CODING CHALLENGE · N°14

Retry with Exponential Backoff

Easy AI EngineeringAgentsReliability

Tools time out and APIs rate-limit — a production agent has to retry without hammering. Exponential backoff waits longer after each failure (1s, 2s, 4s, 8s…) up to a cap, so the system backs off instead of piling on. Compute the delay schedule.

The problem

Implement backoff_delays(retries, base, cap). Return the list of wait times before each retry attempt, where the delay for attempt i (0-indexed) is base × 2ⁱ, capped at cap. There are retries attempts in total, so the list has retries entries.

EXAMPLE 1

Input retries=5, base=1, cap=10

Output [1, 2, 4, 8, 10]

doubles each time, then clamps at the cap

EXAMPLE 2

Input retries=3, base=2, cap=100

Output [2, 4, 8]

never reaches the cap

EXAMPLE 3

Input retries=0, base=1, cap=10

Output []

no retries → no delays

CONSTRAINTS

Delay for attempt i = min(cap, base × 2ⁱ), with i from 0 to retries − 1.
The result always has exactly retries elements.
Pure arithmetic — no sleeping, no randomness (that would be jitter, a separate concern).

SOLVE IT YOURSELF

Your turn — write it

Edit the stub, hit Run (or ⌘/Ctrl + Enter), and watch the hidden tests. Stuck? the hints are right above and Reveal solution is one click away.

YOUR TASK

Implement backoff_delays(retries, base, cap) — the exponential-backoff wait schedule, each delay doubling from base and clamped at cap.

HINTS — 3 IDEAS

Loop i from 0 to retries − 1.
The raw delay is base shifted left i times: base × 2**i.
Clamp each delay with min(cap, …) so it never exceeds the cap.

CPython · WebAssembly

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