[Dynamic Programming] Coin Change - Medium

The Coin Change problem is a well-known challenge in algorithms. It asks how many ways we can make change for a certain amount using different coin types. We can solve this problem in many ways. These include recursive methods and dynamic programming techniques. By using dynamic programming, we can find the number of combinations faster. This approach is better than using simple recursive solutions.

In this article, we will look into the Coin Change problem closely. First, we will understand the problem. Then, we will explore recursive solutions and dynamic programming methods in Java, Python, and C++. We will also talk about ways to make space usage better for the Coin Change problem. Lastly, we will answer common questions to help clear up any confusion.

• Dynamically Solving the Coin Change Problem - Medium Level
• Understanding the Coin Change Problem
• Recursive Solution to Coin Change Problem in Java
• Dynamic Programming Approach for Coin Change in Java
• Recursive Solution to Coin Change Problem in Python
• Dynamic Programming Approach for Coin Change in Python
• Recursive Solution to Coin Change Problem in C++
• Dynamic Programming Approach for Coin Change in C++
• Optimizing Space Complexity in Coin Change Problem
• Frequently Asked Questions

If we want to get better at dynamic programming, we can check out more resources. The articles on Dynamic Programming Fibonacci Number and Dynamic Programming Climbing Stairs will give us more helpful ideas.

Understanding the Coin Change Problem

The Coin Change Problem is a common challenge in dynamic programming. We want to find out how many ways we can make change for a certain amount using a set of coin types. Here is how we can define the problem:

• Input: A list of coin types and a target amount.
• Output: The number of ways to reach the target amount with the coins.

Problem Statement

We are given a list of different integers called coins. These represent the coin types. We also have an integer amount. Our goal is to return how many combinations can make that amount. We can assume we have unlimited coins of each type.

Example

• Input: coins = [1, 2, 5], amount = 5
• Output: 4 (the combinations are: [1,1,1,1,1], [1,1,1,2], [1,2,2], [5])

Key Properties

1. Recursion: The problem has parts that repeat. We can define the solution using smaller parts.
2. Dynamic Programming: When we save results of the smaller parts, we can build the solution faster. This helps us avoid doing the same work again.
3. Combinatorial Nature: The order of the coins does not change the outcome. This is a combination problem, not a permutation problem.

Use Cases

• In finance, where we deal with different types of money.
• In inventory management, where we need to calculate item combinations.
• In any case where we need to reach a specific total using different units.

The Coin Change Problem is a basic example that helps us learn about dynamic programming. It is also linked to other problems like the Fibonacci sequence. We can find more about it in articles such as Dynamic Programming - Fibonacci Number (Easy).

Recursive Solution to Coin Change Problem in Java

We can solve the Coin Change problem using a recursive method. We will look at each coin and make calls to find the number of ways to make the target amount. Here is a simple Java code for the recursive solution.

Java Code Implementation

public class CoinChange {
    public static int countWays(int[] coins, int amount, int index) {
        // Base case: if amount is 0, there's one way to make change (use no coins)
        if (amount == 0) {
            return 1;
        }
        // Base case: if amount is less than 0 or no coins left, no way to make change
        if (amount < 0 || index >= coins.length) {
            return 0;
        }
        
        // Include the coin at index and exclude it
        return countWays(coins, amount - coins[index], index) + countWays(coins, amount, index + 1);
    }

    public static void main(String[] args) {
        int[] coins = {1, 2, 5};
        int amount = 5;
        int result = countWays(coins, amount, 0);
        System.out.println("Number of ways to make change: " + result);
    }
}

Explanation of the Code

• Function countWays:
  • Gets an array of coin types, the target amount, and the current coin index.
  • Returns how many ways we can make change for the amount.
• Base Cases:
  • If the amount is 0, we return 1 because there is one way to make change.
  • If the amount is less than 0 or the index is more than the number of coins, we return 0 because we can’t make change.
• Recursive Calls:
  • We include the current coin by reducing the amount and keep the same index.
  • We also exclude the current coin by moving to the next index.

This recursive way has a time complexity of O(2^n). It explores all subsets of the coin types. This method is simple but not very efficient for big inputs. For better solutions, we can look at dynamic programming methods in the next sections.

Dynamic Programming Approach for Coin Change in Java

The Coin Change problem is a well-known example of dynamic programming. We need to find the least number of coins to make a specific amount with a set of coin values. Here is a simple dynamic programming solution in Java.

