Dynamic programming
Pay with the Fewest Voucher Tokens
01. The question
Voucher tokens have positive integer values and unlimited supply. Return the fewest tokens that pay an exact amount, or -1 when it cannot be paid. Amount 0 needs no tokens.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Inputtokens = [1, 3, 4], amount = 6
Output2
- The table starts with dp[0] = 0; other amounts are unknown.
- At amount 3, one 3-token is enough.
- At amount 6, two 3-tokens beat the greedy 4 + 1 + 1 plan.
Try your code
Define solve(data, parameter="") and return your result. Your code runs locally, not on a server.
Edit your test cases
Use 1-10 objects with data and an optional string parameter. Expected outputs are computed from the lesson. An optional expected field must be correct.
Input contract & reference code
Example data: [1,3,4]. Parameter: "6". Numeric parameters arrive as strings; convert them with int(parameter) when needed.
The example and generated cases use the question’s exact variant. This checks behavior on a small test batch—not performance on every possible input.
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Printed output
Five-second execution limit per batch. Standard-library Python only; additional packages and network access are unavailable. Stop ends execution without blocking the lesson.
03. The idea
Try sketching your own approach first. What would you track, and why would it work?
Reveal the idea
- Let dp[value] be the fewest tokens for that exact value.
- Try each token as the final token of a payment.
- Build smaller amounts before larger amounts and keep unreachable states explicit.
04. Watch it happen
Use Step to move slowly, or Play to follow the whole trace. Change a preset or load your own input.
Input & settings
Red = current item / blue = active window or cells / green = completed work. A value docks at its actual destination; the original input stays unchanged. Positions start at 0.
Working state / what changes
Ready to begin
Enable JavaScript to step through the animation. The worked example and solution below remain readable without it.
Python / why it changes
The line behind the step
def voucher_tokens(tokens, amount):
dp = [amount + 1] * (amount + 1)
dp[0] = 0
for value in range(1, amount + 1):
for token in tokens:
if token <= value:
dp[value] = min(dp[value], dp[value - token] + 1)
return dp[amount] if dp[amount] <= amount else -1How to use this walkthrough
Hover or focus a control for its label. Play forms the input and follows values into the working state. Step shows one change; Back and the slider restore exact snapshots. Reduced motion uses instant updates. Saved snapshots and sorted heap displays are teaching overhead, not algorithm complexity.
05. Python solution
def voucher_tokens(tokens, amount):
dp = [amount + 1] * (amount + 1)
dp[0] = 0
for value in range(1, amount + 1):
for token in tokens:
if token <= value:
dp[value] = min(dp[value], dp[value - token] + 1)
return dp[amount] if dp[amount] <= amount else -106. Cost & edge cases
- Time
- O(Ak), amount A and k token values
- Space
- O(A)
- [2] cannot pay 3.
- Greedy largest-first is not correct for arbitrary denominations.
- An empty token list pays only amount 0.