# Duplicates in an array in O(n) and by using O(1) extra space | Set-2

Given an array of n elements containing elements from 0 to n-1, with any of these numbers appearing any number of times, find these repeating numbers in O(n) and using only constant memory space.

Example:

```Input: n = 7 , array = {1, 2, 3, 1, 3, 6, 6}
Output: 1, 3 and 6.
Explanation: Duplicate element in the array are 1 , 3 and 6

Input: n = 6, array = {5, 3, 1, 3, 5, 5}
Output: 3 and 5.
Explanation: Duplicate element in  the array are 3 and 5```

We have discussed an approach for this question in the below post:
Duplicates in an array in O(n) and by using O(1) extra space | Set-2
But there is a problem in the above approach. It prints the repeated number more than once.

## We strongly recommend that you click here and practice it, before moving on to the solution.

Approach: The basic idea is to use a HashMap to solve the problem. But there is a catch, the numbers in the array are from 0 to n-1, and the input array has length n. So, the input array can be used as a HashMap. While traversing the array, if an element a is encountered then increase the value of a%n‘th element by n. The frequency can be retrieved by dividing the a%n‘th element by n.

Algorithm:

1. Traverse the given array from start to end.
2. For every element in the array increment the arr[i]%n‘th element by n.
3. Now traverse the array again and print all those indices i for which arr[i]/n is greater than 1. Which guarantees that the number n has been added to that index.

Note: This approach works because all elements are in the range from 0 to n-1 and arr[i]/n would be greater than 1 only if a value “i” has appeared more than once.

Below is the implementation of the above approach:

## CPP

 `// C++ program to print all elements that ` `// appear more than once. ` `#include ` `using` `namespace` `std; ` ` `  `// function to find repeating elements ` `void` `printRepeating(``int` `arr[], ``int` `n) ` `{ ` `    ``// First check all the values that are ` `    ``// present in an array then go to that ` `    ``// values as indexes and increment by ` `    ``// the size of array ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``int` `index = arr[i] % n; ` `        ``arr[index] += n; ` `    ``} ` ` `  `    ``// Now check which value exists more ` `    ``// than once by dividing with the size ` `    ``// of array ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `((arr[i] / n) >= 2) ` `            ``cout << i << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 6, 3, 1, 3, 6, 6 }; ` `    ``int` `arr_size = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``cout << ``"The repeating elements are: \n"``; ` ` `  `    ``// Function call ` `    ``printRepeating(arr, arr_size); ` `    ``return` `0; ` `}`

## Java

 `// Java program to print all elements that ` `// appear more than once. ` `import` `java.util.*; ` `class` `GFG { ` ` `  `    ``// function to find repeating elements ` `    ``static` `void` `printRepeating(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``// First check all the values that are ` `        ``// present in an array then go to that ` `        ``// values as indexes and increment by ` `        ``// the size of array ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{ ` `            ``int` `index = arr[i] % n; ` `            ``arr[index] += n; ` `        ``} ` ` `  `        ``// Now check which value exists more ` `        ``// than once by dividing with the size ` `        ``// of array ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{ ` `            ``if` `((arr[i] / n) >= ``2``) ` `                ``System.out.print(i + ``" "``); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `arr[] = { ``1``, ``6``, ``3``, ``1``, ``3``, ``6``, ``6` `}; ` `        ``int` `arr_size = arr.length; ` ` `  `        ``System.out.println(``"The repeating elements are: "``); ` ` `  `        ``// Function call ` `        ``printRepeating(arr, arr_size); ` `    ``} ` `}`

## Python3

 `# Python3 program to ` `# print all elements that ` `# appear more than once. ` ` `  `# function to find ` `# repeating elements ` ` `  ` `  `def` `printRepeating(arr, n): ` ` `  `    ``# First check all the ` `        ``# values that are ` `    ``# present in an array ` `        ``# then go to that ` `    ``# values as indexes ` `        ``# and increment by ` `    ``# the size of array ` `    ``for` `i ``in` `range``(``0``, n): ` `        ``index ``=` `arr[i] ``%` `n ` `        ``arr[index] ``+``=` `n ` ` `  `    ``# Now check which value ` `        ``# exists more ` `    ``# than once by dividing ` `        ``# with the size ` `    ``# of array ` `    ``for` `i ``in` `range``(``0``, n): ` `        ``if` `(arr[i]``/``n) >``=` `2``: ` `            ``print``(i, end``=``" "``) ` ` `  ` `  `# Driver code ` `arr ``=` `[``1``, ``6``, ``3``, ``1``, ``3``, ``6``, ``6``] ` `arr_size ``=` `len``(arr) ` ` `  `print``(``"The repeating elements are:"``) ` ` `  `# Function call ` `printRepeating(arr, arr_size) ` ` `  `# This code is contributed ` `# by Shreyanshi Arun. `

## C#

 `// C# program to print all elements that ` `// appear more than once. ` ` `  `using` `System; ` `class` `GFG { ` ` `  `    ``// function to find repeating elements ` `    ``static` `void` `printRepeating(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``// First check all the values that are ` `        ``// present in an array then go to that ` `        ``// values as indexes and increment by ` `        ``// the size of array ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``int` `index = arr[i] % n; ` `            ``arr[index] += n; ` `        ``} ` ` `  `        ``// Now check which value exists more ` `        ``// than once by dividing with the size ` `        ``// of array ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `((arr[i] / n) >= 2) ` `                ``Console.Write(i + ``" "``); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 1, 6, 3, 1, 3, 6, 6 }; ` `        ``int` `arr_size = arr.Length; ` ` `  `        ``Console.Write(``"The repeating elements are: "` `                      ``+ ``"\n"``); ` ` `  `        ``// Function call ` `        ``printRepeating(arr, arr_size); ` `    ``} ` `}`

## PHP

 `= 2) ` `            ``echo` `\$i` `, ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `\$arr` `= ``array``(1, 6, 3, 1, 3, 6, 6); ` `\$arr_size` `= sizeof(``\$arr``) /  ` `            ``sizeof(``\$arr``[0]); ` ` `  `echo` `"The repeating elements are: \n"``; ` ` `  `// Function call ` `printRepeating( ``\$arr``, ``\$arr_size``); ` ` `  `// This code is contributed by nitin mittal. ` `?> `

## Javascript

 ``

Output

```The repeating elements are:
1 3 6 ```

Complexity Analysis:

• Time Complexity: O(n).
Only two traversals are needed. So the time complexity is O(n)
• Auxiliary Space: O(1).
As no extra space is needed, so the space complexity is constant

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