@atul your understanding of my recurrences are fine but of the question are
not. You cannot have 3 adjacent houses with same colour. GGY is a fine case
for this problem.
On 28 Dec 2013 20:44, "atul anand" wrote:
> @saurabh : i did not get your algo for modified ques i.e "No 3 adjacent
> houses sh
@saurabh : i did not get your algo for modified ques i.e "No 3 adjacent
houses should get same color"
according to your recurrence
answer[i][j][1] = min(answer[i-1][t][0],answer[i-1][t][1]) + colorvalue[j] for
all t from 1 to k except j.
now suppose in some case *answer[i-1][t][0] *is found min
Good solution buddy, thanks for the answer.
On 26 December 2013 17:25, Saurabh Paliwal wrote:
> @atul anand :- no, we need not use all the colors.
>
> @kumar raja :- sorry dude for replying this late. Continuing with the
> previous idea, I extend the solution for the modified problem as
>
> 1. l
@atul anand :- no, we need not use all the colors.
@kumar raja :- sorry dude for replying this late. Continuing with the
previous idea, I extend the solution for the modified problem as
1. let answer[i][j][0] represent minimum cost of coloring i houses when ith
house is colored with jth color and
@kumar : i have one doubt regarding this question.Do we have to use all K
colors[K] to color all building.
for example :-
color[3]={1,2,10};
now if we have to color 6 building then i can use only 1st 2 color to color
all building and never 10 ...is this allowed ???
building[N]={1,2,1,2,1,2}
O
ohh shoots...my bad.got this condition wrong "t varies from 1 to k
except j."
now got it :)..ignore my previous comment :) :)
On Tue, Dec 17, 2013 at 11:47 PM, atul anand wrote:
> @saurabh :
>
> answer[i][j] = min(answer[i-1][t]) + colorvalue[t] where t varies from 1
> to k except j.
>
> h
@saurabh :
answer[i][j] = min(answer[i-1][t]) + colorvalue[t] where t varies from 1 to
k except j.
how this DP is taking care of the case where no adjacent house if of same
color.
what i understand from the DP is the following :-
find minimum cost of coloring previous house with color t (
min(a
Please elaborate the explanation, i want to understand it in detail.
On 17 December 2013 11:36, Saurabh Paliwal wrote:
> I think that can be done by having 2 different values at every position in
> the answer matrix. One when the previous house is of same color and one
> when it does not. Answer
I think that can be done by having 2 different values at every position in
the answer matrix. One when the previous house is of same color and one
when it does not. Answer[n][k][2] will be the new dimensions.
I can explain in detail if you don't get this. ☺
On Dec 15, 2013 4:43 PM, "kumar raja" w
What can be the recurrence relation if the condition is changed to "No 3
adjacent houses should get same color"?
On 15 December 2013 16:26, kumar raja wrote:
> No need for the explanation ,got it man thanks.
>
>
> On 15 December 2013 16:20, kumar raja wrote:
>
>> Saurabh your solutions seems r
No need for the explanation ,got it man thanks.
On 15 December 2013 16:20, kumar raja wrote:
> Saurabh your solutions seems right , but can u explain me how did u arrive
> at the time and space complexity with some proof /pseudocode/ explanation?
>
>
> On 15 December 2013 09:47, Saurabh Paliwal
Saurabh your solutions seems right , but can u explain me how did u arrive
at the time and space complexity with some proof /pseudocode/ explanation?
On 15 December 2013 09:47, Saurabh Paliwal wrote:
> what is the issue with the usual recurrence :-
>
> Let answer[i][j] represent minimum cost of
what is the issue with the usual recurrence :-
Let answer[i][j] represent minimum cost of coloring i houses with ith house
colored with jth color.
answer[i][j] = min(answer[i-1][t]) + colorvalue[t] where t varies from 1 to
k except j.
initial setting :- answer[1][j] = colorvalue[j]
final answe
You have 'n' houses and 'k' colors. The cost of coloring a house with one
of the 'k' colors is different from coloring another house with same color.
The cost of coloring the house with different colors are different. So now
how to colour the 'n' houses such that no two adjcent houses will get the
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