@atul : In the second for loop.......temp will also contain one which is being missing along with the one which is being repeated..... kindly check it once again.
On Mon, Nov 19, 2012 at 11:44 AM, atul anand <atul.87fri...@gmail.com>wrote: > array has all distinct elements ans lie b/w 1 to n , now bcozz they are > all distinct except 1 element means it should have all element with range 1 > to n...except 1 element , which can be any element b/w 1 to n. > temp=arr[0] > *for i=1 to n > temp=temp^arr[i]; * > //now temp will contain all distinct elements except one which is > repeated(they cancel out) > *for i=1 to n > temp=temp ^ i; * > // now this will cancel out distinct elements excluding one which is > repeated. > temp will contain that repeated element > > On Sun, Nov 18, 2012 at 7:31 PM, shady <sinv...@gmail.com> wrote: > >> Given an array of size n, which has all distinct elements between 1 to n >> with one element repeating, which also implies that one element is missing. >> How to find the repeating element without using extra space and linear >> time complexity ? >> >> Any way to do it with exor ? :P >> >> -- >> >> >> > > -- > > > --