Dynamic Programming Solution

We use a 1D array to keep track of the minimum number of coins needed for each amount up to our target amount.

Code Implementation

public class CoinChange {
    public static int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];
        // Set dp array with a big number
        Arrays.fill(dp, amount + 1);
        dp[0] = 0; // Base case
        
        for (int coin : coins) {
            for (int x = coin; x <= amount; x++) {
                dp[x] = Math.min(dp[x], dp[x - coin] + 1);
            }
        }
        
        return dp[amount] > amount ? -1 : dp[amount];
    }

    public static void main(String[] args) {
        int[] coins = {1, 2, 5};
        int amount = 11;
        int result = coinChange(coins, amount);
        System.out.println(result); // Output: 3
    }
}

Explanation of Code

• Initialization: We start the dp array with amount + 1. This means it is an impossible number of coins.
• Base Case: We set dp[0] to 0 because we do not need any coins to make the amount 0.
• Dynamic Programming Loop: For each coin, we update the dp array for all amounts that can be made with that coin.
• Result: If dp[amount] is still more than amount, then it means we cannot make that amount with the coins we have. So we return -1.

Time Complexity

The time complexity for this solution is O(n * m). Here n is the number of coins and m is the target amount. The space complexity is O(m) because of the dp array.

For more problems about dynamic programming, we can check these articles: - Dynamic Programming - Fibonacci Number - Dynamic Programming - Climbing Stairs

Recursive Solution to Coin Change Problem in Python

The Coin Change problem is about finding how many ways we can make a certain amount using some coin types. We can use a recursive method to look at all possible ways to use coins.

Recursive Function

This recursive function takes the target amount and a list of coin types. It gives back the number of ways to make that amount.

def coin_change_recursive(coins, amount):
    # Base case: if amount is 0, there is one way to make that amount (using no coins)
    if amount == 0:
        return 1
    # If amount is less than 0, there are no ways to make that amount
    if amount < 0:
        return 0
    
    ways = 0
    for coin in coins:
        # Recursively call for the remaining amount
        ways += coin_change_recursive(coins, amount - coin)
    
    return ways

# Example usage
coins = [1, 2, 5]
amount = 5
result = coin_change_recursive(coins, amount)
print(f"Number of ways to make {amount} using coins {coins}: {result}")

Explanation

• Base Case: If amount is 0, we return 1 because there is one way to do that (using no coins).
• Recursive Case: For each coin, we reduce the amount and add the results together.
• This method takes a lot of time because it has overlapping problems.

This recursive solution is easy to understand but not fast for big amounts or many coins. If we want to make it better, we can use the dynamic programming method we will talk about next.

For more information on dynamic programming methods, you can check the Dynamic Programming - Fibonacci Number.

Dynamic Programming Approach for Coin Change in Python

The Coin Change problem helps us find the fewest coins needed to make a certain amount using a set of coins. The dynamic programming approach is quick. It avoids repeating calculations like in the recursive method.

Dynamic Programming Solution

We can use a 1D array to keep track of the minimum number of coins needed for each amount from 0 to our target amount. We build the solution by using values we already calculated.

Implementation in Python

Here is how we can use the dynamic programming method for the Coin Change problem in Python:

def coinChange(coins, amount):
    # Initialize the dp array with infinity
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0  # Base case: 0 coins are needed to make the amount 0

    # Fill the dp array
    for coin in coins:
        for x in range(coin, amount + 1):
            dp[x] = min(dp[x], dp[x - coin] + 1)

    # If we have found a solution, return it; otherwise, return -1
    return dp[amount] if dp[amount] != float('inf') else -1

# Example usage
coins = [1, 2, 5]
amount = 11
print(coinChange(coins, amount))  # Output: 3 (11 = 5 + 5 + 1)

Explanation of the Code

• Initialization: We create an array called dp. It is the size of amount + 1, and we set all values to infinity (float('inf')). The first element dp[0] is set to 0 because we need 0 coins to make 0 amount.
• Filling the dp array: For every coin, we check all amounts from the coin value to the target amount. We update the dp array with the minimum coins we need to reach each amount.
• Result: If dp[amount] is still infinity, it means we cannot make that amount with the coins we have. So we return -1. If we can make it, we return dp[amount].

