__Questions__: All numbers in an array with length n+1 are in range from 1 to n, so there is at least one duplication in the array. How to find any a duplication? Please don't modify the input array.

__The simple solution is to utilize hash tables. When scanning the array, array elements are inserted into the hash table one by one. In this way, it's easy to find a duplication with the hash table, which costs O(n) space.__

**Analysis**:
Let's try not to
employ extra space. Why there are duplications in the array? If there are no
duplications, the count of numbers ranging from 1 to n is n. Since the array
contains more than n numbers, there
should be duplications. It looks like it's important to count numbers in
ranges.

Let's divide numbers
from 1 to n into two ranges, split with
the number in the middle (denoted as m),
and then count the numbers of the two subranges. If the count of numbers from 1
to m is greater than m, the duplication is in the range from 1 to m. Otherwise, there should be at least one
duplication in the range from m+1 to n. And then we continue the recursive process
until we find the duplication.

The Java code is
listed below:

static int
countRange(int[] numbers, int start, int
end)

{

int count = 0;

for(int
i = 0; i < numbers.length; i++)

if(numbers[i] >= start &&
numbers[i] <= end)

++count;

return count;

}

static int
getDuplication(int[] numbers)

{

int start = 1;

int end = numbers.length;

while(end >= start)

int middle = ((end - start) >>
1) + start;

int count = countRange(numbers,
start, middle);

if(end == start) {

if(count > 1)

return start;

else

break;

}

if(count > (middle - start + 1))

end = middle;

else

start = middle + 1;

}

return -1;

}

The code with unit
tests is shared at http://ideone.com/lhV22m.

More coding interview questions are discussed
in my book< Coding Interviews: Questions, Analysis & Solutions>. You
may find the details of this book on Amazon.com, or Apress.

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