L3 · FAANGDynamic programming~18 min · 5 tests

Coin Change: Minimum Number of Coins

Find the fewest coins that make an amount with bottom-up dynamic programming in O(amount × coins), and see why greedy fails. FAANG DP staple, run live.

The problem

Return the fewest coins from coins (unlimited supply of each) that add up to amount, or -1 if it can't be made.

Examples

Example 1
Input
```
coin_change([1, 2, 5], 11)
```
Expected output
```
3
```
Example 2
Input
```
coin_change([2], 3)
```
Expected output
```
-1
```

+ 3 hidden tests on Submit — greedy would say 3.

Edge cases to ask about

amount = 0
Impossible amount
Greedy counter-example

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · coin_change

def coin_change(coins, amount):
    """Fewest coins summing to amount, or -1."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

Your code runs in real CPython inside your browser — nothing is sent anywhere. The first run downloads the interpreter (about 6 MB, once). Your code is saved on this device as you type.

Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and measure your code against the optimal one at growing input sizes.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

Measure it

Runs the function on inputs of size 250 up to 16,000 and records the time and peak memory. Slow solutions stop early — a short curve is itself the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
Greedy (largest coin first)	O(amount)	O(1)	WRONG for coins like [1, 3, 4] and amount 6.
Recursion over every choice	O(coinsᵃᵐᵒᵘⁿᵗ)	O(amount)
bestBottom-up DP	O(amount · k)	O(amount)	dp[v] = 1 + min(dp[v − c]).

Walkthrough of the optimal approach (try it yourself first)

Greedy fails ([1, 3, 4], amount 6: greedy gives 4+1+1, the best is 3+3). DP: dp[v] is the fewest coins for value v; the last coin used is some c, so dp[v] = 1 + min(dp[v − c]). Fill from 0 up; amount + 1 works as "infinity" because no answer can need more coins than that.

Name the state (dp[v]), the transition and the base case (dp[0] = 0) — that's what an interviewer grades in any DP answer.

Complexity: O(amount · k) time, O(amount) space. For each of the amount + 1 values, every one of the k coins is tried once.

Reveal the reference solution

def coin_change(coins, amount):
    INF = amount + 1
    dp = [0] + [INF] * amount
    for value in range(1, amount + 1):
        for c in coins:
            if c <= value and dp[value - c] + 1 < dp[value]:
                dp[value] = dp[value - c] + 1
    return dp[amount] if dp[amount] != INF else -1

Follow-ups interviewers ask

Count the NUMBER of ways (Coin Change II).
Return the coins used.

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Coin Change: Minimum Number of Coins in Python?

The optimal solution runs in O(amount · k) time and O(amount) auxiliary space. For each of the amount + 1 values, every one of the k coins is tried once.

What is the brute-force approach, and how do you optimise it?

Greedy (largest coin first): O(amount) time, O(1) space. WRONG for coins like [1, 3, 4] and amount 6. Recursion over every choice: O(coinsᵃᵐᵒᵘⁿᵗ) time, O(amount) space. Bottom-up DP: O(amount · k) time, O(amount) space. dp[v] = 1 + min(dp[v − c]).

What follow-up questions do interviewers ask about Coin Change: Minimum Number of Coins?

Count the NUMBER of ways (Coin Change II). Return the coins used.