Coin Change - Find minimum number of coins

Given a list of N coins, their values (V1, V2, ... , VN), and the total sum S. Find the minimum number of coins the sum of which is S (we can use as many coins of one type as we want), or report that it's not possible to select coins in such a way that they sum up to S. 

 

For a better understanding let's take this example:
Given coins with values 1, 3, and 5.
And the sum S is set to be 11. 

First of all we mark that for state 0 (sum 0) we have found a solution with a minimum number of 0 coins. We then go to sum 1. First, we mark that we haven't yet found a solution for this one (a value of Infinity would be fine). Then we see that only coin 1 is less than or equal to the current sum. Analyzing it, we see that for sum 1-V1= 0 we have a solution with 0 coins. Because we add one coin to this solution, we'll have a solution with 1 coin for sum 1. It's the only solution yet found for this sum. We write (save) it. Then we proceed to the next state - sum 2. We again see that the only coin which is less or equal to this sum is the first coin, having a value of 1. The optimal solution found for sum (2-1) = 1 is coin 1. This coin 1 plus the first coin will sum up to 2, and thus make a sum of 2 with the help of only 2 coins. This is the best and only solution for sum 2. Now we proceed to sum 3. We now have 2 coins which are to be analyzed - first and second one, having values of 1 and 3. Let's see the first one. There exists a solution for sum 2 (3 - 1) and therefore we can construct from it a solution for sum 3 by adding the first coin to it. Because the best solution for sum 2 that we found has 2 coins, the new solution for sum 3 will have 3 coins. Now let's take the second coin with value equal to 3. The sum for which this coin needs to be added to make 3 , is 0. We know that sum 0 is made up of 0 coins. Thus we can make a sum of 3 with only one coin - 3. We see that it's better than the previous found solution for sum 3 , which was composed of 3 coins. We update it and mark it as having only 1 coin. The same we do for sum 4, and get a solution of 2 coins - 1+3. And so on. 

Pseudocode: 

Set Min[i] equal to Infinity for all of i
Min[0]=0

For i = 1 to S
For j = 0 to N - 1
   If (Vj<=i AND Min[i-Vj]+1<Min[i])
Then Min[i]=Min[i-Vj]+1

Output Min[S]

Here are the solutions found for all sums: 

Sum	Min. nr. of coins	Coin value added to a smaller sum to obtain this sum (it is displayed in brackets)
0	0	-
1	1	1 (0)
2	2	1 (1)
3	1	3 (0)
4	2	1 (3)
5	1	5 (0)
6	2	3 (3)
7	3	1 (6)
8	2	3 (5)
9	3	1 (8)
10	2	5 (5)
11	3	1 (10)

As a result we have found a solution of 3 coins which sum up to 11. 

public int findMinCoins(int[] coins, int S) {
	int N = coins.length;
	int[] f = new int[S+1];
	Arrays.fill(f, Integer.MAX_VALUE);
	f[0] = 0;
	for(int i=0; i<=S; i++) {
		for(int j=0; j<N; j++) {
			if(coins[j] <= i) {
				f[i] = Math.min(f[i-coins[j]] + 1, f[i]);
			}
		}
	}
	return f[S];
}

 

Reference:

http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=dynProg