# Find duplicates in an Array with values 1 to N using counting sort

Given a constant array of N elements which contain elements from 1 to N – 1, with any of these numbers appearing any number of times.
Examples:

Input: N = 5, arr[] = {1, 3, 4, 2, 2}
Output:
Explanation:
2 is the number occurring more than once.
Input: N = 5, arr[] = {3, 1, 3, 4, 2}
Output:
Explanation:
3 is the number occurring more than once.

Naive Approach: The naive method is to first sort the given array and then look for adjacent positions of the array to find the duplicate number.

Below is the implementation of the approach:

## C++

 `// C++ code for the approach` `#include ``using` `namespace` `std;` `// Function to find duplicate``int` `findDuplicates(``int` `arr[], ``int` `n) {``      ``// sort the array``    ``sort(arr, arr + n);``      ` `      ``// boolean flag to check if duplicate exists or not``    ``bool` `flag = ``false``;``      ` `      ``// find duplicates by checking adjacent elements``    ``for` `(``int` `i = 0; i < n - 1; i++) {``        ``if` `(arr[i] == arr[i + 1]) {``            ``flag = ``true``;``              ``return` `arr[i];``        ``}``    ``}``      ` `      ``// if there is no duplicate``    ``if` `(!flag) {``        ``return` `-1;``    ``}``}` `// Driver code``int` `main() {``      ``// Input array``    ``int` `arr[] = { 1, 3, 4, 2, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``  ` `      ``// Function call``    ``cout << findDuplicates(arr, n);``  ` `    ``return` `0;``}`

## Java

 `// Java code for the approach` `import` `java.util.Arrays;` `public` `class` `GFG {``    ``// Function to find duplicate``    ``public` `static` `int` `findDuplicates(``int``[] arr, ``int` `n)``    ``{``        ``// sort the array``        ``Arrays.sort(arr);``        ``// boolean flag to check if duplicate exists or not``        ``boolean` `flag = ``false``;` `        ``// find duplicates by checking adjacent elements``        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {``            ``if` `(arr[i] == arr[i + ``1``]) {``                ``flag = ``true``;``                ``return` `arr[i];``            ``}``        ``}` `        ``// if there is no duplicate``        ``if` `(!flag) {``            ``return` `-``1``;``        ``}``      ` `          ``return` `-``1``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Input array``        ``int``[] arr = { ``1``, ``3``, ``4``, ``2``, ``2` `};``        ``int` `n = arr.length;` `        ``// Function call``        ``System.out.println(findDuplicates(arr, n));``    ``}``}`

## Python3

 `# Python code for the approach` `# Function to find duplicate``def` `find_duplicates(arr):``    ``# sort the array``    ``arr.sort()``    ` `    ``# find duplicates by checking adjacent elements``    ``for` `i ``in` `range``(``len``(arr)``-``1``):``        ``if` `arr[i] ``=``=` `arr[i``+``1``]:``            ``return` `arr[i]``    ``return` `-``1``# Drive function``arr ``=` `[``1``, ``3``, ``4``, ``2``, ``2``]``n ``=` `len``(arr)``print``(find_duplicates(arr))` `# This code is contributed by shiv1o43g`

## C#

 `using` `System;` `public` `class` `GFG``{``    ``// Function to find duplicate``    ``public` `static` `int` `FindDuplicates(``int``[] arr, ``int` `n)``    ``{``        ``// Sort the array``        ``Array.Sort(arr);``        ``// Boolean flag to check if duplicate exists or not``        ``bool` `flag = ``false``;` `        ``// Find duplicates by checking adjacent elements``        ``for` `(``int` `i = 0; i < n - 1; i++)``        ``{``            ``if` `(arr[i] == arr[i + 1])``            ``{``                ``flag = ``true``;``                ``return` `arr[i];``            ``}``        ``}` `        ``// If there is no duplicate``        ``if` `(!flag)``        ``{``            ``return` `-1;``        ``}` `        ``return` `-1;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``// Input array``        ``int``[] arr = { 1, 3, 4, 2, 2 };``        ``int` `n = arr.Length;` `        ``// Function call``        ``Console.WriteLine(FindDuplicates(arr, n));``    ``}``}`

## Javascript

 `function` `findDuplicates(arr) {``    ``// sort the array``    ``arr.sort();` `    ``// find duplicates by checking adjacent elements``    ``for` `(let i = 0; i < arr.length - 1; i++) {``        ``if` `(arr[i] === arr[i + 1]) {``            ``return` `arr[i];``        ``}``    ``}``    ``return` `-1;``}` `// Drive function``let arr = [1, 3, 4, 2, 2];``let n = arr.length;``console.log(findDuplicates(arr));` `// This code is contributed by Dwaipayan Bandyopadhyay`

