By O(1) we mean that we can use variables, can't use something that equals
size of the i/p array.

Use Quick sort, sorting is done in-place which O(n log n) in time, and O(1)
in space.
Sanju
:)



On Sun, Aug 21, 2011 at 7:31 AM, sarvesh saran
<[email protected]>wrote:

> Hi Sanjay,
>
> *Because of O(1) complexity you cannot copy and sort the input array.
>
> *thanks,
> Sarvesh
>  *
> *
>
> On Sun, Aug 21, 2011 at 8:00 PM, Sanjay Rajpal <[email protected]> wrote:
>
>>  Simply sort the array and check for a[i]+a[i+1] > a[i+2] for i=0 to n-3.
>>
>>
>> Sanju
>> :)
>>
>>
>>
>>   On Sun, Aug 21, 2011 at 7:28 AM, sarvesh saran <
>> [email protected]> wrote:
>>
>>>    No idea how to solve the problem except the brute force O(n3)
>>> approach.I am looking for a better solution than this. Because of O(1)
>>> complexity you cannot copy and sort the input.
>>>
>>>
>>> A zero-indexed array A consisting of N integers is given. A triplet (P,
>>> Q, R) is triangular if and
>>> A[P] + A[Q] > A[R],
>>> A[Q] + A[R] > A[P],
>>> A[R] + A[P] > A[Q].
>>>
>>> For example, consider array A such that
>>>
>>> A[0] = 10 A[1] = 2 A[2] = 5
>>> A[3] = 1 A[4] = 8 A[5] = 20
>>> Triplet (0, 2, 4) is triangular.
>>>
>>> public int triangle(int[] A)
>>>
>>> that, given a zero-indexed array A consisting of N integers, returns 1 if
>>> there exists a triangular triplet for this array and returns 0 otherwise.
>>>
>>> Assume that:
>>>
>>> N is an integer within the range [0..100,000];
>>> each element of array A is an integer within the
>>> range[-2,147,483,648..2,147,483,647].
>>> For example, given array A such that
>>>
>>> A[0] = 10 A[1] = 2 A[2] = 5
>>> A[3] = 1 A[4] = 8 A[5] = 20
>>> the function should return 1, as explained above. Given arrayA such that
>>>
>>> A[0] = 10 A[1] = 50 A[2] = 5
>>> A[3] = 1
>>> the function should return 0.
>>> Expected worst-case time complexity: O(n log n)
>>> Expected worst-case space complexity: O(1)
>>>
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