Two odd occurring elements in an array where all other occur even times

Given an array where all elements appear even number of times except two, print the two odd occurring elements. It may be assumed that the size of array is at-least two.

Examples:

Input : arr[] = {2, 3, 8, 4, 4, 3, 7, 8}
Output : 2 7

Input : arr[] = {15, 10, 10, 50 7, 5, 5, 50, 50, 50, 50, 50}
Output : 7 15

Simple solution :

Approach :

A simple solution is to use two nested loops. The outer loop traverses through all elements. The inner loop counts occurrences of the current element. We print the elements whose counts of occurrences are odd.

Below is the code of the given approach :

C++

// CPP code to find two odd occurring elements
// in an array where all other elements appear
// even number of times.
#include <bits/stdc++.h>
using namespace std;

void printOdds(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 += 1;
            }
        }
        if (count % 2 != 0) {
            cout << arr[i]
                 << " "; // Print the elements that occur
                         // odd number of times in an array
        }
    }
    cout << endl;
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 3, 4, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    // function call
    printOdds(arr, n);
    return 0;
}

// This code is contributed by Suruchi Kumari

C

// C code to find two odd occurring elements
// in an array where all other elements appear
// even number of times.
#include <stdio.h>

void printOdds(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 += 1;
            }
        }
        if (count % 2 != 0) {
            printf(
                "%d ",
                arr[i]); // Print the elements that occur
                         // odd number of times in an array
        }
    }
    printf("\n");
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 3, 4, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    // function call
    printOdds(arr, n);
    return 0;
}

// This code is contributed by Suruchi Kumari

Java

// Java program to find the maximum stolen value
public class GFG {

  // calculate the maximum stolen value
  static void printOdds(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 += 1;
        }
      }
      if (count % 2 != 0) {
        System.out.print(arr[i]+" ");// Print the elements that occur
        // odd number of times in an array
      }
    }
    System.out.print("\n");
  }

  // Driver Code
  public static void main (String[] args)
  {
    int arr[] = { 2, 3, 3, 4, 4, 5 };
    int n = arr.length;

    // function call
    printOdds(arr, n);
  }
}
// This code is contributed by adityapatil12

Python3

# python3 code for the above approach

# Function to find and Replace in String
def printOdds(arr, n) :

    for i in range(0,n) :
        count = 0
        for j in range(0,n) :
            if arr[i] == arr[j] :
                count += 1
        if count % 2 != 0 :
            print(arr[i],end=' ') # Print the elements that occur
                                   # odd number of times in an array

# Driver code
if __name__ == "__main__" :

    arr = [ 2, 3, 3, 4, 4, 5 ]
    n = len(arr)

    #Function call
    printOdds(arr,n)

# This code is contributed by adityapatil12

C#

// Include namespace system
using System;

// C# program to find the maximum stolen value
public class GFG
{
    // calculate the maximum stolen value
    public static void printOdds(int[] arr, int n)
    {
        for (int i = 0; i < n; i++)
        {
            var count = 0;
            for (int j = 0; j < n; j++)
            {
                if (arr[i] == arr[j])
                {
                    count += 1;
                }
            }
            if (count % 2 != 0)
            {
                Console.Write(arr[i].ToString() + " ");
            }
        }
        Console.Write("\n");
    }
    // Driver Code
    public static void Main(String[] args)
    {
        int[] arr = {2, 3, 3, 4, 4, 5};
        var n = arr.Length;
        // function call
        GFG.printOdds(arr, n);
    }
}

// This code is contributed by aadityaburujwale.

Javascript

// JavaScript code to find two odd occurring elements
// in an array where all other elements appear
// even number of times.
function printOdds(arr, n)
{
    for (var i = 0; i < n; i++) {
        let count = 0;
        for (var j = 0; j < n; j++) {
            if (arr[i] == arr[j]) {
                count += 1;
            }
        }
        if (count % 2 != 0) {
            process.stdout.write(arr[i] + " ");
                        // Print the elements that occur
                         // odd number of times in an array
        }
    }
    process.stdout.write("\n");
}

// Driver code
let arr = [ 2, 3, 3, 4, 4, 5 ];
let n = arr.length;

// function call
printOdds(arr, n);


// This code is contributed by phasing17

Output

2 5 

Complexity Analysis:

• Time complexity : O(n^2)
• Auxiliary space : O(n)

A better solution is to use hashing. Time complexity of this solution is O(n) but it requires extra space.