Output
```2

```

Time Complexity: O(N*log N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above method the idea is to use the concept of Counting Sort. Since the range of elements in the array is known, hence we could use this sorting technique to improvise the time complexity.
The idea is to initialize another array(say count[]) with the same size N and initialize all the elements as 0. Then count the occurrences of each element of the array and update the count in the count[]. Print all the element whose count is greater than 1.
Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the duplicate``// number using counting sort method``int` `findDuplicate(``int` `arr[], ``int` `n)``{``    ``int` `countArr[n + 1], i;` `    ``// Initialize all the elements``    ``// of the countArr to 0``    ``for` `(i = 0; i <= n; i++)``        ``countArr[i] = 0;` `    ``// Count the occurrences of each``    ``// element of the array``    ``for` `(i = 0; i <= n; i++)``        ``countArr[arr[i]]++;` `    ``bool` `a = ``false``;` `    ``// Find the element with more``    ``// than one count``    ``for` `(i = 1; i <= n; i++) {` `        ``if` `(countArr[i] > 1) {``            ``a = ``true``;``            ``cout << i << ``' '``;``        ``}``    ``}` `    ``// If unique elements are there``    ``// print "-1"``    ``if` `(!a)``        ``cout << ``"-1"``;``}` `// Driver Code``int` `main()``{``    ``// Given N``    ``int` `n = 4;` `    ``// Given array arr[]``    ``int` `arr[] = { 1, 3, 4, 2, 2 };` `    ``// Function Call``    ``findDuplicate(arr, n);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG {` `    ``// Function to find the duplicate number``    ``// using counting sort method``    ``public` `static` `int``    ``findDuplicate(``int` `arr[], ``int` `n)``    ``{``        ``int` `countArr[] = ``new` `int``[n + ``1``], i;` `        ``// Initialize all the elements of the``        ``// countArr to 0``        ``for` `(i = ``0``; i <= n; i++)``            ``countArr[i] = ``0``;` `        ``// Count the occurrences of each``        ``// element of the array``        ``for` `(i = ``0``; i <= n; i++)``            ``countArr[arr[i]]++;` `        ``bool a = ``false``;` `        ``// Find the element with more``        ``// than one count``        ``for` `(i = ``1``; i <= n; i++) {` `            ``if` `(countArr[i] > ``1``) {``                ``a = ``true``;``                ``System.out.print(i + ``" "``);``            ``}``        ``}``        ``if` `(!a)``            ``System.out.println(``"-1"``);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``4``;``        ``int` `arr[] = { ``1``, ``3``, ``4``, ``2``, ``2` `};` `        ``// Function Call``        ``findDuplicate(arr, n);``    ``}``}`

## Python3

 `# Python3 program for the above approach` `# Function to find the duplicate``# number using counting sort method ``def` `findDuplicate(arr, n):` `    ``# Initialize all the elements``    ``# of the countArr to 0``    ``countArr ``=` `[``0``] ``*` `(n ``+` `1``)` `    ``# Count the occurrences of each``    ``# element of the array``    ``for` `i ``in` `range``(n ``+` `1``):``        ``countArr[arr[i]] ``+``=` `1` `    ``a ``=` `False` `    ``# Find the element with more``    ``# than one count``    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``if``(countArr[i] > ``1``):``            ``a ``=` `True``            ``print``(i, end ``=` `" "``)` `    ``# If unique elements are there``    ``# print "-1"``    ``if``(``not` `a):``        ``print``(``-``1``)` `# Driver code``if` `__name__ ``=``=` `'__main__'``:` `    ``# Given N``    ``n ``=` `4` `    ``# Given array arr[]``    ``arr ``=` `[ ``1``, ``3``, ``4``, ``2``, ``2` `]` `    ``# Function Call``    ``findDuplicate(arr, n)` `# This code is contributed by Shivam Singh`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to find the duplicate number``// using counting sort method``static` `void` `findDuplicate(``int` `[]arr, ``int` `n)``{``    ``int` `[]countArr = ``new` `int``[n + 1];``    ``int` `i;` `    ``// Initialize all the elements of the``    ``// countArr to 0``    ``for``(i = 0; i <= n; i++)``       ``countArr[i] = 0;` `    ``// Count the occurrences of each``    ``// element of the array``    ``for``(i = 0; i <= n; i++)``       ``countArr[arr[i]]++;` `    ``bool` `a = ``false``;` `    ``// Find the element with more``    ``// than one count``    ``for``(i = 1; i <= n; i++) ``    ``{``       ``if` `(countArr[i] > 1)``       ``{``           ``a = ``true``;``           ``Console.Write(i + ``" "``);``        ``}``    ``}``    ``if` `(!a)``        ``Console.WriteLine(``"-1"``);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 4;``    ``int` `[]arr = { 1, 3, 4, 2, 2 };` `    ``// Function Call``    ``findDuplicate(arr, n);``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output
```2

```

Time Complexity: O(N)
Auxiliary Space: O(N)
Related articles:

Approch 3 : The given code use a hashing technique to locate duplicates in an array. To be more precise, the frequency of each element in the array is stored using an unordered map. An element is considered to be duplicated if its frequency is more than 1, in which case the method returns that element.

