Q21: How do you solve the Coin Change problem using Bottom-Up Dynamic Programming?

Find the fewest coins needed to make up amount. Define dp[i] as the minimum coins needed for amount i. Initialize dp array of size amount + 1 filled with amount + 1 (infinity placeholder), with dp[0] = 0. Outer loop: i from 1 to amount. Inner loop: each coin in coins. If i - coin >= 0, update dp[i] = Math.min(dp[i], 1 + dp[i - coin]). Return dp[amount] > amount ? -1 : dp[amount].

Bottom-Up Coin Change DP Solution

import java.util.Arrays;

public class CoinChange {
    public static int coinChange(int[] coins, int amount) {
        int max = amount + 1;
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, max);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int coin : coins) {
                if (i - coin >= 0) {
                    dp[i] = Math.min(dp[i], 1 + dp[i - coin]);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
}

💡 Computes the minimum coin count for target amounts up to amount.

Key Takeaways

Bottom-Up 1D Dynamic Programming.
State Relation: $dp[i] = min(dp[i], 1 + dp[i - coin])$.

Knowledge Check

What is the time complexity of the Bottom-Up Coin Change Dynamic Programming solution?

Q21: How do you solve the Coin Change problem using Bottom-Up Dynamic Programming?

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