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[i];
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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