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Problem Statement:

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.
You may assume that you have an infinite number of each kind of coin.

Example 1:
Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

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

Example 3:
Input: coins = [1], amount = 0
Output: 0

Example 4:
Input: coins = [1], amount = 1
Output: 1

Example 5:
Input: coins = [1], amount = 2
Output: 2

Solution:

This problem has striking similarity with Unbounded Knapsack Concept. (1) We have unlimited supply of each items (coins) and (2) we need to meet exactly the amount specified.

For every coin denomination we compute the coin count to make up the amount (1) by INCLUDING one or more instances of that coin denomination, and (2) by NOT INCLUDING an instance. We take the minimum of these two computations. The code below would clarify the idea.

Top-Down Dynamic programming Solution with 2D Tabulation:

Java Code:

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Python Code:

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Bottom-Up Solution with Space Optimization:

We would be using the template discussed in Unbounded Knapsack Concept chapter to implement an efficient Dynamic Programming solution, as shown below:

Java Code:

public int coinChangeSpaceOptimized(int[] coins, int amount) {
    int[] dp = new  int[amount + 1];
        
    Arrays.fill(dp, Integer.MAX_VALUE);
    dp[0] = 0;
    for (int coinIndex = 0; coinIndex < coins.length; coinIndex++) {
        if (coins[coinIndex] <= amount) {
            dp[coins[coinIndex]] = 1;
        }
    }
        
    for (int i = 1; i <= amount; i++) {
        for (int j = 0; j < coins.length; j++) {
            if (coins[j] <= i && dp[i - coins[j]] != Integer.MAX_VALUE) {
                dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
            }
        }
    }
    return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
}

Python Code:

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Tip:

Often times if a problem mentions that you can include an element as many time as you want (or, infinitely), there is a chance that the problem could be solved using Unbounded Knapsack Concept.

Instructor: