*NO, u r getting it wrong*
*given a value x, we can find how many contiguous sums are lesser than x
using the above mentioned algorithm in O(N)*
*so we are searching a x in range 0-S such that, x has exactly k-1 sums
lesser than x and x is kth*
*
*
*Algorithm: *
*for
*
On Wed, Feb 22, 2012 at 10:41 AM, atul anand <atul.87fri...@gmail.com>wrote:
> /*
> 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
> */
> do a binary search in range 0-S--> to search what??
> acc to the complexity , O(N *log S) it seems that you are searching each
> element in given input array from range 0-S
> for given input = 1,2,3,4,5
> S= 15
>
> please clarify ..... sorry but i am not getting it ...
>
>
>
>
>
>
> On Wed, Feb 22, 2012 at 10:25 AM, sunny agrawal
> <sunny816.i...@gmail.com>wrote:
>
>> Yes, read my first post
>>
>>
>> On Wed, Feb 22, 2012 at 10:19 AM, atul anand <atul.87fri...@gmail.com>wrote:
>>
>>> 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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>>>>>>
>>>>>>
>>>>>>
>>>>>> --
>>>>>> 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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>
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Sunny Aggrawal
B.Tech. V year,CSI
Indian Institute Of Technology,Roorkee
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