It is just integer knapsack.

On Fri, Jul 29, 2011 at 11:34 AM, Arun Vishwanathan
<[email protected]>wrote:

> 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 <[email protected]>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 
>> <[email protected]>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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>
>
>
>
>
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-- 
*With Regards :*

Ravinder Kumar
B.Tech Final Year
Computer Science and Engineering
MNNIT Allahabad

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