0039MediumArrayDynamic ProgrammingBreadth-First SearchKnapsack ProblemComplete Knapsack

Fewest Tokens for the Charge

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A toll booth accepts plastic tokens. The integer array values lists the token types the booth recognises, where each entry is the credit one token of that type carries, and the booth's dispenser holds an unlimited number of tokens of every listed type.

A driver owes charge credits and the booth gives no change, so the tokens dropped in must add up to charge exactly. Return the smallest number of tokens that does it. If no combination of the listed types adds up to charge, return -1.

A charge of 0 credits is settled by dropping in nothing at all, so the answer there is 0.

Examples

Example 1

Input: values = [1, 4, 6], charge = 8
Output: 2

Two 4-credit tokens settle the charge. Reaching for the 6 first would force two 1s after it, which costs three tokens.

Example 2

Input: values = [7, 5], charge = 3
Output: -1

Every token carries more credit than the charge, and the booth gives no change, so the charge cannot be settled.

Example 3

Input: values = [3, 10], charge = 0
Output: 0

Nothing is owed, so no token is needed.

Constraints

1 <= values.length <= 12
1 <= values[i] <= 2^31 - 1
0 <= charge <= 10^4
Every token type may be used any number of times

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def fewest_tokens(values: list[int], charge: int) -> int:

Java

public int fewestTokens(int[] values, int charge)

The intended approach

Time O(charge * len(values))Space O(charge)

Build up the cheapest settlement for every credit total from `0` to `charge`. A total is settled by dropping in one token of some type on top of an already-settled smaller total, so the best count for a total is one more than the best among those predecessors. Working upward means every predecessor is final by the time it is read, and a total left unreachable stays marked as impossible.

Pythonentry point: fewest_tokens

def fewest_tokens(values, charge):
    unreachable = charge + 1
    best = [0] + [unreachable] * charge
    for total in range(1, charge + 1):
        for credit in values:
            if credit <= total and best[total - credit] + 1 < best[total]:
                best[total] = best[total - credit] + 1
    return -1 if best[charge] == unreachable else best[charge]

The obvious version

Breadth-first search over reachable credit totals. Level k holds every total settleable with k tokens, so the first time the charge shows up gives the fewest tokens. Exponential-looking but bounded by the number of totals below the charge.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_fewest_tokens(values, charge):
    if charge == 0:
        return 0
    from collections import deque
    seen = {0}
    queue = deque([(0, 0)])
    while queue:
        total, dropped = queue.popleft()
        for credit in values:
            nxt = total + credit
            if nxt == charge:
                return dropped + 1
            if nxt < charge and nxt not in seen:
                seen.add(nxt)
                queue.append((nxt, dropped + 1))
    return -1

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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