@Umer: I say that you forgot to increment i and decrement j in the loop.
But you don't need any loop-counting comparisons at all.
Dave
On Tuesday, October 2, 2012 3:36:58 AM UTC-5, Umer Farooq wrote:
> why don't we try it from both ends ... something like this:
>
> int i = 0; j = size-1;
>
> while (i != j)
> {
> if (arr[i] == elem)
> return arr[i];
> if (arr[j] == elem)
> return arr[j];
> }
>
> this way, we have eliminated half of the comparisons in for loop? What do
> you guys say?
>
> On Tue, Oct 2, 2012 at 12:18 PM, rafi <[email protected] <javascript:>>wrote:
>
>> Vikas is right!
>> while (n) is equal to (while n!=0)
>> you have 2n compares!
>>
>> בתאריך יום שני, 1 באוקטובר 2012 12:12:21 UTC+2, מאת vikas:
>>
>>> still there is no improvement, compiler will generate the code to
>>> compare with zero here. what you have accomplished is , hide it from human
>>> eyes
>>>
>>> On Monday, 1 October 2012 15:25:09 UTC+5:30, Navin Kumar wrote:
>>>>
>>>> @atul:
>>>> still it won't compare 0 th element. Slight modification in your code:
>>>>
>>>> n=*sizeof(arr)*;
>>>> do
>>>> {
>>>> if(elem==arr[*--n*])
>>>> print found;
>>>>
>>>> }while(n);
>>>>
>>>> On Mon, Oct 1, 2012 at 9:50 AM, atul anand <[email protected]> wrote:
>>>>
>>>>> 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 <[email protected]>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 <[email protected]>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 <[email protected]>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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>
>
> --
> Umer
>
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