First element that appears even number of times in an array

Given an array, find the first element that appears even number of times in the array. It returns the element if exists otherwise returns 0.

Examples:

Input : arr[] = {1, 5, 4, 7, 4, 1, 5, 7, 1, 5}; 
Output : 4 
Explanation, 4 is the first element that appears even number of times.

Input : arr[] = {2, 4, 6, 8, 1, 6}; 
Output : 6 
Explanation, 6 is the first element that appears even number of times

A Simple solution is to consider every element one by one. For every element, start counting the frequency, the first element whose count is even gives the result. 

C++

// C++ code to find the first element
// that appears even number times
#include <bits/stdc++.h>
using namespace std;
int firstEven(int arr[], int n)
{
    // Counting the frequency of every
    // element in array.
    for (int i=0;i<n;i++)
    {
        // Count variable to store the
        // frequency of arr[i]
        int count=0;
        for(int j=0;j<n;j++)
        {
            if(arr[i]==arr[j])
            {
                count++;
            }
        }
        // checking if frequency is even or not
        if(count%2==0)
        {
            // Returning the first element
            // which appears even times
            return arr[i];
        }
    }
    return 0;
}

// Driver code
int main()
{
    int arr[] = { 2, 4, 6, 8, 1, 6 };
    cout << firstEven(arr, 6);
    return 0;
}

// This code is contributed by Utkarsh Kumar.

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    static int firstEven(int[] arr, int n) {
        for (int i = 0; i < n; i++) {
            int count = 0;
            for (int j = 0; j < n; j++) {
                if (arr[i] == arr[j]) {
                    count++;
                }
            }
            if (count % 2 == 0) {
                return arr[i];
            }
        }
        return 0;
    }

    public static void main(String[] args) {
        int[] arr = {2, 4, 6, 8, 1, 6};
        System.out.println(firstEven(arr, arr.length));
    }
}
//This code is contributed by Nikhil Garg

Python3

def firstEven(arr, n):
    # Counting the frequency of every
    # element in array.
    for i in range(n):
        # Count variable to store the
        # frequency of arr[i]
        count = 0
        for j in range(n):
            if arr[i] == arr[j]:
                count += 1
        # checking if frequency is even or not
        if count % 2 == 0:
            # Returning the first element
            # which appears even times
            return arr[i]
    return 0

# Driver code
arr = [2, 4, 6, 8, 1, 6]
print(firstEven(arr, 6))

Javascript

// Javascript code to find the first element
// that appears even number times
function firstEven(arr) {
  // Counting the frequency of every element in array.
  for (let i = 0; i < arr.length; i++) {
    // Count variable to store the frequency of arr[i]
    let count = 0;
    for (let j = 0; j < arr.length; j++) {
      if (arr[i] == arr[j]) {
        count++;
      }
    }
    // checking if frequency is even or not
    if (count % 2 == 0) {
      // Returning the first element which appears even times
      return arr[i];
    }
  }
  return 0;
}

// Driver code
const arr = [2, 4, 6, 8, 1, 6];
console.log(firstEven(arr));

C#

using System;

class Program
{
    static void Main(string[] args)
    {
        int[] arr = {2, 4, 6, 8, 1, 6};
        Console.WriteLine(FirstEven(arr));
    }

    static int FirstEven(int[] arr)
    {
        for (int i = 0; i < arr.Length; i++)
        {
            int count = 0;
            for (int j = 0; j < arr.Length; j++)
            {
                if (arr[i] == arr[j])
                {
                    count++;
                }
            }
            if (count % 2 == 0)
            {
                return arr[i];
            }
        }
        return 0;
    }
}

Output

6

Time Complexity : O(n² ) 

Space Complexity : O(1)

An Efficient solution can solve this problem using hash map having O(n) time and O(n) extra space as:  

• Just store the frequency of all elements into the set
• Iterate the array again and check frequency and as soon as you get a element with even frequency return it, else return 0.

