Dynamic Programming – Coin Change Problem
Code Review Stack Exchange is a question and answer site for peer programmer code reviews. It only takes a minute to sign up. I'm trying to learn dynamic programming and the most popular one is to find the minimum number of coins for the change.
I come up with this but I just want to improve this code which I think it could be shorter in Python but I'm new to Python too. This is already pretty short. Making it shorter doesn't seem a good objective, and risks hurting readability.
That being said, there are a few things you could do. While at it, why are there 20 elements? Your program will find the minimum number of coins up to 19, and I have a feeling that you actually want it for In which case you would need:. And then use range 21 in the loop. One more optimization is possible by flipping the order of the nested loops. In the current code you need an if statement to skip the amounts that are too small for the coin.
If you make the loop over coins to be the outer loop, then you can set a starting value for the inner value to make the if statement unnecessary. To be honest, I don't understand what your program does, even the hints that it has something to do with coins and change don't help me. Here is the same program with some variables renamed. However, it is inelegant to have to use sys.
You can solve both problems by using min with a generator expression. First of all, that's not dynamic programming as I understand the term.
You haven't broken the problem down into simpler sub-problems, and in fact you have used an algorithm so convoluted in its logic that it's inordinately difficult to understand. Divide the problem into two parts: 1 a function to compute the number of coins needed to achieve a given total. Here it is:. You want the answers for the first 20 integers, so I can only guess whether that qualifies as dynamic programming, but I think it's a far better approach.
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Asked 4 years, 10 months ago. Active 4 years, 10 months ago. Viewed 10k times. Active Oldest Votes. So, one major step to improve the code would be to add some documentation: What does the program do? What is its input and in what form do I have to give it to the program? What is its output and how is it formatted?
Paul Cornelius Paul Cornelius 1 1 silver badge 1 1 bronze badge. Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password.Please read the Help Documents before posting. Hello There, Guest! Login Register. Login Username: Password: Lost Password? Remember me. Thread Rating: 0 Vote s - 0 Average 1 2 3 4 5. Thread Modes. Gateux Programmer named Tim. Write a program that reads an input representing a change which is an amount less than 1 dollar.
The program calculates the change into 50, 10, 5 and 1 cent coins. The program then displays the number of each coin required for that change. Need some guidance on writing the correct calculations, thanks.
You need to subtract out each denomination before going on to the next one. In your example, after you give them 4 dimes, you only need to give them 7 cents more.
But you try to give them 9 nickels 45 cents because you haven't subtracted out the 40 cents from the 4 dimes. Craig "Ichabod" O'Brien - xenomind. Recommended Tutorials: BBCodefunctionsclassestext adventures. Website Find.
It is always good idea to separate how from what. How to get coins quantity and what to do with these quantities.
How to get coins quantity? One needs to have quotient and reminder. There is function for that in Python: divmod. Quotient is quantity of coins and reminder becomes new amount. One can create function which iterates over coins values and assigns quantity to each coin value and returns dictionary NB! Print out in any format you like? ThomasL and saravanatn like this post. I'm not 'in'-sane. Indeed, I am so far 'out' of sane that you appear a tiny blip on the distant coast of sanity.
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The Coin Change Problem — Explained
I am in a beginner programming course. We must do an exercise where we make a change maker program. The input has to be between and must be represented in quarters, dimes, nickles, and pennies when the input is divided down between the four. I wrote a code that involved loops and whiles, but he wants something more easy and a smaller code.
He gave me this as a way of helping me along:. He told us that this was basically all we needed and just needed to add in the dimes, nickles, and pennies. I try it multiple ways with the dimes, nickles, and pennies, but I cannot get the output right. Whenever I enter '99', I get 3 for quarters, 2 for dimes, 1 for nickles, and 0 for pennies. If anyone would be able to help me, that would be wonderful! I'm now sure about what you want to achieve.
Using the modulo operator you could easily find out how many quarters, dimes, nickles and pennies. I hope this helps? Something like this but don't just copy it. Everytime you divide by 25 10 5 you must lose that part because it's already counted. At the end print what ever you want :. The actual trick is knowing that because each coin is worth at least twice of the next smaller denomination, you can use a greedy algorithm.
The rest is just implementation detail. Here's a slightly DRY'er but possibly, uh, more confusing implementation. All I'm really doing differently is using a list to store my results, and taking advantage of tuple unpacking and divmod. And so on. Learn more. Asked 4 years, 7 months ago. Active 5 months ago. Viewed 18k times. Eric Renouf There's an interesting wikipedia page on this algorithm: en. The tricky part is to know that you by taking as many quarters as possible without exceeding the sum, then as many dimes and then nickels and last pennies will actually result in the minimal amount of coins.
And of course you should present the minimal amount or you could just anser to use c pennies. Active Oldest Votes. Let's just say you input Saimu Saimu 1 1 gold badge 4 4 silver badges 9 9 bronze badges.
Thank you so much for your response!Objective: Given a set of coins and amount, Write an algorithm to find out how many ways we can make the change of the amount using the coins given. This is another problem in which i will show you the advantage of Dynamic programming over recursion. Code: Run This Code. Run This Code Time Complexity : 2 n. So including a simple explanation. For every coin we have 2 options, either we include it or exclude it so if we think in terms of binary, its 0 exclude or 1 include.
In all these options we will be checking whether that selection has made the change which is required. Dynamic Programming: Bottom-up —. This problem is slightly different than that but approach will be bit similar. Code : Run This Code. If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment. Tags: Intermediate. September 5, November 16, August 26, Enter your email address to subscribe to this blog and receive notifications of new posts by email.