We can construct a frequency hashmap, then iterate over all of its key – value pairs, and print all keys whose values (frequency) are odd.

Implementation:

C++

// CPP code to find two odd occurring elements
// in an array where all other elements appear
// even number of times.
#include <bits/stdc++.h>
using namespace std;


void printOdds(int arr[], int n)
{
    //declaring an unordered map
    unordered_map<int, int> freq;
    //building the frequency table
    for (int i = 0; i < n; i++)
        freq[arr[i]]++;

    //iterating over the map
    for (auto& it: freq) {
        //if the frequency is odd
        //print the element
        if (it.second % 2)
            cout << it.first << " ";
    }
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 3, 4, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    //function call
    printOdds(arr, n);
    return 0;
}

//this code is contributed by phasing17

Java

/*package whatever //do not write package name here */
import java.util.*;

class GFG {
    static void printOdds(int arr[], int n)
    {
        // declaring an unordered map
        TreeMap<Integer, Integer> freq = new TreeMap<>(Collections.reverseOrder());

        // building the frequency table
        for (int i = 0; i < n; i++) {
            freq.put(arr[i],
                     freq.getOrDefault(arr[i], 0) + 1);
        }

        // iterating over the map
        for (int i : freq.keySet()) {
            // if the frequency is odd
            // print the element
            if (freq.get(i) % 2 == 1)
                System.out.print(i + " ");
        }
    }
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, 3, 4, 4, 5 };
        int n = arr.length;

        // function call
        printOdds(arr, n);
    }
}

// This code is contributed by aadityaburujwale.

Python3

# Python code to find two odd occurring elements
# in an array where all other elements appear
# even number of times.
from collections import defaultdict

def printOdds(arr, n):
    freq = defaultdict(int)

    # building the frequency table
    for i in range(n):
        freq[arr[i]] += 1

    # iterating over the map
    for key, value in freq.items():

        # if the frequency is odd
        # print the element
        if value % 2:
            print(key,end=" ")

# Driver code
arr = [2, 3, 3, 4, 4, 5]
n = len(arr)

# function call
printOdds(arr, n)

# this code is contributed by akashish__

C#

// C# code to find two odd occurring elements
// in an array where all other elements appear
// even number of times.
using System;
using System.Collections.Generic;

public class GFG {
  static void printOdds(int[] arr, int n)
  {

    // declaring an unordered map
    Dictionary<int, int> freq
      = new Dictionary<int, int>();

    // building the frequency table
    for (int i = 0; i < n; i++) {
      if (freq.ContainsKey(arr[i])) {
        int val = freq[arr[i]];
        freq.Remove(arr[i]);
        freq.Add(arr[i], val + 1);
      }
      else {
        freq.Add(arr[i], 1);
      }
    }

    // iterating over the map
    foreach(KeyValuePair<int, int> entry in freq)
    {

      // if the frequency is odd
      // print the element
      if (entry.Value % 2 != 0)
        Console.Write(entry.Key + " ");
    }
  }

  // Driver code
  static public void Main()
  {
    int[] arr = { 2, 3, 3, 4, 4, 5 };
    int n = 6;

    // function call
    printOdds(arr, n);
  }
}

// This code is contributed by garg28harsh.

Javascript

// function code to find two odd occurring elements
// in an array where all other elements appear
// even number of times.
function printOdds(arr, n)
{
    // declaring an unordered map
    var mp = new Map();

    // building the frequency table
    for (var i = 0; i < n; i++)
    {
        if(mp.has(arr[i]))
            mp.set(arr[i], mp.get(arr[i])+1);
        else
            mp.set(arr[i], 1);
    }
    let el=[];
    mp.forEach(( key,value) => {
        if(key%2!=0)
        {
            el.push(value);
        }
    });
     console.log(el);
}

// Driver code

    let arr = [2, 3, 3, 4, 4, 5 ];
    let n = arr.length;

    // function call
    printOdds(arr, n);

// This code is contributed by garg28harsh.

