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

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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 <bits/stdc++.h>
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 <bits/stdc++.h>
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

<script>
 
// JavaScript program for the above approach
 
    // Function to find the duplicate number
    // using counting sort method
    function
    findDuplicate(arr, n)
    {
        let countArr = Array.from({length: n+1}, (_, i) => 0), 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]]++;
 
        let a = false;
 
        // Find the element with more
        // than one count
        for (i = 1; i <= n; i++) {
 
            if (countArr[i] > 1) {
                a = true;
                document.write(i + " ");
            }
        }
        if (!a)
            document.write("-1");
    }
 
// Driver Code
 
        let n = 4;
        let arr = [ 1, 3, 4, 2, 2 ];
 
        // Function Call
        findDuplicate(arr, n);
 
</script>

                    

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 <bits/stdc++.h>
using namespace std;
 
int findDuplicates(int arr[], int n) {
    unordered_map<int, int> freq;
    for(int i=0; i<n; i++) {
        if(++freq[arr[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<Integer, Integer> 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).

  1. Find duplicates in O(n) time and O(1) extra space | Set 1
  2. Duplicates in an array in O(n) and by using O(1) extra space | Set-2
     


Last Updated : 07 Nov, 2023
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