Dynamic Programming Coin Change Problems

Question

I am having issues with understanding dynamic programming solutions to various problems, specifically the coin change problem:

"Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn’t matter.

For example, for N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}. So output should be 4. For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}. So the output should be 5."

There is another variation of this problem where the solution is the minimum number of coins to satisfy the amount.

These problems appear very similar, but the solutions are very different.

Number of possible ways to make change: the optimal substructure for this is DP(m,n) = DP(m-1, n) + DP(m, n-Sm) where DP is the number of solutions for all coins up to the mth coin and amount=n.

Minimum amount of coins: the optimal substructure for this is DP[i] = Min{ DP[i-d1], DP[i-d2],...DP[i-dn] } + 1 where i is the total amount and d1..dn represent each coin denomination.

Why is it that the first one required a 2-D array and the second a 1-D array? Why is the optimal substructure for the number of ways to make change not "DP[i] = DP[i-d1]+DP[i-d2]+...DP[i-dn]" where DP[i] is the number of ways i amount can be obtained by the coins. It sounds logical to me, but it produces an incorrect answer. Why is that second dimension for the coins needed in this problem, but not needed in the minimum amount problem?

LINKS TO PROBLEMS:

http://comproguide.blogspot.com/2013/12/minimum-coin-change-problem.html http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/

Thanks in advance. Every website I go to only explains how the solution works, not why other solutions do not work.

Answer 1

1. Lets first talk about the number of ways, DP(m,n) = DP(m-1, n) + DP(m, n-Sm). This in indeed correct because either you can use the mth denomination or you can avoid it. Now you say why don't we write it as DP[i] = DP[i-d1]+DP[i-d2]+...DP[i-dn]. Well this will lead to over counting , lets take an example where n=4 m=2 and S={1,3}. Now according to your solution dp[4]=dp[1]+dp[3]. ( Assuming 1 to be a base case dp[1]=1 ) .Now dp[3]=dp[2]+dp[0]. ( Again dp[0]=1 by base case ). Again applying the same dp[2]=dp[1]=1. Thus in total you get answer as 3 when its supposed to be just 2 ( (1,3) and (1,1,1,1) ). Its so because your second method treats (1,3) and (3,1) as two different solution.Your second method can be applied to case where order matters, which is also a standard problem.

2. Now to your second question you say that minimum number of denominations can be found out by DP[i] = Min{ DP[i-d1], DP[i-d2],...DP[i-dn] } + 1. Well this is correct as in finding minimum denominations, order or no order does not matter. Why this is linear / 1-D DP , well although the DP array is 1-D each state depends on at most m states unlike your first solution where array is 2-D but each state depends on at most 2 states. So in both case run time which is ( number of states * number of states each state depends on ) is the same which is O(nm). So both are correct, just your second solution saves memory. So either you can find it by 1-D array method or by 2-D by using the recurrence dp(n,m)=min(dp(m-1,n),1+dp(m,n-Sm)). (Just use min in your first recurrence)

   
   Hope I cleared the doubts , do post if still something is unclear.

Answer 2

This is a very good explanation of the coin change problem using Dynamic Programming.

The code is as follows:

public static int change(int amount, int[] coins){
    int[] combinations = new int[amount + 1];

    combinations[0] = 1;

    for(int coin : coins){
        for(int i = 1; i < combinations.length; i++){
            if(i >= coin){
                combinations[i] += combinations[i - coin];
                //printAmount(combinations);
            }
        }
        //System.out.println();
    }

    return combinations[amount];
}