The minimum number of coins the sum of which is S

Question

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 a

Answer 1

The main idea is - for each coin j, value[j] <= i (i.e sum) we look at the minimum number of coins found for i-value[j] (let say m) sum (previously found). If m+1 is less than the minimum number of coins already found for current sum i then we update the number of coins in the array.

For ex - sum = 11 n=3 and value[] = {1,3,5}
Following is the output we get

i- 1  mins[i] - 1  
i- 2  mins[i] - 2  
i- 3  mins[i] - 3  
i- 3  mins[i] - 1  
i- 4  mins[i] - 2  
i- 5  mins[i] - 3  
i- 5  mins[i] - 1  
i- 6  mins[i] - 2  
i- 7  mins[i] - 3  
i- 8  mins[i] - 4  
i- 8  mins[i] - 2  
i- 9  mins[i] - 3  
i- 10 mins[i] - 4  
i- 10 mins[i] - 2  
i- 11 mins[i] - 3 

As we can observe for sum i = 3, 5, 8 and 10 we improve upon from our previous minimum in following ways -

sum = 3, 3 (1+1+1) coins of 1 to one 3 value coin  
sum = 5, 3 (3+1+1) coins to one 5 value coin  
sum = 8, 4 (5+1+1+1) coins to 2 (5+3) coins  
sum = 10, 4 (5+3+1+1) coins to 2 (5+5) coins.  

So for sum=11 we will get answer as 3(5+5+1).

Here is the code in C. Its similar to pseudocode given in topcoder page whose reference is provided in one of the answers above.

int findDPMinCoins(int value[], int num, int sum)
{
    int mins[sum+1];
    int i,j;

   for(i=1;i<=sum;i++)
       mins[i] = INT_MAX;

    mins[0] = 0;
    for(i=1;i<=sum;i++)
    {
        for(j=0;j