sum[0-1] = 3 --> (1,2)
sum[0-2] = 6 --> (1,2,3)
sum[1-2] = 5 --> (2,3)
ok...so we can consider 3 , (1,2) as different contiguous.
how did you choose candidate sum for the given input ?? will it not add to
the complexity
On Wed, Feb 22, 2012 at 9:59 AM, sunny agrawal <sunny816.i...@gmail.com>wrote:
> @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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