Output

5 2 

Complexity Analysis:

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

An efficient solution is to use bitwise operators. The idea is based on approach used in two missing elements and two repeating elements.  

Implementation:

C++

// CPP code to find two odd occurring elements
// in an array where all other elements appear
// even number of times.
#include <bits/stdc++.h>
using namespace std;

void printOdds(int arr[], int n)
{
    // Find XOR of all numbers
    int res = 0;
    for (int i = 0; i < n; i++)
        res = res ^ arr[i];

    // Find a set bit in the XOR (We find
    // rightmost set bit here)
    int set_bit = res & (~(res - 1));

    // Traverse through all numbers and
    // divide them in two groups
    // (i) Having set bit set at same
    //     position as the only set bit
    //     in set_bit
    // (ii) Having 0 bit at same position
    //      as the only set bit in set_bit
    int x = 0, y = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] & set_bit)
            x = x ^ arr[i];
        else
            y = y ^ arr[i];
    }

    // XOR of two different sets are our
    // required numbers.
    cout << x << " " << y;
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 3, 4, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printOdds(arr, n);
    return 0;
}

Java

// Java code to find two
// odd occurring elements
// in an array where all
// other elements appear
// even number of times.

class GFG
{
static void printOdds(int arr[],
                      int n)
{
    // Find XOR of
    // all numbers
    int res = 0;
    for (int i = 0; i < n; i++)
        res = res ^ arr[i];

    // Find a set bit in the
    // XOR (We find rightmost
    // set bit here)
    int set_bit = res &
                  (~(res - 1));

    // Traverse through all
    // numbers and divide them
    // in two groups (i) Having
    // set bit set at same position
    // as the only set bit in
    // set_bit (ii) Having 0 bit at
    // same position as the only
    // set bit in set_bit
    int x = 0, y = 0;
    for (int i = 0; i < n; i++)
    {
        if ((arr[i] & set_bit) != 0)
            x = x ^ arr[i];
        else
            y = y ^ arr[i];
    }

    // XOR of two different
    // sets are our required
    // numbers.
    System.out.println( x + " " + y);
}

// Driver code
public static void main(String [] args)
{
    int arr[] = { 2, 3, 3,
                  4, 4, 5 };
    int n = arr.length;
    printOdds(arr, n);
}
}

// This code is contributed by
// Smitha Dinesh Semwal

Python3

# Python 3 code to find two
# odd occurring elements in
# an array where all other
# elements appear even number
# of times.
def printOdds(arr, n):

    # Find XOR of all numbers
    res = 0
    for i in range(0, n):
        res = res ^ arr[i]

    # Find a set bit in
    # the XOR (We find
    # rightmost set bit here)
    set_bit = res & (~(res - 1))

    # Traverse through all numbers
    # and divide them in two groups
    # (i) Having set bit set at
    # same position as the only set
    # bit in set_bit
    # (ii) Having 0 bit at same
    # position as the only set
    # bit in set_bit
    x = 0
    y = 0
    for i in range(0, n):
        if (arr[i] & set_bit):
            x = x ^ arr[i]
        else:
            y = y ^ arr[i]

    # XOR of two different
    # sets are our
    # required numbers.
    print(x , y, end = "")

# Driver code
arr = [2, 3, 3, 4, 4, 5 ]
n = len(arr)
printOdds(arr, n)

# This code is contributed
# by Smitha

C#

// C# code to find two
// odd occurring elements
// in an array where all
// other elements appear
// even number of times.
using System;

class GFG
{
static void printOdds(int []arr,
                      int n)
{
    // Find XOR of
    // all numbers
    int res = 0;
    for (int i = 0; i < n; i++)
        res = res ^ arr[i];

    // Find a set bit in the
    // XOR (We find rightmost
    // set bit here)
    int set_bit = res &
               (~(res - 1));

