Find duplicate element in array in time O(n)

Question

I have been asked this question in a job interview and I have been wondering about the right answer.

You have an array of numbers from 0 to n-1, one o

Answer 1

public class FindDuplicate {
    public static void main(String[] args) {
        // assume the array is sorted, otherwise first we have to sort it.
        // time efficiency is o(n)
        int elementData[] = new int[] { 1, 2, 3, 3, 4, 5, 6, 8, 8 };
        int count = 1;
        int element1;
        int element2;

        for (int i = 0; i < elementData.length - 1; i++) {
            element1 = elementData[i];
            element2 = elementData[count];
            count++;
            if (element1 == element2) {
                System.out.println(element2);
            }
        }
    }
}

Answer 2

As described,

You have an array of numbers from 0 to n-1, one of the numbers is removed, and replaced with a number already in the array which makes a duplicate of that number.

I'm assuming elements in the array are sorted except the duplicate entry. If this is the scenario , we can achieve the goal easily as below :

        public static void main(String[] args) {
    //int arr[] = { 0, 1, 2, 2, 3 };
    int arr[] = { 1, 2, 3, 4, 3, 6 };
    int len = arr.length;
    int iMax = arr[0];
    for (int i = 1; i < len; i++) {
        iMax = Math.max(iMax, arr[i]);
        if (arr[i] < iMax) {
            System.out.println(arr[i]);
            break;
        }else if(arr[i+1] <= iMax) {
            System.out.println(arr[i+1]);
            break;
        }
    }
}

• O(n) time and O(1) space ;please share your thoughts.

Answer 3

1. sort the array O(n ln n)

2. using the sliding window trick to traverse the array O(n)

   Space is O(1)

   Arrays.sort(input);
   for(int i = 0, j = 1; j < input.length ; j++, i++){
       if( input[i] == input[j]){
           System.out.println(input[i]);
           while(j < input.length && input[i] == input[j]) j++;
           i = j - 1;
       }
   }
   

Test case int[] { 1, 2, 3, 7, 7, 8, 3, 5, 7, 1, 2, 7 }

output 1, 2, 3, 7

Answer 4

One working solution:

asume number are integers

create an array of [0 .. N]

int[] counter = new int[N];

Then iterate read and increment the counter:

 if (counter[val] >0) {
   // duplicate
 } else {
   counter[val]++;
 }

Answer 5

Use a HashSet to hold all numbers already seen. It operates in (amortized) O(1) time, so the total is O(N).

Answer 6

Scan the array 3 times:

1. XOR together all the array elements -> A. XOR together all the numbers from 0 to N-1 -> B. Now A XOR B = X XOR D, where X is the removed element, and D is the duplicate element.
2. Choose any non-zero bit in A XOR B. XOR together all the array elements where this bit is set -> A1. XOR together all the numbers from 0 to N-1 where this bit is set -> B1. Now either A1 XOR B1 = X or A1 XOR B1 = D.
3. Scan the array once more and try to find A1 XOR B1. If it is found, this is the duplicate element. If not, the duplicate element is A XOR B XOR A1 XOR B1.