Coin changing using Dynamic Programming

Question

Coin changing is a popular interview question. Essentially the question means given a set of coin denominations and a total how many ways can one get the total provided there's an infinite supply of each denomination of coin.

This is my code. The logic is every time I pick a coin the problem reduces to solving for the total minus the coin.

public static int numberOfWays(int total, int[] options){

        int[][] memo = new int[options.length][total+1];

        for (int i = 0; i <memo.length ; i++) {

            for (int j = 0; j <memo[i].length ; j++) {

                if(i == 0)  memo[i][j] = 1;
                else if(options[i] > j ) memo[i][j] = memo[i-1][j];
                else memo[i][j] = memo[i-1][j] + memo[i][j - options[i]];
            }
        }
        return memo[options.length-1][total];
    }

This works on a test case of total = 5 and options = 1, 2, 3 But fails total = 10 and options = 2, 5, 3, 6

Can someone help me understand what I'm doing wrong.

Answer 1

First, it's good to write out a statement of what each array element represents:

memo[i][j] represents how many ways to make the total amount j given only coins in denominations options[0] , options[1] , ..., options[i] .

Now, from that you appear to have derived a few laws:

1. memo[0][j] is 1 for all j
2. for i greater than 0, memo[i][j] is the same as memo[i-1][j] whenever options[i] > j
3. for i greater than 0, memo[i][j] is memo[i-1][j] + memo[i][j - options[i]] whenever options[i] <= j

Your problem is that the first of these laws isn't true. (the second two are)

The statement " memo[0][j] is 1 for all j " only holds if options[0] is 1 . If options[0] isn't 1 , then memo[0][j] is 1 when j is a multiple of options[0] , and 0 when it isn't. Using only coins of denomination 2 , you can't make 5 cents, so you should have (with the second set of data) memo[0][5] == 0 , but your program says memo[0][5] == 1 . This then throws off all your subsequent calculations.

So I'd modify your program to say:

            if(i == 0) { if (j % options[i] == 0) memo[i][j] = 1;
                         else memo[i][j] = 0; }
            else if(options[i] > j ) memo[i][j] = memo[i-1][j];
            else memo[i][j] = memo[i-1][j] + memo[i][j - options[i]];

(Though on a purely stylistic note, I find that if / else statements that don't use braces even for a single statement are asking for errors)