I can tell u the logic if the minimum existing coin denomination is 1 cent
for instance.If the minimum denomination is not 1 u might need to modify the
algorithm I guess

let v1,v2.....vn be the values of the coin denominations u have in the array

dynamic programming solution would be as follows:

we need  another array to store all possible values using these coins that
is say S[i] is the value i that is obtained using minimum number of coins

so initialisation is S[0]=0;
S[1]=1( if minimum denomination is 1 in given array)

every other value S[i]= min( S[i-1] + 1, 1  if either of v1 to vn equals i)


check and see if it is ok...

arun

On Fri, Jul 29, 2011 at 10:59 AM, AMAN AGARWAL <mnnit.a...@gmail.com> wrote:

> what do u mean by
>
> memo[i]=memo[i-denom[j]] +1
>
> memo[i]   will be containing infinite value.
>
> memo[i-denom[j]]   will also contain infinite value
>
>
>
>
>  On Fri, Jul 29, 2011 at 11:32 AM, ankit sambyal 
> <ankitsamb...@gmail.com>wrote:
>
>> Let S be the exact amount for which minimum coins are to found. And
>> denom[] be the array which contains different denominations. Let N be the
>> size of the denom[] array.
>>
>> Algo:
>> 1. int memo[S]
>> 2. initialize all its elements to infinite
>> 3.for i=1 to S
>>       for j=0 to N-1
>>          if(denom[j] < i  &&  memo[i-denom[j]] +1 < memo[i])
>>               memo[i]=memo[i-denom[j]] +1
>> 4. return memo[S]
>>
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>
>
>
> --
> AMAN AGARWAL
> "Success is not final, Failure is not fatal: It is the courage to continue
> that counts!"
>
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