Coin Change (Minimum Coins) šŸŽÆ

beginner
22 min

Coin Change (Minimum Coins) šŸŽÆ

Welcome to this comprehensive guide on the Coin Change problem! In this lesson, we'll explore how to determine the minimum number of coins required to make change for a given amount using different types of coins. This practical and real-world problem is an excellent way to understand and master the fundamental concepts of dynamic programming and data structures.

Table of Contents

  1. Introduction
  2. Understanding the Problem
  3. Dynamic Programming Approach
  4. Step-by-Step Solution
  5. Example: Making Change with Different Coin Denominations
  6. Efficiency Analysis
  7. Implementing the Solution in Python
  8. Advanced Example: Making Change with Unlimited Coins
  9. Quiz

1. Introduction šŸ“

In the Coin Change problem, we aim to find the minimum number of coins required to make change for a given amount using a set of coin denominations. This problem is a classic example of dynamic programming, which is a powerful algorithmic technique used to solve optimization problems.

2. Understanding the Problem šŸ’”

Let's break down the Coin Change problem:

  • We have a set of coins with denominations C = [C1, C2, ..., Cn].
  • We need to make change for an amount A.
  • The goal is to find the minimum number of coins needed to make change for the amount A.
  • It's important to note that we can use each coin multiple times, and the coins are unsorted.

3. Dynamic Programming Approach šŸ’”

We'll solve the Coin Change problem using dynamic programming, a technique that solves problems by breaking them down into smaller subproblems. In the Coin Change problem, we'll create a table to store the minimum number of coins required to make change for amounts from 0 to A.

4. Step-by-Step Solution šŸ“

  1. Initialize a table dp[A+1] to store the minimum number of coins required to make change for amounts from 0 to A. Set dp[0] to 0 since no coins are needed to make change for 0.

  2. Iterate through the coin denominations and for each denomination Ci, calculate the minimum number of coins required to make change for amounts from 1 to A.

  3. For each amount i, calculate the minimum number of coins needed by considering the following cases:

    • If i is equal to the denomination Ci, we can make change for i using one coin of denomination Ci. So, set dp[i] to 1.
    • Otherwise, we can make change for i using coins of smaller denominations. In this case, we find the minimum number of coins required to make change for amounts from 0 to i-1 and add the minimum number of coins required to make change for i-Ci (since we can subtract Ci from i using one coin of denomination Ci). Set dp[i] to the minimum of these two values.

5. Example: Making Change with Different Coin Denominations šŸ’”

Let's work through an example:

Suppose we have coins with denominations C = [1, 5, 10] and we need to make change for an amount A = 17. Here's how we can solve the problem:

dp[0] = 0 # No coins needed for 0 dp[1] = 1 # One coin of denomination 1 is needed for 1 dp[2] = 1 # One coin of denomination 1 is needed for 2 dp[3] = 1 # One coin of denomination 1 is needed for 3 dp[4] = 1 # One coin of denomination 1 is needed for 4 dp[5] = 1 # One coin of denomination 1 is needed for 5 dp[6] = 1 # One coin of denomination 1 is needed for 6 dp[7] = 1 # One coin of denomination 1 is needed for 7 dp[8] = 1 # One coin of denomination 1 is needed for 8 dp[9] = 2 # One coin of denomination 1 for 9, and one coin of denomination 5 for 4 dp[10] = 2 # One coin of denomination 1 for 10, and one coin of denomination 5 for 5 dp[11] = 2 # One coin of denomination 1 for 11, and one coin of denomination 5 for 6 dp[12] = 2 # One coin of denomination 1 for 12, and one coin of denomination 5 for 7 dp[13] = 3 # One coin of denomination 1 for 13, one coin of denomination 5 for 8, and one coin of denomination 10 for 5 dp[14] = 3 # One coin of denomination 1 for 14, one coin of denomination 5 for 9, and one coin of denomination 10 for 5 dp[15] = 3 # One coin of denomination 1 for 15, one coin of denomination 5 for 10, and one coin of denomination 10 for 5 dp[16] = 3 # One coin of denomination 1 for 16, one coin of denomination 5 for 11, and one coin of denomination 10 for 5 dp[17] = 3 # One coin of denomination 1 for 17, one coin of denomination 5 for 12, and one coin of denomination 10 for 5

In this example, the minimum number of coins required to make change for A = 17 is 3.

6. Efficiency Analysis šŸ“

The time complexity of the Coin Change algorithm is O(Am), where A is the amount and m is the number of coin denominations. The space complexity is also O(Am) due to the table used to store the minimum number of coins required to make change for amounts from 0 to A.

7. Implementing the Solution in Python šŸ’”

python
def min_coins(A, C): dp = [0] + [float('inf')] * A for Ci in C: for i in range(Ci, A+1): dp[i] = min(dp[i], dp[i-Ci] + 1) return dp[A] # Example usage: coins = [1, 5, 10] amount = 17 print(min_coins(amount, coins))

8. Advanced Example: Making Change with Unlimited Coins šŸ’”

In some cases, we may have an unlimited supply of coins for each denomination. In such scenarios, we can solve the problem by iterating through the coins in decreasing order and keeping track of the number of coins used for each denomination. Here's an example:

python
def min_coins_unlimited(A, C): coins, count = sorted(C, reverse=True), 0 for coin in coins: count += A // coin A %= coin return count # Example usage: coins = [1, 5, 10] amount = 17 print(min_coins_unlimited(amount, coins))

9. Quiz šŸŽÆ

Quick Quiz
Question 1 of 1

What is the time complexity of the Coin Change algorithm with an unlimited number of coins for each denomination?

Now that you've learned the basics of the Coin Change problem, you can apply this concept to various real-world projects and further explore dynamic programming and data structures. Happy coding! šŸš€šŸ’»