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:

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:

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:

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.

If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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