Find a duplicate in array of integers

Question

This was an interview question.

I was given an array of n+1 integers from the range [1,n]. The property of the array is that it has k (k&

Answer 1

Here is a possible implementation:

function checkDuplicate(arr) {
  console.log(arr.join(", "));
  let  len = arr.length
      ,pos = 0
      ,done = 0
      ,cur = arr[0]
      ;
  while (done < len) {
    if (pos === cur) {
      cur = arr[++pos];
    } else {
      pos = cur;
      if (arr[pos] === cur) {
        console.log(`> duplicate is ${cur}`);
        return cur;
      }
      cur = arr[pos];
    }
    done++;
  }
  console.log("> no duplicate");
  return -1;
}

for (t of [
     [0, 1, 2, 3]
    ,[0, 1, 2, 1]
    ,[1, 0, 2, 3]
    ,[1, 1, 0, 2, 4]
  ]) checkDuplicate(t);

It is basically the solution proposed by @maraca (typed too slowly!) It has constant space requirements (for the local variables), but apart from that only uses the original array for its storage. It should be O(n) in the worst case, because as soon as a duplicate is found, the process terminates.

Answer 2

• It depends what tools are you(your app) can use. Currently a lot of frameworks/libraries exists. For exmaple in case of C++ standart you can use std::map<> ,as maraca mentioned.

• Or if you have time you can made your own implementation of binary tree, but you need to keep in mind that insert of elements differs in comarison with usual array. In this case you can optimise search of duplicates as it possible in your particular case.

binary tree expl. ref: https://www.wikiwand.com/en/Binary_tree

Answer 3

1. Take the very last element (x).

2. Save the element at position x (y).

3. If x == y you found a duplicate.

4. Overwrite position x with x.

5. Assign x = y and continue with step 2.

You are basically sorting the array, it is possible because you know where the element has to be inserted. O(1) extra space and O(n) time complexity. You just have to be careful with the indices, for simplicity I assumed first index is 1 here (not 0) so we don't have to do +1 or -1.

Edit: without modifying the input array

This algorithm is based on the idea that we have to find the entry point of the permutation cycle, then we also found a duplicate (again 1-based array for simplicity):

Example:

2 3 4 1 5 4 6 7 8

Entry: 8 7 6

Permutation cycle: 4 1 2 3

As we can see the duplicate (4) is the first number of the cycle.

1. Finding the permutation cycle

   1. x = last element
   2. x = element at position x
   3. repeat step 2. n times (in total), this guarantees that we entered the cycle

2. Measuring the cycle length

   1. a = last x from above, b = last x from above, counter c = 0
   2. a = element at position a, b = elment at position b, b = element at position b, c++ (so we make 2 steps forward with b and 1 step forward in the cycle with a)
   3. if a == b the cycle length is c, otherwise continue with step 2.

3. Finding the entry point to the cycle

   1. x = last element
   2. x = element at position x
   3. repeat step 2. c times (in total)
   4. y = last element
   5. if x == y then x is a solution (x made one full cycle and y is just about to enter the cycle)
   6. x = element at position x, y = element at position y
   7. repeat steps 5. and 6. until a solution was found.

The 3 major steps are all O(n) and sequential therefore the overall complexity is also O(n) and the space complexity is O(1).

Example from above:

1. x takes the following values: 8 7 6 4 1 2 3 4 1 2

2. a takes the following values: 2 3 4 1 2
   b takes the following values: 2 4 2 4 2
   therefore c = 4 (yes there are 5 numbers but c is only increased when making steps, not initially)

3. x takes the following values: 8 7 6 4 | 1 2 3 4
   y takes the following values: | 8 7 6 4
   x == y == 4 in the end and this is a solution!

Example 2 as requested in the comments: 3 1 4 6 1 2 5

1. Entering cycle: 5 1 3 4 6 2 1 3

2. Measuring cycle length:
   a: 3 4 6 2 1 3
   b: 3 6 1 4 2 3
   c = 5

3. Finding the entry point:
   x: 5 1 3 4 6 | 2 1
   y: | 5 1
   x == y == 1 is a solution

Answer 4

If you are allowed to non-destructively modify the input vector, then it is pretty easy. Suppose we can "flag" an element in the input by negating it (which is obviously reversible). In that case, we can proceed as follows:

Note: The following assume that the vector is indexed starting at 1. Since it is probably indexed starting at 0 (in most languages), you can implement "Flag item at index i" with "Negate the item at index i-1".

1. Set i to 0 and do the following loop:
   1. Increment i until item i is unflagged.
   2. Set j to i and do the following loop:
      1. Set j to vector[j].
      2. if the item at j is flagged, j is a duplicate. Terminate both loops.
      3. Flag the item at j.
      4. If j != i, continue the inner loop.
2. Traverse the vector setting each element to its absolute value (i.e. unflag everything to restore the vector).