/*
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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>>>
>>>
>>>
>>> --
>>> 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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