shashi_genius wrote:
> There are n numbers with all (n-1) unique numbers, except for one which
> is repeating.
> Find out the repeating number in O(n) complexity.
I would use a simple distribution count.
for (i = 0; i < n; i++) {
dcount[i]++;
if (dcount[i] > 1)
printf( "\n The repeating number is: %d", i );
}
This is way faster than sorting. I believe it is exactly O(n).
Adak
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