some thing like would work.
int find(int x ) {
int p = sqrt(x/2);
int total=0;
float y;
for( int i=0;i<=p;i++) {
y=sqrt(x*x-i*i);
if( y(int) == y) total++;
}
return total;
}
find(
}
On Thu, Feb 14, 2013 at 10:22 AM, Shachindra A C <[email protected]>wrote:
> Guneesh,
>
> Thanks for the reply.
> I interpret ur answer as follows:
>
> If N = say, 50
>
> the two combinations are (1,7) and (5,5).
>
> Acc to you, first find sqrt(50) = 7
>
> fill in 1,4,9,16,25,36,49 in an array.
>
> Then have min = 0, max = 6 and get all combinations in one pass over the
> array, right?
>
> So, the complexity would be O(n^0.5), right?
>
> Can you think of any solution making use of complex domain? Just curious
> to know...
>
> On Wed, Feb 13, 2013 at 10:28 PM, Guneesh Paul Singh
> <[email protected]>wrote:
>
>> Replace all elements of array by its square.and sort it
>> Now question is to find two nos in the array such that their sum is N.
>> For this take two pointers min and max
>> Initialize min to 0 and max to n-1(n-size of array)
>> 1.find the sum a[min]+a[max]
>> 2.if sum>N max=max-1
>> if sum<N min=min+1
>> if sum==N we have a sol
>>
>> Now in case of nonunique values all possible pairs must be displayed
>> eg for 2 2 3 3 N =5
>> min=0,max 3 is a sol but u need to display all combination of 2 and 3 for
>> that i used tempmin=position till which we have a value a[min] and temp max
>> postion till which we have a value a[max] and display
>> all possible combinations.
>>
>> P.S. This was asked to me in Amazon
>>
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>
>
>
> --
> Regards,
> Shachindra A C
>
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