### Ways to check Array for duplicate elements Java

In this tutorial, we will see couple of ways to find out the duplicate or repeated elements in an array. Some of the ways are given below:#### Solution 1

public void findDuplicate(int arr[]) { System.out.println("The repeating elements are: "); for (int index = 0; index < arr.length; index++) { if (arr[Math.abs(arr[index])] >= 0) arr[Math.abs(arr[index])] = -arr[Math.abs(arr[index])]; else { System.out.println(Math.abs(arr[index])); } } }

#### Solution 2

public void findDuplicate(int[] arr) { for (int index = 0; index < arr.length; index++) { for (int index1 = 0; index1 < arr.length; index1++) { if (arr[index] == arr[index1] && index != index1) { System.out.println(arr[index]); } } } }

#### Solution 3 (Using Set)

public void findDuplicate(Integer[] arr) { Set hashSet = new HashSet(); for (Integer num : arr) { if (!hashSet.add(num)) { System.out.println(num); } } }

#### Solution 4 (Using List)

public int findDuplicate(List<Integer> arr){ int length = arr.size() - 1; int total = getTotal(numbers); int duplicate = total - (length*(length + 1)/2); return duplicate; } public int getTotal(List<Integer> arr){ int sum = 0; for (int num : arr) { sum += num; } return sum; }

I tried my best to provide possible solutions to find out the duplicate elements in an array. Please provide your valuable comments, suggestions and feedback to correct any mistakes made or to make content valuable for other users.

Solution 2 takes more than required iterations, you can actually reduce the outer loop iterations.

ReplyDeleteCould you please explain how first method is working?

ReplyDeletetime complexity of second solution is O(n^2), and for the third solution is O(n).

ReplyDelete