I dnt get the iterative version. Can u explain it. I can do it top down in
O(n) with state index
I am at.
On Thu, Jan 20, 2011 at 11:30 PM, Avi Dullu <avi.du...@gmail.com> wrote:
> @^ Just check ur solution for boundary case ( n = 2) .. M$ is *generally*
> very strict about such mistakes :)
>
>
> Programmers should realize their critical importance and responsibility in
> a world gone digital. They are in many ways similar to the priests and monks
> of Europe's Dark Ages; they are the only ones with the training and insight
> to read and interpret the "scripture" of this age.
>
>
>
> On Thu, Jan 20, 2011 at 11:05 PM, Davin <dkthar...@googlemail.com> wrote:
>
>> static int MaxLoot(int[] A, int n)
>> {
>> int[] S = new int[n];
>> S[0] = A[0];
>> S[1] = A[1];
>> S[2] = S[0] + A[2];
>>
>>
>> int maxloot = Math.Max(S[1], S[2]);
>>
>>
>>
>> for (int i = 3; i < n; i++)
>> {
>> S[i] = Math.Max(S[i - 2], S[i - 3]) + A[i];
>> maxloot = Math.Max(maxloot, S[i]);
>> }
>>
>>
>>
>> return maxloot;
>> }
>>
>> On Jan 20, 7:30 pm, Manmeet Singh <mans.aus...@gmail.com> wrote:
>> > It can be done in O(n) space too using DP :) :).
>> > U dont need that flag, but that solution u said is absolutely correct.
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> > On Thu, Jan 20, 2011 at 7:27 PM, Decipher <ankurseth...@gmail.com>
>> wrote:
>> > > There is a row of houses in which each house contains some amount of
>> money.
>> > > Write an algorithm that loots the maximum amount of money from these
>> houses.
>> > > The only restriction is that you cannot loot two houses that are
>> directly
>> > > next to each other.
>> >
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