The problem we will address today is the Coin Change Problem<\/strong>. This problem involves finding the least number of coins needed to make a specific amount using various coins that we often encounter in real life.<\/p>\n Write a function This function should return the minimum number of coins needed to make the given amount. If it is not possible to use the coins, it should return To understand the problem more deeply, let’s look at a few examples.<\/p>\n To solve this problem, we will use a greedy algorithm. The greedy algorithm works by making the best choice in the current situation. Therefore, we will start by using the largest coins sequentially to try to form the target amount.<\/p>\n The specific steps of the algorithm are as follows:<\/p>\n Now, let’s write the code based on this approach:<\/p>\n Now, let’s test the function we have written.<\/p>\n We have solved the Coin Change Problem using the greedy algorithm. Through this problem, we learned the fundamental approach of the greedy algorithm and studied a common type of problem seen in coding tests.<\/p>\n I hope to deepen my understanding of the greedy algorithm by practicing more problems. Like the problem above, it is essential to practice how to sort data and use loops to find solutions. The greedy algorithm can be a useful tool for solving various problems.<\/p>\n Thank you!<\/p>\n <\/body><\/p>\n","protected":false},"excerpt":{"rendered":"Problem Definition<\/h3>\n
min_coins(coins, amount)<\/code> that satisfies the following conditions:<\/p>\n\n
coins<\/code>: A list of available coins (e.g., [1, 5, 10, 25])<\/li>\namount<\/code>: The target amount to be formed<\/li>\n<\/ul>\n-1<\/code>.<\/p>\nUnderstanding the Problem<\/h2>\n
\n Input: coins = [1, 5, 10, 25], amount = 63<\/code>
\n Output: 6<\/code>
\n Explanation: 25 + 25 + 10 + 1 + 1 + 1 = 63\n <\/div>\n
\n Input: coins = [2, 4], amount = 5<\/code>
\n Output: -1<\/code>
\n Explanation: There are no ways to form 5.\n <\/div>\nApproach<\/h2>\n
\n
-1<\/code>.<\/li>\n<\/ol>\nCode Implementation<\/h2>\n
\ndef min_coins(coins, amount):\n # Sort the coins in descending order\n coins.sort(reverse=True)\n \n count = 0 # Number of coins used\n \n for coin in coins:\n # Ignore if the current coin is greater than amount\n while amount >= coin:\n amount -= coin # Subtract the coin value from amount\n count += 1 # Increase the coin count\n \n # Check if the final amount is 0\n return count if amount == 0 else -1\n <\/code><\/pre>\nTesting<\/h2>\n
\n# Test cases\nprint(min_coins([1, 5, 10, 25], 63)) # 6\nprint(min_coins([2, 4], 5)) # -1\nprint(min_coins([5, 2, 1], 11)) # 3 (5 + 5 + 1)\n <\/code><\/pre>\nConclusion<\/h2>\n