I totally agree with Umesh's algo which gives O(K+1) time and an inplace
solution. The only point is the first K+1 numbers may get negated and the
array is modified. In that case after finding the duplicate we can traverse
the array again for the first k+1 no.s and make the negative numbers
positive.
code is -(considering array contains only positive numbers)
for(i=0;i<k+1;i++){
if(A[abs(A[i])])<0) return abs(A[i]); //1st duplicate, abs is absolute
function
A[abs(A[i])]*=(-1);
}
I am explaining Umesh's solution with an example
let k=6 , k<n
A= 1,6,4,5,2,6,3,..........
A[0]-1st element, a[k]-k+1 element in array
now A[0]=1; and A[1]=6 now A[1]= -6
A[1]=6 and A[6]=3 now A[6]=-3
A[2]=4 and A[4]=2 now A[4]=-2
A[3]=5 and A[5]=6 now A[5]=-6
A[4]=-2 and A[abs(-2)]=4 now A[2]=-4
A[5]=-6 and A[abs(-6]<0 therefore return 6
time complexity=O(k+1) and very much inplace solution .
--
Thanks & Regards,
Priyanka Chatterjee
Final Year Undergraduate Student,
Computer Science & Engineering,
National Institute Of Technology,Durgapur
India
http://priyanka-nit.blogspot.com/
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