__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.

The author Harry He owns all the rights of
this post. If you are going to use part of or the whole of this article in your
blog or webpages, please add a reference to http://codercareer.blogspot.com/. If you are going to use it in your books,
please contact the author via zhedahht@gmail.com . Thanks.

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looks like a typo, the statement "int end = numbers.length;" should be above the while loop.

ReplyDeleteThanks for your good finding. It's been fixed.

DeleteThis is purely wrong, counting numbers does not work.

ReplyDeletetry: int[] nums = new int[] { 9, 8, 7, 2, 2, 3, 4, 6, 1, 5 }

Just sum all the numbers, subtract n*(n-1)/2 (sum of 1..n) and you get your answer

that's what I would think.

Deletethat's what I would think.

DeleteThe solution works only if there is exactly one duplication in the array. This is a special case for the problem here. It doesn't limit the number of duplication.

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ReplyDeleteanother solution ( I assume the elements in array are all integers): a = sum of n+1 elements, b = sum of n continuous integers, thus the duplicated integer is equal to b-a.

ReplyDeleteThat won't work.

Delete1+5+5+2 = 13

1+5+5 = 11

13-11 = 2: therefore, according to your method, the duplicated int is 2.

I think he requires contiguous integers, so for your example 1+2+3+4+5+5 = 20 while 1+2+3+4+5 should equal 15, so 20 - 15 = 5

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ReplyDeleteWhat is the time complexity of this solution? I guess it is O(nlogn). Complexity of countRange() is O(n) and it is invoked at most log(n) time. Is that correct?

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ReplyDeleteSince you're only looking for duplicates you only need to store if they number has been seen before or not, so you could initialize a suitably large numbers of bits and use the value of the number as your offset. Initialize all the bits to zero and then scroll though your list checking each bit before you set it and if it was already been set then you have your duplicate and this won't require that the numbers be sorted and avoid large sums.

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