Implementation:

C++

// C++ code to find the first element
// that appears even number times
#include <bits/stdc++.h>
using namespace std;
int firstEven(int arr[], int n)
{

    unordered_map<int, int> map1;

    for (int i = 0; i < n; i++)
    {
          //storing frequency
          map1[arr[i]]++;
    }

    int j = 0;
    for (j = 0; j < n; j++)
    {
          //checking the frequency is even or not
          if(map1[arr[j]]%2==0)return arr[j];
    }

    return 0;
}

// Driver code
int main()
{
    int arr[] = { 2, 4, 6, 8, 1, 6 };
    cout << firstEven(arr, 6);
    return 0;
}

Java

// JAVA code to find the first element
// that appears even number times
import java.util.*;
class GFG {
    public static int firstEven(int arr[], int n)
    {

        HashMap<Integer, Boolean> map =
                 new HashMap<Integer, Boolean>();

        for (int i = 0; i < n; i++) {

            // first time occurred
            if (map.get(arr[i]) == null)
                map.put(arr[i], false);

            // toggle for repeated occurrence
            else {
                boolean val = map.get(arr[i]);
                if (val == true)
                    map.put(arr[i], false);
                else
                    map.put(arr[i], true);
            }
        }

        int j = 0;
        for (j = 0; j < n; j++) {

            // first integer with true value
            if (map.get(arr[j]) == true)
                break;
        }

        return arr[j];
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 4, 6, 8, 1, 6 };
        int n = arr.length;
        System.out.println(firstEven(arr, n));
    }
}

Python3

# Python3 code to find the first element
# that appears even number times
def firstEven(arr, n):

    map1 = {}
    for i in range(0, n):

        # first time occurred
        if arr[i] not in map1:
            map1[arr[i]] = False

        # toggle for repeated occurrence
        else:
            map1[arr[i]] = not map1[arr[i]]

    for j in range(0, n):

        # first integer with true value
        if map1[arr[j]] == True:
            break

    return arr[j]

# Driver code
if __name__ == "__main__":

    arr = [2, 4, 6, 8, 1, 6]
    print(firstEven(arr, 6))

# This code is contributed
# by Rituraj Jain

C#

// C# code to find the first element
// that appears even number times
using System;
using System.Collections.Generic;

class GFG
{
    static int firstEven(int []arr,
                         int n)
    {
        var map = new Dictionary<int,
                                 string>();

        var hash = new HashSet<int>(arr);

        foreach (int a in hash)
            map.Add(a, "null");

        for (int i = 0; i < n; i++)
        {

            // first time occurred
            if (map[arr[i]].Equals("null"))
                map[arr[i]] = "false";

            // toggle for repeated
            // occurrence
            else
            {
                string val = map[arr[i]];
                if (val.Equals("true"))
                    map[arr[i]] = "false";
                else
                    map[arr[i]] = "true";
            }
        }

        int j = 0;
        for (j = 0; j < n; j++)
        {

            // first integer with
            // true value
            if (map[arr[j]].Equals("true"))
                break;
        }
        return arr[j];
    }

    // Driver code
    static void Main()
    {
        int []arr = new int[]{ 2, 4, 6,
                               8, 1, 6 };
        int n = arr.Length;
        Console.Write(firstEven(arr, n));
    }
}

// This code is contributed by
// Manish Shaw(manishshaw1)

Javascript

<script>
// Javascript code to find the first element
// that appears even number times


function firstEven(arr, n)
{

    let map1 = new Map();

    for (let i = 0; i < n; i++)
    {

        // first time occurred
        if (!map1.has(arr[i]))
            map1.set(arr[i],false);

        // toggle for repeated occurrence
        else
        {
            let val = map1.get(arr[i]);
            if (val == true)
                map1.set(arr[i], false);
            else
                map1.set(arr[i], true);
        }
    }

    let j = 0;
    for (j = 0; j < n; j++)
    {

        // first integer with true value
        if (map1.get(arr[j]) == true)
            break;
    }

    return arr[j];
}

// Driver code

    let arr = [ 2, 4, 6, 8, 1, 6 ];
    document.write(firstEven(arr, 6));

    // This code is contributed by gfgking.
</script>

Output

6

Complexity Analysis:

• Time Complexity: O(n) 
• Auxiliary Space: O(n)