How to find a duplicate element in an array of shuffled consecutive integers?

Question

I recently came across a question somewhere:

Suppose you have an array of 1001 integers. The integers are in random order, but you know each of the integers is between 1 and 1000 (inclusive). In addition, each number appears only once in the array, except for one number, which occurs twice. Assume that you can access each element of the array only once. Describe an algorithm to find the repeated number. If you used auxiliary storage in your algorithm, can you find an algorithm that does not require it?

What I am interested in to know is the second part, i.e., without using auxiliary storage. Do you have any idea?

Answer 1

Just add them all up, and subtract the total you would expect if only 1001 numbers were used from that.

Eg:

Input: 1,2,3,2,4 => 12
Expected: 1,2,3,4 => 10

Input - Expected => 2

Answer 2

Add all the numbers together. The final sum will be the 1+2+...+1000+duplicate number.

Answer 3

Add all numbers. The sum of integers 1..1000 is (1000*1001)/2. The difference from what you get is your number.

Answer 4

If you know that we have the exact numbers 1-1000, you can add up the results and subtract 500500 (sum(1, 1000)) from the total. This will give the repeated number because sum(array) = sum(1, 1000) + repeated number.

Answer 5

Well, there is a very simple way to do this... each of the numbers between 1 and 1000 occurs exactly once except for the number that is repeated.... so, the sum from 1....1000 is 500500. So, the algorithm is:

sum = 0
for each element of the array:
   sum += that element of the array
number_that_occurred_twice = sum - 500500

Answer 6

No extra storage requirement (apart from loop variable).

int length = (sizeof array)/(sizeof array[0]);
for(int i=1; i < length; i++){
   array[0] += array[i];
}

printf("Answer : %d\n", array[0] - (length*(length+1))/2);

Answer 7

Do arguments and callstacks count as auxiliary storage?

int sumRemaining(int* remaining, int count) {
    if (!count) {
        return 0;
    }
    return remaining[0] + sumRemaining(remaining + 1, count - 1);
}

 

printf("duplicate is %d", sumRemaining(array, 1001) - 500500);

Edit: tail call version

int sumRemaining(int* remaining, int count, int sumSoFar) {
    if (!count) {
        return sumSoFar;
    }
    return sumRemaining(remaining + 1, count - 1, sumSoFar + remaining[0]);
}
printf("duplicate is %d", sumRemaining(array, 1001, 0) - 500500);

Answer 8

Update 2: Some people think that using XOR to find the duplicate number is a hack or trick. To which my official response is: "I am not looking for a duplicate number, I am looking for a duplicate pattern in an array of bit sets. And XOR is definitely suited better than ADD to manipulate bit sets". :-)

Update: Just for fun before I go to bed, here's "one-line" alternative solution that requires zero additional storage (not even a loop counter), touches each array element only once, is non-destructive and does not scale at all :-)

printf("Answer : %d\n",
           array[0] ^
           array[1] ^
           array[2] ^
           // continue typing...
           array[999] ^
           array[1000] ^
           1 ^
           2 ^
           // continue typing...
           999^
           1000
      );

Note that the compiler will actually calculate the second half of that expression at compile time, so the "algorithm" will execute in exactly 1002 operations.

And if the array element values are know at compile time as well, the compiler will optimize the whole statement to a constant. :-)

Original solution: Which does not meet the strict requirements of the questions, even though it works to find the correct answer. It uses one additional integer to keep the loop counter, and it accesses each array element three times - twice to read it and write it at the current iteration and once to read it for the next iteration.

Well, you need at least one additional variable (or a CPU register) to store the index of the current element as you go through the array.

Aside from that one though, here's a destructive algorithm that can safely scale for any N up to MAX_INT.

for (int i = 1; i < 1001; i++)
{
   array[i] = array[i] ^ array[i-1] ^ i;
}

printf("Answer : %d\n", array[1000]);

I will leave the exercise of figuring out why this works to you, with a simple hint :-):

a ^ a = 0
0 ^ a = a

Answer 9

To paraphrase Francis Penov's solution.

The (usual) problem is: given an array of integers of arbitrary length that contain only elements repeated an even times of times except for one value which is repeated an odd times of times, find out this value.

The solution is:

acc = 0
for i in array: acc = acc ^ i

Your current problem is an adaptation. The trick is that you are to find the element that is repeated twice so you need to adapt solution to compensate for this quirk.

acc = 0
for i in len(array): acc = acc ^ i ^ array[i]

Which is what Francis' solution does in the end, although it destroys the whole array (by the way, it could only destroy the first or last element...)

But since you need extra-storage for the index, I think you'll be forgiven if you also use an extra integer... The restriction is most probably because they want to prevent you from using an array.

It would have been phrased more accurately if they had required O(1) space (1000 can be seen as N since it's arbitrary here).

Answer 10

A non destructive version of solution by Franci Penov.

This can be done by making use of the XOR operator.

Lets say we have an array of size 5: 4, 3, 1, 2, 2
Which are at the index:                        0, 1, 2, 3, 4

Now do an XOR of all the elements and all the indices. We get 2, which is the duplicate element. This happens because, 0 plays no role in the XORing. The remaining n-1 indices pair with same n-1 elements in the array and the only unpaired element in the array will be the duplicate.

int i;
int dupe = 0;
for(i=0;i<N;i++) {
 dupe = dupe ^ arr[i] ^ i;
}
// dupe has the duplicate.

The best feature of this solution is that it does not suffer from overflow problems that is seen in the addition based solution.

Since this is an interview question, it would be best to start with the addition based solution, identify the overflow limitation and then give the XOR based solution :)

This makes use of an additional variable so does not meet the requirements in the question completely.

Answer 11

One line solution in Python

arr = [1,3,2,4,2]
print reduce(lambda acc, (i, x): acc ^ i ^ x, enumerate(arr), 0)
# -> 2

Explanation on why it works is in @Matthieu M.'s answer.

Answer 12

A triangle number T(n) is the sum of the n natural numbers from 1 to n. It can be represented as n(n+1)/2. Thus, knowing that among given 1001 natural numbers, one and only one number is duplicated, you can easily sum all given numbers and subtract T(1000). The result will contain this duplicate.

For a triangular number T(n), if n is any power of 10, there is also beautiful method finding this T(n), based on base-10 representation:

n = 1000
s = sum(GivenList)
r = str(n/2)
duplicate = int( r + r ) - s