This dynamic programming way is much faster than the simple recursive way. It works better for bigger inputs. For more practice on dynamic programming, we can check out dynamic programming Fibonacci number or dynamic programming climbing stairs.

Recursive Solution to Coin Change Problem in C++

We can solve the Coin Change problem using a recursive method. This method looks at all the ways to use coins to get a certain amount. The recursive function checks each coin. It calls itself with the new amount until it reaches zero or less.

C++ Implementation

Here is a simple C++ code for the recursive solution of the Coin Change problem:

#include <iostream>
#include <vector>

using namespace std;

int countWays(int coins[], int n, int amount) {
    // Base case: if amount is 0, there's one way to make change
    if (amount == 0) {
        return 1;
    }
    // Base case: if amount is less than 0, there's no way to make change
    if (amount < 0) {
        return 0;
    }
    // If no coins left
    if (n <= 0) {
        return 0;
    }
    
    // Count the solutions including the last coin + excluding the last coin
    return countWays(coins, n - 1, amount) + countWays(coins, n, amount - coins[n - 1]);
}

int main() {
    int coins[] = {1, 2, 3};
    int amount = 4;
    int n = sizeof(coins) / sizeof(coins[0]);
    
    cout << "Number of ways to make change: " << countWays(coins, n, amount) << endl;
    return 0;
}

Explanation of the Code

• Function Signature: countWays(int coins[], int n, int amount) takes an array of coin types, the number of coins, and the target amount.
• Base Cases:
  • If amount is 0, we return 1. This means there is one way to make change.
  • If amount is negative, we return 0. This means there is no way to make change.
  • If we have no coins left (n <= 0), we return 0.
• Recursive Calls: The function returns the total of two cases:
  • We do not use the last coin and solve the same amount.
  • We use the last coin and solve for the new amount (amount - last coin).

This recursive way is simple. But it has a problem with time because it takes too long for big inputs. So for bigger numbers, we usually use a dynamic programming method to make it faster.

For more learning about dynamic programming, we can check out articles like the Dynamic Programming Fibonacci Number and Dynamic Programming Climbing Stairs.

Dynamic Programming Approach for Coin Change in C++

We can solve the Coin Change problem well with a Dynamic Programming approach in C++. Our goal is to find the least number of coins needed to make a target amount from a set of coin denominations.

Problem Statement

We have an array of coin denominations and a target amount. We need to find the minimum number of coins to make that amount. If we can’t make the amount, we will return -1.

Dynamic Programming Solution

1. Create a DP Array: We will define a DP array. In this array, dp[i] is the minimum number of coins to get the amount i. We set dp[0] = 0 because we need 0 coins to make amount 0. We set all other values to INT_MAX to show they are unreachable.

2. Fill the DP Array: For each coin in our denominations, we will update the DP array for all amounts from the coin value to the target amount.

3. Return Result: The result will be in dp[amount]. If this value is still INT_MAX, we return -1.

C++ Code Implementation

#include <iostream>
#include <vector>
#include <climits>

int coinChange(std::vector<int>& coins, int amount) {
    std::vector<int> dp(amount + 1, INT_MAX);
    dp[0] = 0; // Base case

    for (const auto& coin : coins) {
        for (int j = coin; j <= amount; ++j) {
            if (dp[j - coin] != INT_MAX) {
                dp[j] = std::min(dp[j], dp[j - coin] + 1);
            }
        }
    }

    return dp[amount] == INT_MAX ? -1 : dp[amount];
}

int main() {
    std::vector<int> coins = {1, 2, 5};
    int amount = 11;
    int result = coinChange(coins, amount);
    std::cout << "Minimum coins needed: " << result << std::endl;
    return 0;
}

Time Complexity

The time complexity for this method is O(n * amount). Here, n is the number of coin denominations.

Space Complexity

The space complexity is O(amount) because we use the DP array to store results.

This dynamic programming way helps to solve the Coin Change problem well. It is good for cases where we need to use coins in the best way. If you want to learn more about dynamic programming, you can check out related topics like Dynamic Programming - Fibonacci Number or Dynamic Programming - Unique Paths in a Grid.

Optimizing Space Complexity in Coin Change Problem

To make space better in the Coin Change problem, we can use a one-dimensional array. This is better than using a two-dimensional array for our dynamic programming solution. We see that the current state only needs the previous state. So, we can save space by storing fewer results.

