yes, but there no need of checking outside the loop
n=sizeof(arr)-1;
do
{
if(elem==arr[n])
print found;
n--;
}while(n);
On Mon, Oct 1, 2012 at 9:33 AM, Navin Kumar <algorithm.i...@gmail.com>wrote:
> @atul: keep one more checking outside loop for element at 0 th index.
> Because when n = 0 the your loop come out from the loop without comparing
> it.
>
>
> On Mon, Oct 1, 2012 at 8:55 AM, atul anand <atul.87fri...@gmail.com>wrote:
>
>> n=sizeof(arr);
>> n--;
>>
>> while(n)
>> {
>> if(elem=arr[n])
>> print found;
>>
>> n--;
>>
>> }
>>
>> On Sun, Sep 30, 2012 at 2:56 PM, רפי וינר <rafiwie...@gmail.com> wrote:
>>
>>> Hi
>>> i was in an interview and was given a simple function
>>> int arrExsits(int* arr,int size,int elem){
>>> for (int i=0;i<size;++i)
>>> if(elem==arr[i])
>>> return i;
>>> return -1;
>>> }
>>> this function does 2n compares
>>> n- the if statment
>>> n-check that i is smaller then size
>>> i was suppose to give an optimal (less compares) solution so i gave
>>>
>>> int arrExsits(int* arr,int size,int elem){
>>> if (arr[size-1]==elem)
>>> return size-1;
>>> arr[size-1]=elem]
>>> for (int i=0;;++i)
>>> if(elem==arr[i]){
>>> if (i!=size-1)
>>> return i;
>>> return -1;
>>> }
>>> this solution works and it has n+2 compares the first one another n and
>>> the second inner if.
>>> they told me it's good (and I've passed) but they told just for my
>>> knowledge that there is a better N compare solution.
>>> I've searched the web but couldn't find it.
>>> anybody knows?
>>> Thanks
>>>
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