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Coin Change

Medium Dynamic Programming Knapsack Original on LeetCode

Coin Change is a medium dynamic programming problem solved with the knapsack pattern. The best approach, optimal (unbounded knapsack dp), runs in O(amount · k) time and O(amount) space. Below are 2 approaches in Java, from recursion up.

Problem

Given coin values and a target amount, return the fewest coins needed to make exactly that amount, or -1 if it cannot be made. You have unlimited coins of each value.

Examples

Example 1

Input
coins = [1, 2, 5], amount = 11
Output
3
Why
5 + 5 + 1.

Example 2

Input
coins = [2], amount = 3
Output
-1

Constraints

  • 1 <= coins.length <= 12; 0 <= amount <= 10^4.
  • Unlimited coins of each type.

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
RecursionO(k^(amount / min coin))O(amount)
Optimal (unbounded knapsack DP)O(amount · k)O(amount)

1Recursion

TimeO(k^(amount / min coin))
SpaceO(amount)

Try every coin as the last one and recurse on the remaining amount.

  1. f(0) = 0; f(a < 0) = infinity; f(a) = 1 + min f(a - c).
Java
class Solution {
    public int coinChange(int[] coins, int amount) {
        int r = f(coins, amount);
        return r == Integer.MAX_VALUE ? -1 : r;
    }

    private int f(int[] coins, int a) {
        if (a == 0) return 0;
        if (a < 0) return Integer.MAX_VALUE;
        int best = Integer.MAX_VALUE;
        for (int c : coins) {
            int sub = f(coins, a - c);
            if (sub != Integer.MAX_VALUE) best = Math.min(best, sub + 1);
        }
        return best;
    }
}

2Optimal (unbounded knapsack DP)

TimeO(amount · k)
SpaceO(amount)

Build dp from 0 up to amount. Each amount takes the best of using each coin last. Unreachable amounts stay at a sentinel larger than any real answer.

  1. dp[0] = 0, dp[1..amount] = amount + 1.
  2. For a in 1..amount, for c in coins with c <= a: dp[a] = min(dp[a], dp[a - c] + 1).
  3. Return dp[amount] > amount ? -1 : dp[amount].
Java
class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, amount + 1);
        dp[0] = 0;
        for (int a = 1; a <= amount; a++)
            for (int c : coins)
                if (c <= a) dp[a] = Math.min(dp[a], dp[a - c] + 1);
        return dp[amount] > amount ? -1 : dp[amount];
    }
}

Edge cases to test

  • amount = 0 (answer 0)
  • Impossible amounts
  • Greedy fails, e.g. coins [1, 3, 4] for amount 6 (3 + 3 beats 4 + 1 + 1)

Hints

Hint 1

dp[a] = fewest coins that make amount a = 1 + min over coins c of dp[a - c].

FAQ

What is the best time complexity for Coin Change?

Optimal (unbounded knapsack DP) runs in O(amount · k) time and O(amount) extra space.

Which pattern does Coin Change use?

It is a dynamic programming problem that uses the knapsack pattern. Other problems with the same pattern: 0 - 1 Knapsack Problem, Fractional Knapsack, Knapsack with Duplicate Items.

Is there a brute force solution for Coin Change?

Yes. Recursion takes O(k^(amount / min coin)) time and O(amount) space. Try every coin as the last one and recurse on the remaining amount.

Which edge cases should I test for Coin Change?

amount = 0 (answer 0); Impossible amounts; Greedy fails, e.g. coins [1, 3, 4] for amount 6 (3 + 3 beats 4 + 1 + 1).