Space Optimization Technique

1. 1D Array Utilization: Instead of using a 2D array dp[i][j] where i is the number of coins and j is the amount, we use a single array dp[j]. This array keeps track of the number of ways to make change for amounts from 0 to n.

Dynamic Programming Approach in Java

Here is how we can do this space saving in Java:

public class CoinChange {
    public static int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];
        dp[0] = 1; // Base case

        for (int coin : coins) {
            for (int j = coin; j <= amount; j++) {
                dp[j] += dp[j - coin];
            }
        }

        return dp[amount];
    }

    public static void main(String[] args) {
        int[] coins = {1, 2, 5};
        int amount = 5;
        System.out.println("Number of ways to make change: " + coinChange(coins, amount));
    }
}

Dynamic Programming Approach in Python

The same way in Python would look like this:

def coin_change(coins, amount):
    dp = [0] * (amount + 1)
    dp[0] = 1  # Base case

    for coin in coins:
        for j in range(coin, amount + 1):
            dp[j] += dp[j - coin]

    return dp[amount]

if __name__ == "__main__":
    coins = [1, 2, 5]
    amount = 5
    print("Number of ways to make change:", coin_change(coins, amount))

Dynamic Programming Approach in C++

In C++, we can write the solution like this:

#include <iostream>
#include <vector>

using namespace std;

int coinChange(vector<int>& coins, int amount) {
    vector<int> dp(amount + 1, 0);
    dp[0] = 1; // Base case

    for (int coin : coins) {
        for (int j = coin; j <= amount; j++) {
            dp[j] += dp[j - coin];
        }
    }

    return dp[amount];
}

int main() {
    vector<int> coins = {1, 2, 5};
    int amount = 5;
    cout << "Number of ways to make change: " << coinChange(coins, amount) << endl;
    return 0;
}

Key Benefits

• Reduced Space Complexity: The space needed for the algorithm is less now. It goes from O(n * m) to O(n). Here n is the amount and m is the types of coins.
• Improved Efficiency: The time complexity stays O(n * m). But now with less space, we can manage larger amounts without using too much memory.

For more information about dynamic programming techniques, you can check the Dynamic Programming Fibonacci Number article.

Frequently Asked Questions

What is the Coin Change Problem in Dynamic Programming?

The Coin Change Problem is a well-known challenge in algorithms. We need to find out how many ways we can make change for a certain amount using different coins. We can solve this problem well with dynamic programming. We break the problem into smaller parts. Then we build the solution step by step. This method is much faster than just using a simple recursive way.

How do I implement the Coin Change Problem in Java?

To implement the Coin Change Problem in Java, we will make a dynamic programming table. Each spot in the table shows how many ways we can make change for that amount. We go through each coin and update the table using earlier values we calculated. This way, we consider all combinations without doing extra work. For a detailed guide, check our Dynamic Programming Approach for Coin Change in Java.

What are the differences between recursive and dynamic programming solutions for Coin Change?

The main difference between the recursive way and the dynamic programming way for the Coin Change Problem is how fast they work. The recursive way can take a long time because it has repeating problems. The dynamic programming way uses a table to keep track of results. This makes it much faster and reduces the time to a polynomial level. If you want to see how they compare, look at our sections on Recursive Solution to Coin Change Problem in Java and Dynamic Programming Approach for Coin Change in Java.

Is there a space-efficient way to solve the Coin Change Problem?

Yes, we can save space in the Coin Change Problem. Instead of using a two-dimensional table, we can use a one-dimensional array. This works because each step only depends on the last steps. This change brings the space needed down from O(n*m) to O(n). Here, n is the target amount and m is the number of coin types. For more details, check our section on Optimizing Space Complexity in Coin Change Problem.

Can the Coin Change Problem be solved in Python?

Yes, for sure! We can solve the Coin Change Problem in Python using both recursive and dynamic programming methods. The dynamic programming way uses a list to keep track of how many ways we can make change for each amount up to the target. You can read more about how to do this in our Dynamic Programming Approach for Coin Change in Python section.

These FAQs give a short overview of the Coin Change Problem and its solutions. They help learners and developers who want to understand dynamic programming better. For more topics like this, you can check out our articles on Dynamic Programming Fibonacci Number and Dynamic Programming Climbing Stairs.