It is funny but right input is as mentioned earlier to rahul. 0,2,3,8,
10, 12, 14., 15 :).. Sorry for unnecessarily flooding your mail
accounts. Please ignore previous post
thanks,
ankit!!
On Thu, Mar 3, 2011 at 8:15 PM, rajul jain <rajuljain...@gmail.com> wrote:
> i think he is wrong bcoz this array in not sorted one.
> so solution of Ankit is right.
>
> On Thu, Mar 3, 2011 at 7:33 PM, nishaanth <nishaant...@gmail.com> wrote:
>>
>> Ignore the previous post..there is a small error in the code..
>> @Ankit..your algm is O(n)...you should split the problem size to n/2 at
>> every stage...rather you are again computing both the subarrays..
>> Here is the correct code...
>> int BsearchElemEqualIndex (int *a, int start, int end)
>> {
>> int mid = (((end - start) >> 1) + start);
>> if (a[mid] == mid)
>> return a[mid];
>> else if (a[mid] != mid)
>> {
>> if (mid == start || mid == end)
>> {
>> return -1;
>> }
>> else
>> {
>> if(a[mid] < mid )
>> BsearchElemEqualIndex (a, start, mid);
>> else
>> BsearchElemEqualIndex (a, mid + 1, end);
>> }
>> }
>> }
>>
>> int _tmain(int argc, _TCHAR* argv[])
>> {
>> int a[SIZE] = {5,9,3,8,1,2,6,7};
>> int x = BsearchElemEqualIndex (a, 0, SIZE);
>> printf ("%d", x);
>> system ("PAUSE");
>> return 0;
>> }
>> S.Nishaanth,
>> Computer Science and engineering,
>> IIT Madras.
>>
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