# Find duplicates in an given array in O(n) time and O(1) extra space.

Objective: Given an array of integers, find out duplicates in it.

Example:

```int [] a = {4, 6, 2, 1, 2, 5};
Output: Array has duplicates : 2

int a [] = {1, 6, 5, 2, 3, 3, 2};
Output: Array has duplicates : 2 , 3
```

Approach:

Naive approach: Use 2 loops. Check each element in the array with all other elements in the array and check if it has the same value.

Time Complexity : O(n^2) Space Complexity: O(1)

Code:

 import java.util.Arrays; public class CheckDuplicatesUsingTwoLoops { public void hasDuplicates2ForLoops(int [] arrA) { for (int i = 0; i < arrA.length; i++) { for (int j = i+1; j < arrA.length; j++) { if(arrA[i]==arrA[j]){ System.out.println("Array has duplicates : " + arrA[i]); } } } } public static void main(String[] args) { int a [] = {1, 6, 5, 2, 3, 3, 2}; new CheckDuplicatesUsingTwoLoops().hasDuplicates2ForLoops(a); } }

Sorting approach: Sort the array, this will bring all the duplicates together if present. Now navigate the array and check the adjacent elements to check for duplicates.

Time Complexity : O(nlogn) Space Complexity: O(n) by using merge sort.

Code:

 import java.util.Arrays; public class CheckDuplicatesUsingSorting { public void hasDuplicatesUsingSorting(int [] arrA) { Arrays.sort(arrA); for (int i = 0; i < arrA.length–1; i++) { if(arrA[i]==arrA[i+1]){ System.out.println("Array has duplicates : " + arrA[i]); } } } public static void main(String[] args) { int a [] = {1, 6, 5, 2, 3, 3, 2}; new CheckDuplicatesUsingSorting().hasDuplicatesUsingSorting(a); } }

Better Solution : Use Hash map. Store the count of each element of array in a hash table and later check in Hash table if any element has count more than 1. ( Similar approach is used in problem – Find the first non repeating character in a given string

Time Complexity : O(n) and Space Complexity: O(n).

Code:

 import java.util.Arrays; import java.util.HashMap; public class CheckDuplicatesUsingHashMap { public void hasDuplicatesUsingMap(int [] arrA){ HashMap map = new HashMap<>(); for (int i = 0; i

Better Solution (Conditional) : O(n) time and O(1) extra space.

• This solution works only if array has positive integers and all the elements in the array are in range from 0 to n-1 where n is the size of the array.
• Nav­i­gate the array.
• Update the array as for ith index :- arrA[arrA[i]] = arrA[arrA[i]]*-1 (if it already not negative).
• If is already neg­a­tive, we have dupli­cates, return false.

Note:

• The code given below does not handle the case when 0 is present in the array.
• To handle 0 in array, while navigating the array, when 0 is encountered, replace it with INT_MIN and if INT_MIN is encountered while traversing, this means 0 is repeated in the array.

Similar approach used in problem : If array has all consecutive numbers.

Code:

 public class CheckDuplicates { public void hasDuplicates(int[] arrA) { for (int i = 0; i < arrA.length; i++) { //check if element is negative, if yes the we have found the duplicate if (arrA[Math.abs(arrA[i])] < 0) { System.out.println("Array has duplicates : " + Math.abs(arrA[i])); } else { arrA[Math.abs(arrA[i])] = arrA[Math.abs(arrA[i])] * –1; } } } public static void main(String[] args) { int a[] = {1, 6, 5, 2, 3, 3, 2}; new CheckDuplicates().hasDuplicates(a); } }

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CheckDuplicates.java
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Output:

```Array has duplicates : 3
Array has duplicates : 2
```