The main difference between the standard CC algo, and the one with limited
coin supply is the maximum value that we can form with the given coins.
For example :
In stranded CC
coins[] = {1,2,3}
then we can can form denominations to any value(assuming that coin with 1
value will always be the
Shashwat, very nice.
Don
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In my solution, the line which reads:
if (sum < result) sum = result;
Should read
if (sum < result) result = sum;
Don
On Friday, May 16, 2014 11:51:52 PM UTC-4, Shashwat Anand wrote:
>
> @Don
> int coins [] = {1, 3, 5};
> int cnt [] = {7, 3, 1};
> int S = 9;
> Your code returns 9, for the
@Don
int coins [] = {1, 3, 5};
int cnt [] = {7, 3, 1};
int S = 9;
Your code returns 9, for the aforementioned test case. Should not it return 3 ?
Here is my take which takes O (|number of denominators| x |S| x
|maximum count for any coin|) time and
O (|number of denominators| x |S|) time. It is
If you find a way to do that for more than a few coins I'd be interested in
seeing it too.
Don
On Thursday, May 15, 2014 3:00:04 PM UTC-4, atul007 wrote:
>
> @Don : i am intersted in DP bottom up approach with less time complexity.
> solving it recursively is a simpler approach...
> On 15 May 201
@Don : i am intersted in DP bottom up approach with less time complexity.
solving it recursively is a simpler approach...
On 15 May 2014 22:25, "Don" wrote:
> How about this:
>
> const int N = 6;
> unsigned int coins[N] = {100,50,25,10,5,1};
> unsigned int count[N] = {2, 1, 1, 5, 4, 10};
> int be
How about this:
const int N = 6;
unsigned int coins[N] = {100,50,25,10,5,1};
unsigned int count[N] = {2, 1, 1, 5, 4, 10};
int best = 10;
int minCoins(int s, int f=0, int d=0)
{
if (d == 0) best = s; // It can never take more than s coins
if (s < 1) return 0; // No coins required
