no need to addi all element if overflow is feared , subtract values
for given index in the loop
something like this
result = 0 ;
for(count = 0; count < n ;count ++){
result += (B[count]-A[count]);
}
result += B[count] ; //! this is the answer...
-Manish
On Sep 21, 9:44 am, Baljeet Kumar <[email protected]> wrote:
> There can be overflow in case of adding up all the elements. Use Xor
> instead.
>
>
>
>
>
> > int result = 0;
> > for (int i = 0; i < n ;i++){
> > result ^= A[i]^B[i];
> > }
>
> > result ^= B[n]; <=== (correction)
> > result is the number we need.
>
> > On Tue, Sep 21, 2010 at 9:48 AM, vishal raja <[email protected]>wrote:
>
> >> add up all the elements in array A say sumA and array B say sumB
> >> ,substract the sumA from sumB... You'll get the element.
>
> >> On Tue, Sep 21, 2010 at 5:36 AM, Anand <[email protected]> wrote:
>
> >>> Two unsorted arrays are given A[n] and B[n+1]. Array A contains n
> >>> integers and B contains n+1 integers of which n are same as in array B but
> >>> in different order and one extra element x. Write an optimized algorithm
> >>> to
> >>> find the value of element x. Use only one pass of both arrays A and B.
>
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> > --
> > Baljeet Kumar
>
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> Baljeet Kumar
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