@ankit : sorry dude , its not working for given input
{2,1,3,4,5} k=12,
ans=12, sub-array={3,4,5}
your algo will give ans=10 sub-array{0 to 3}
On Thu, Dec 1, 2011 at 11:33 PM, Ankit Sinha <akki12...@gmail.com> wrote:
> A little variation of kadane's algo will do as written below: -
>
> #include "stdafx.h"
> #include "stdlib.h"
>
> int _tmain(int argc, _TCHAR* argv[])
> {
> int a[5] = {-1,3,1,2,-3};
> int max_so_far= 0, max_ending_here= 0,max_limit_here=0,startIndex=0
> ,
> endIndex = 0, itr = 0;
> int k=12;
> for (itr = 0; itr<5;itr++)
> {
> max_ending_here +=a[itr];
> if (max_ending_here < 0 && max_limit_here <=k)
> {
> max_ending_here = 0;
> startIndex++;
> }
> else if (max_so_far <max_ending_here)
> {
> if (max_ending_here <= k)
> {
> max_so_far = max_ending_here;
> endIndex = itr;
> }
> else
> {
> max_limit_here = max_ending_here;
> max_ending_here = 0;
>
> }
> }
>
> }
>
> printf("%d%d%d", max_so_far, startIndex, endIndex);
> system("PAUSE");
> return 0;
> }
> Complexity O(n)
>
> Cheers,
> Ankit Sinha
> On Thu, Dec 1, 2011 at 4:58 PM, sourabh <sourabhd2...@gmail.com> wrote:
> > @atul...
> > thanks dude for ur thorough screening of the algo and pointing out the
> > mistakes... I think that's y its always said that until and unless we
> > don't turn an algo to a working code we are never really sure whether
> > its perfect and covers all the cases.
> >
> > On Dec 1, 4:23 pm, atul anand <atul.87fri...@gmail.com> wrote:
> >> yup i made some calculation error....
> >>
> >> Now this algo works perfectly :) :)
> >>
> >> Thanks for your help and explanation :) :)
> >>
> >>
> >>
> >>
> >>
> >>
> >>
> >> On Thu, Dec 1, 2011 at 4:26 PM, sourabh <sourabhd2...@gmail.com> wrote:
> >> > @atul ...
> >>
> >> > Reply 1:
> >> > Yes, you are correct.. i missed it... Correct statement is as follows:
> >>
> >> > 12 + 6 = 18 , closest element found = 15 , closest to X = 15 - 6 =9 ,
> >> > i = 3, j = 4
> >> > 12 + 10 = 22 , closest element found = 15 , closest to X = 15 - 10
> >> > =5 , i = 4, j = 4
> >>
> >> > -------------------------------------------------------------
> >>
> >> > Reply 2:
> >> > I might be wrong in calculating 12 + 2 = 14.... but i guess you are
> >> > not getting my point .... even if 14 is the search element, still the
> >> > element smaller than equal to 14 in array B is 10 and not 15...
> >>
> >> > Hence, the above calculation that you have made are incorrect.
> >>
> >> > If you look at the problem statement it says that we have to find the
> >> > sum which is smaller than equal to X.
> >> > Now, if you look ta ur calculations you will see that your 'closest to
> >> > X' search space contains elements 13 which is invalid as it is greater
> >> > than 12...
> >>
> >> > Hence, i m re-calculating the values based on the above given
> >> > algorithm...
> >>
> >> > 12 + 2 = 14, closest element found = 10 , closest to X = 10 - 2 =
> >> > 8 , i = 1, j = 3
> >>
> >> > 12 + 3 = 15 , closest element found = 15 , closest to X = 15 - 3
> >> > =12 , i = 2, j = 4
> >>
> >> > 12 + 6 = 18 , closest element found = 15 , closest to X = 15 - 6 =
> >> > 9 , i = 3 , j = 4
> >>
> >> > 12 + 10 = 22 , closest element found = 15 , closest to X = 15 - 10 =
> >> > 5 , i = 4 , j = 4
> >>
> >> > Also, as calculated in the previous post the corner case gives 10 as
> >> > the closest to X.
> >>
> >> > Hence, max of all closest values to X = max { 8, 12, 9, 5, 10 } =
> >> > 12.
> >>
> >> > On Dec 1, 2:52 pm, atul anand <atul.87fri...@gmail.com> wrote:
> >> > > and you made mistake above in calculating 12 + 2 = *12* , its 14
> >>
> >> > > 12 + 2 = 14, closest element found = 15 , closest to X = 15 - 2 =
> 13 ,
> >> > i
> >> > > = 1, j = 4
> >> > > 12 + 3 = 15 , closest element found = 15 , closest to X = 15 - 3
> =12 ,
> >> > i
> >> > > = 2, j = 4
> >> > > 12 + 6 = 18 , closest element found = 15 , closest to X = 15 - 6 =
> 11
> >> > , i
> >> > > = 3 , j = 4
> >> > > 12 + 10 = 22 , closest element found = 15 , closest to X = 15 - 10
> = 5 ,
> >> > i
> >> > > = 4 , j = 4
> >>
> >> > > out of {10,13,12,11,5 } , element 12 is closest to X ( which is
> 12) .
> >> > > So basically among this we have to find element closest X ( where
> >> > element <
> >> > > = X )
> >> > > hence 12 is the answer.
> >>
> >> > > On Thu, Dec 1, 2011 at 3:11 PM, atul anand <atul.87fri...@gmail.com
> >
> >> > wrote:
> >> > > > @sourabh
> >>
> >> > > > i guess you need to modify this statement :-
> >>
> >> > > > 3) Now for each element B[i][0] , do a (modified) binary *search
> for
> >> > > > the closest value smaller than equal to (X + B[i][0])* in array
> B[i
> >> > > > +1... n][0] ,
> >> > > > Say the found index after binary search is j ( which is > i)...
> >>
> >> > > > 12 + 2 = 12 , closest element found = 10 , closest to X = 10 - 2
> = 8 ,
> >> > i
> >> > > > = 1, j = 3
> >> > > > 12 + 3 = 15 , closest element found = 15 , closest to X = 15 - 3
> =12 ,
> >> > i =
> >> > > > 2, j = 4
> >> > > > 12 + 6 = 18 , closest element found = *no element found ??? how *
> >>
> >> > > > *Cumulative SUM*
> >>
> >> > > > *i*
> >>
> >> > > > 2
> >>
> >> > > > 0
> >>
> >> > > > 3
> >>
> >> > > > 1
> >>
> >> > > > 6
> >>
> >> > > > 2
> >>
> >> > > > 10
> >>
> >> > > > 3
> >>
> >> > > > *15*
> >>
> >> > > > 4
> >>
> >> > > > for 12 + 6 =18 , closest element less than equal to 18 = 15 (15 <
> 18 )
> >> > > > ..right ??
> >>
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