    // Traverse through all
    // numbers and divide them
    // in two groups (i) Having
    // set bit set at same position
    // as the only set bit in
    // set_bit (ii) Having 0 bit at
    // same position as the only
    // set bit in set_bit
    int x = 0, y = 0;
    for (int i = 0; i < n; i++)
    {
        if ((arr[i] & set_bit) != 0)
            x = x ^ arr[i];
        else
            y = y ^ arr[i];
    }

    // XOR of two different
    // sets are our required
    // numbers.
    Console.WriteLine(x + " " + y);
}

// Driver code
public static void Main()
{
    int []arr = { 2, 3, 3,
                  4, 4, 5 };
    int n = arr.Length;
    printOdds(arr, n);
}
}

// This code is contributed by
// Akanksha Rai(Abby_akku)

PHP

<?php
// PHP code to find two odd
// occurring elements in an
// array where all other elements
// appear even number of times.
function printOdds($arr, $n)
{
    // Find XOR of all numbers
    $res = 0;
    for ($i = 0; $i < $n; $i++)
        $res = $res ^ $arr[$i];

    // Find a set bit in the
    // XOR (We find rightmost
    // set bit here)
    $set_bit = $res & (~($res - 1));

    // Traverse through all numbers
    // and divide them in two groups
    // (i) Having set bit set at same
    // position as the only set bit
    // in set_bit
    // (ii) Having 0 bit at same position
    // as the only set bit in set_bit
    $x = 0;
    $y = 0;
    for ($i = 0; $i < $n; $i++)
    {
        if ($arr[$i] & $set_bit)
            $x = $x ^ $arr[$i];
        else
            $y = $y ^ $arr[$i];
    }

    // XOR of two different sets
    // are our required numbers.
    echo($x . " " . $y);
}

// Driver code
$arr = array( 2, 3, 3, 4, 4, 5 );
$n = sizeof($arr);
printOdds($arr, $n);

// This code is contributed by Smitha
?>

Javascript

<script>

// Javascript code to find two odd
// occurring elements in an array
// where all other elements appear
// even number of times.
function printOdds(arr, n)
{

    // Find XOR of all numbers
    let res = 0;
    for(let i = 0; i < n; i++)
        res = res ^ arr[i];

    // Find a set bit in the XOR (We find
    // rightmost set bit here)
    let set_bit = res & (~(res - 1));

    // Traverse through all numbers and
    // divide them in two groups
    // (i) Having set bit set at same
    //     position as the only set bit
    //     in set_bit
    // (ii) Having 0 bit at same position
    //      as the only set bit in set_bit
    let x = 0, y = 0;
    for(let i = 0; i < n; i++)
    {
        if (arr[i] & set_bit)
            x = x ^ arr[i];
        else
            y = y ^ arr[i];
    }

    // XOR of two different sets are our
    // required numbers.
    document.write(x + " " + y);
}

// Driver code
let arr = [ 2, 3, 3, 4, 4, 5 ];
let n = arr.length;

printOdds(arr, n);

// This code is contributed by subhammahato348

</script>

Output

5 2

Complexity Analysis:

• Time Complexity : O(n) 
• Auxiliary Space : O(1)

Another Efficient Solution (Using binary search) : Sort the array for binary search . Then we can find frequency of all array elements using binary search function . Then we can check if frequency of array element is odd or not , If frequency is odd , then print that element .

Below is the implementation of above approach:

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find element that occurs
// odd times in the array
void printOdds(int arr[], int n)
{
   int count = 0;
   sort(arr,arr+n);//sort array for binary search

   for(int i = 0 ; i < n ;i++)
   {
     //index of first and last occ of arr[i]
     int first_index = lower_bound(arr,arr+n,arr[i])- arr;
     int last_index = upper_bound(arr,arr+n,arr[i])- arr-1;
     i = last_index; // assign i to last_index to avoid printing
                     // same element multiple time

     int fre = last_index-first_index+1;//finding frequency
     //( occurences of arr[i] in the array )

     if(fre % 2 != 0)
     { // if element occurs odd times then print that number
          cout << arr[i]<<" ";
     }
   }

}

// Drive code
int main()
{
    int arr[] = { 2, 3, 3, 4, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function call
    printOdds(arr, n);
    return 0;
}

