@atul there are 8 sums less than 7
sum[0 - 0] = 1
sum[1-1] = 2
sum[2 - 2] = 3
sum[3-3] = 4
sum[4-4] = 5
sum[0-1] = 3
sum[0-2] = 6
sum[1-2] = 5
contiguous sum (1,2) , (2,3) --> these contiguous sum has already been
counted ??? where ?
Read problem statement carefully !!
On Wed, Feb 22, 2012 at 9:39 AM, atul anand <atul.87fri...@gmail.com> wrote:
> @sunny : before moving to your algorithm , i can see wrong output in your
> example:-
>
> in you example dere are 8 sums less than 7.
> but for given input contiguous sum less than 7 are
> 1,2,3,4,5 = 4
> so output is 4.
>
> correct me if i am wrong...
>
>
> On Wed, Feb 22, 2012 at 12:41 AM, sunny agrawal
> <sunny816.i...@gmail.com>wrote:
>
>> we need to find how many sums are less than candidate Sum chosen in one
>> iteration of binary search in range 0-S
>> To count this, for each i we try to find how many sums ending at i are
>> lesser than candidate sum !!
>>
>> lets say for some i-1 sum[0 - i-1] < candidate sum then we can say that
>> i*(i-1)/2 sums are less than candidate sum.
>> now lets say after adding a[i] again sum[0 - i] < candidateSum then u can
>> add (i+1) to previous count because all sums [0 - i], sum[1 - i],
>> ............. sum[i - i] will be lesser than candidate sum
>> or if adding a[i] causes sum[0 - i] > candidateSum then u have to find a
>> index g such that sum[g - i] < candidate sum, and increase the count by
>> ((i)-(g) +1).
>>
>> eg lets say your candidate sum is 7 (for the given example{1,2,3,4,5}) k
>> = 3 n = 5
>> initially g = 0
>> sum = 0;
>> candidateSum = 7;
>> count = 0
>> iteration one:
>> sum[0 - 0] = 1 < 7 so count += 0-0+1;
>>
>> iteration 2
>> sum[0-1] = 3 < 7, count += 1-0+1
>>
>> iteration 3
>> sum[0-2] = 6 < 7 count += 2-0+1;
>>
>> iteration 4
>> sum[0,3] = 10 > 7 so now increment g such that sum[g,i] < 7
>> so g = 3 count += 3-3+1;
>>
>> iteration 5
>> sum[3 - 4] = 9 > 7
>> new g = 4 count += 4-4+1
>>
>> final count = 8, so there are 8 sums less than 7
>>
>>
>>
>> On Wed, Feb 22, 2012 at 12:16 AM, shady <sinv...@gmail.com> wrote:
>>
>>> didn't get you, how to check for subsequences which doesn't start from
>>> the beginning ? can you explain for that same example... should we check
>>> for all contiguous subsequences of some particular length?
>>>
>>>
>>> On Tue, Feb 21, 2012 at 11:15 PM, sunny agrawal <sunny816.i...@gmail.com
>>> > wrote:
>>>
>>>> i dont know if a better solution exists
>>>> but here is one with complexity O(N*logS)...
>>>> N = no of elements in array
>>>> S = max sum of a subarray that is sum of all the elements as all are
>>>> positive
>>>>
>>>> algo goes as follows
>>>> do a binary search in range 0-S, for each such candidate sum find how
>>>> many sums are smaller than candidate sum
>>>>
>>>> there is also need to take care of some cases when there are exactly
>>>> k-1 sums less than candidate sum, but there is no contigious where sum =
>>>> candidate sum.
>>>>
>>>>
>>>> On Tue, Feb 21, 2012 at 11:02 PM, shady <sinv...@gmail.com> wrote:
>>>>
>>>>> Problem link <http://www.spoj.pl/ABACUS12/status/ABA12E/>
>>>>>
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>>>>
>>>>
>>>>
>>>> --
>>>> Sunny Aggrawal
>>>> B.Tech. V year,CSI
>>>> Indian Institute Of Technology,Roorkee
>>>>
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>>>
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>>
>>
>>
>> --
>> Sunny Aggrawal
>> B.Tech. V year,CSI
>> Indian Institute Of Technology,Roorkee
>>
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>
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--
Sunny Aggrawal
B.Tech. V year,CSI
Indian Institute Of Technology,Roorkee
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