Email Address. Print Paths from root to all leaf nodes in a binary tree. Check if one string is Rotation of another string. Find Kth Smallest or Largest element in an Array.Many programs in computer science are written to optimize some value; for example, find the shortest path between two points, find the line that best fits a set of points, or find the smallest set of objects that satisfies some criteria. There are many strategies that computer scientists use to solve these problems.
One of the goals of this book is to expose you to several different problem solving strategies. Dynamic programming is one strategy for these types of optimization problems. A classic example of an optimization problem involves making change using the fewest coins.
Suppose you are a programmer for a vending machine manufacturer. Your company wants to streamline effort by giving out the fewest possible coins in change for each transaction. Suppose a customer puts in a dollar bill and purchases an item for 37 cents. What is the smallest number of coins you can use to make change? The answer is six coins: two quarters, one dime, and three pennies. How did we arrive at the answer of six coins?Coin Changing Minimum Number of Coins Dynamic programming
We start with the largest coin in our arsenal a quarter and use as many of those as possible, then we go to the next lowest coin value and use as many of those as possible. This first approach is called a greedy method because we try to solve as big a piece of the problem as possible right away. The greedy method works fine when we are using U. In this instance our greedy method fails to find the optimal solution for 63 cents in change. With the addition of the 21 cent coin the greedy method would still find the solution to be six coins.
However, the optimal answer is three 21 cent pieces. Since this section is about recursion, you may have guessed that we will use a recursive solution. If we are trying to make change for the same amount as the value of one of our coins, the answer is easy, one coin. If the amount does not match we have several options. What we want is the minimum of a penny plus the number of coins needed to make change for the original amount minus a penny, or a nickel plus the number of coins needed to make change for the original amount minus five cents, or a dime plus the number of coins needed to make change for the original amount minus ten cents, and so on.
So the number of coins needed to make change for the original amount can be computed according to the following:. The algorithm for doing what we have just described is shown in Listing 7.I hope to provide a step-by-step walkthrough of the Dynamic Programming solution to this problem.
This article is assuming that the reader is already versed in the recursive solution. If there are many questions on that one, I will follow up by editing this post. Or I might just come back to edit it anyway. I will be using Python and be breaking down what happens in the key parts of the script.
This is called a bottom-up approach because we are working our way up the indexes instead of recursion which is done by looking to decrement indexes until termination. The function takes S the coins listm the length of the coins list and n the change we want to make up.
We are first initialising all values in the individual lists to be 0. Why is this so? The variables we will be working with are as follows: S is the input list, j is the index tracking each S. We will be iterating through each list in table starting from index 1, and within each iteration we will be iterating through S. In this way, we accumulatively compute and would be able to avoid duplicated computations moving forward. By implementing it in nested for loops and incrementing the indexes, we call it a bottom up approach.
Now comes the tricky part at least for me when I was studying it. Similar to the recursive solution, for each iteration of the inner loop, we want to get the solutions which compulsorily includes S[ j ] and store it in x, and also get the solutions which compulsorily excludes S[ j ] and store it in y. In doing so, we would be able to reference earlier solutions to avoid duplicate computations.
Why do it this way of including and excluding the coin S[ j ] for each computation? The explanation is simple. For each coin, there can be only two possibilities; to include it or exclude it in consideration with that value of n. And we know that if the coin which is S[ j ] is larger than n, then x would return 0 since including it into consideration would be impossible. The value of previous i to reference after S[ j ] is taken from i.
Including S[ j ] is compulsory. We want to reference the same position in S, which is S[ j ], just at a different n.Coin change problem is the last algorithm we are going to discuss in this section of dynamic programming.
Now, we have to make an amount by using these coins such that a minimum number of coins are used. Like the rod cutting problemcoin change problem also has the property of the optimal substructure i.
For example, we are making an optimal solution for an amount of 8 by using two values - 5 and 3. So, the optimal solution will be the solution in which 5 and 3 are also optimally made, otherwise, we can reduce the total number of coins of optimizing the values of 5 and 8.
The reason we are checking if the problem has optimal substructure or not because if there is optimal substructure, then the chances are quite high that using dynamic programming will optimize the problem.
Python Program for Coin Change
Let's start by picking up the first coin i. We can repeat the process with all the k coins and then the minimum value of all these will be our answer. The above process can also be understood in a different way.
Suppose, we have picked a coin with value x and we know that this coin is going to be in the solution. So, our next task is to find the minimum number of coins needed to make the change of value n-x i. Also, by choosing the coin with value x, we have already increased the total number of coins needed by 1. But the real problem is that we don't know the value of x. So, we will try every coin for the value of x and then we will select the minimum among those.
You can see that there are overlapping subproblems in our solution and thus, we can use dynamic programming to optimize this. So, let's look at the coding implementation of the formula we derived above. Our function is going to need the denomination vectors of coin dthe value for which change has to be made n and number of denominations we have k or number of elements in the array d i. Let's start by making an array to store the minimum number of coins needed for each value i.
Now, we need to iterate for each value in array the M and fill it up. Let's suppose we know the first coin needed to get the optimal solution of each value. So, the point is we only need to store the first coin needed for each value and then we can get the optimal coins for any value. In this case, we are also assigning the value of i to the variable 'coin' because this coin 'i' is giving us the minimum value and thus, it will be the coin to be stored in the array S.
As said above, we are first printing the value of the coin S[l] will give us the coin and d[S[l]] is the value of that coin.
Coin Change | DP-7
After this, our initial value has decreased by the value of the coin i. The last loop will also run a total of n times in worst case it can run faster depending upon the value of l-d[S[l]]. Now, there is also a nested loop. All the other statements are constant time taking statements.