.
const maxCount = values.filter((c) => c === max).length4. Slot formula
Frame the schedule around the busiest task's cooldown gaps.
const slots = (max - 1) * (n + 1) + maxCount5. Lower bound
If tasks are dense enough there are no idles, so the answer is just the count.
return Math.max(slots, tasks.length)A complete learning loop
01
Understand
Read the concept map and follow the code breakdown.
02
Reconstruct
Type the implementation, repair mistakes and work through every recall exercise.
03
Verify
Use the final check without hints; revisit weak concepts in short practice sessions.