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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