// This Code is contributed by nikhilsainiofficial546

Python3

# Function to find element that occurs
# odd times in the array
def printOdds(arr, n):
    count = 0
    arr.sort()  # sort array for binary search
    i = 0
    while i < n:
        # index of first and last occ of arr[i]
        first_index = arr.index(arr[i])
        last_index = len(arr) - arr[::-1].index(arr[i]) - 1
        i = last_index  # assign i to last_index to avoid printing
        # same element multiple time

        fre = last_index - first_index + 1  # finding frequency
        # (occurrences of arr[i] in the array)

        if fre % 2 != 0:
            # if element occurs odd times then print that number
            print(arr[i], end=' ')
        i += 1


# Drive code
arr = [2, 3, 3, 4, 4, 5]
n = len(arr)

# Function call
printOdds(arr, n)

Javascript

// Function to find element that occurs odd times in the array
function printOdds(arr, n) {
    let count = 0;
    arr.sort(); // sort array for binary search
    let i = 0; let ans="";
    while (i < n) {

        // index of first and last occ of arr[i]
        let first_index = arr.indexOf(arr[i]);
        let last_index = arr.lastIndexOf(arr[i]);
        i = last_index; // assign i to last_index to avoid printing same element multiple times

        let fre = last_index - first_index + 1; // finding frequency (occurrences of arr[i] in the array)

        if (fre % 2 !== 0) {

            // if element occurs odd times then print that number
            ans = ans + arr[i] + " ";
        }
        i += 1;
    }console.log(ans);
}

// Drive code
let arr = [2, 3, 3, 4, 4, 5];
let n = arr.length;

// Function call
printOdds(arr, n);

C#

// C# Equivalent
using System;

public class GFG {

    // Function to find element that occurs odd times in the
    // array
    static void printOdds(int[] arr, int n)
    {
        int count = 0;
        Array.Sort(arr); // sort array for binary search
        int i = 0;
        string ans = "";
        while (i < n) {

            // index of first and last occ of arr[i]
            int first_index = Array.IndexOf(arr, arr[i]);
            int last_index = Array.LastIndexOf(arr, arr[i]);
            i = last_index; // assign i to last_index to
                            // avoid printing same element
                            // multiple times

            int fre = last_index - first_index
                      + 1; // finding frequency (occurrences
                           // of arr[i] in the array)

            if (fre % 2 != 0) {

                // if element occurs odd times then print
                // that number
                ans = ans + arr[i] + " ";
            }
            i += 1;
        }
        Console.WriteLine(ans);
    }

    public static void Main()
    {
        // Driver's code
        int[] arr = new int[] { 2, 3, 3, 4, 4, 5 };
        int n = arr.Length;

        // Function call
        printOdds(arr, n);
    }
}

Java

import java.util.*;

class Main {
    // Function to find element that occurs
    // odd times in the array
    static void printOdds(int[] arr, int n)
    {
        int count = 0;
        Arrays.sort(arr); // sort array for binary search

        for (int i = 0; i < n; i++) {
            // index of first and last occ of arr[i]
            int first_index
                = Arrays.binarySearch(arr, arr[i]);
            int last_index = first_index;

            // handling case where element is not found by
            // binary search
            if (first_index < 0) {
                continue;
            }

            // loop to find last index of arr[i] in the
            // array
            while (last_index + 1 < n
                   && arr[last_index + 1] == arr[i]) {
                last_index++;
            }
            i = last_index; // assign i to last_index to
                            // avoid printing same element
                            // multiple times

            int fre = last_index - first_index
                      + 1; // finding frequency
                           // (occurrences of arr[i] in the
                           // array)

            if (fre % 2 != 0) {
                // if element occurs odd times then print
                // that number
                System.out.print(arr[i] + " ");
            }
        }
    }

    // Drive code
    public static void main(String[] args)
    {
        int[] arr = { 2, 3, 3, 4, 4, 5 };
        int n = arr.length;

        // Function call
        printOdds(arr, n); // Output: 2 5
    }
}

Output

2 5 

Time Complexity: O(n*log₂n), take log₂n time for binary search function 
Auxiliary Space: O(1)