Algorithm Steps of given approch :

• Create an unordered map called freq to store the frequency of each element in the array.
• Iterate over the input array using a for loop.
• For each element in the array, increment its frequency in the freq map using the ++freq[arr[i]] notation.
• If the frequency of the current element is greater than 1, return the element as it is a duplicate.
• If no duplicates are found, return -1.
• End of the program.

## C++

 `// c++ code implemenation for above approach ``#include ``using` `namespace` `std;` `int` `findDuplicates(``int` `arr[], ``int` `n) {``    ``unordered_map<``int``, ``int``> freq;``    ``for``(``int` `i=0; i 1) {``            ``return` `arr[i];``        ``}``    ``}``    ``return` `-1;``}` `int` `main() {``    ``int` `arr[] = {1, 3, 4, 2, 2};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``// Fucntion calling ``    ``cout << findDuplicates(arr, n);``    ``return` `0;``}`

## Java

 `import` `java.util.HashMap;``import` `java.util.Map;` `public` `class` `Main {``    ``public` `static` `int` `findDuplicates(``int``[] arr) {``        ``// Create a HashMap to store the frequency of each element in the array.``        ``Map freq = ``new` `HashMap<>();``        ` `        ``// Iterate through the array.``        ``for` `(``int` `i = ``0``; i < arr.length; i++) {``            ``// Check if the element is already in the HashMap.``            ``if` `(freq.containsKey(arr[i])) {``                ``// If it's already in the HashMap, increment its frequency.``                ``freq.put(arr[i], freq.get(arr[i]) + ``1``);``                ` `                ``// If the frequency becomes greater than 1, this is a duplicate, so return it.``                ``if` `(freq.get(arr[i]) > ``1``) {``                    ``return` `arr[i];``                ``}``            ``} ``else` `{``                ``// If the element is not in the HashMap, add it with a frequency of 1.``                ``freq.put(arr[i], ``1``);``            ``}``        ``}``        ` `        ``// If no duplicates are found, return -1.``        ``return` `-``1``;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int``[] arr = {``1``, ``3``, ``4``, ``2``, ``2``};``        ` `        ``// Call the findDuplicates function and print the result.``        ``System.out.println(findDuplicates(arr));``    ``}``}`

## Python

 `def` `find_duplicates(arr):``    ``# Create a dictionary to store the frequency of each element``    ``freq ``=` `{}``    ` `    ``# Iterate through the elements in the array``    ``for` `num ``in` `arr:``        ``# If the element is already in the dictionary, it's a duplicate``        ``if` `num ``in` `freq:``            ``return` `num``        ``# Otherwise, add it to the dictionary with a frequency of 1``        ``else``:``            ``freq[num] ``=` `1``    ` `    ``# If no duplicates are found, return -1``    ``return` `-``1` `def` `main():``    ``arr ``=` `[``1``, ``3``, ``4``, ``2``, ``2``]``    ``# Call the function and print the result``    ``result ``=` `find_duplicates(arr)``    ``print``(result)` `if` `__name__ ``=``=` `"__main__"``:``    ``main()`

## C#

 `using` `System;``using` `System.Collections.Generic;` `class` `Program``{``    ``static` `int` `FindDuplicates(``int``[] arr)``    ``{``        ``Dictionary<``int``, ``int``> freq = ``new` `Dictionary<``int``, ``int``>();``        ``foreach` `(``int` `num ``in` `arr)``        ``{``            ``if` `(freq.ContainsKey(num))``            ``{``                ``freq[num]++;``                ``if` `(freq[num] > 1)``                ``{``                    ``return` `num;``                ``}``            ``}``            ``else``            ``{``                ``freq[num] = 1;``            ``}``        ``}``        ``return` `-1;``    ``}` `    ``static` `void` `Main()``    ``{``        ``int``[] arr = { 1, 3, 4, 2, 2 };``        ``int` `result = FindDuplicates(arr);``        ``Console.WriteLine(result);``    ``}``}`

## Javascript

 `function` `findDuplicates(arr) {``    ``// Create an object to store the frequency of each element in the array.``    ``const freq = {};` `    ``// Iterate through the array.``    ``for` `(let i = 0; i < arr.length; i++) {``        ``// Check if the element is already in the object.``        ``if` `(freq.hasOwnProperty(arr[i])) {``            ``// If it's already in the object, increment its frequency.``            ``freq[arr[i]]++;` `            ``// If the frequency becomes greater than 1, this is a duplicate, so return it.``            ``if` `(freq[arr[i]] > 1) {``                ``return` `arr[i];``            ``}``        ``} ``else` `{``            ``// If the element is not in the object, add it with a frequency of 1.``            ``freq[arr[i]] = 1;``        ``}``    ``}` `    ``// If no duplicates are found, return -1.``    ``return` `-1;``}` `const arr = [1, 3, 4, 2, 2];` `// Call the findDuplicates function and print the result.``console.log(findDuplicates(arr));`

Output
```2

```

Time Complexity : O(n), where n is the size of the input array.
Auxiliary Space :  O(n).

Previous
Next