Find missing number in another array which is shuffled copy

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Given an array 'arr1' of n positive integers. Contents of arr1[] are copied to another array 'arr2', but numbers are shuffled and one element is removed. Find the missing element(without using any extra space and in O(n) time complexity).

Examples : 

Input : arr1[] = {4, 8, 1, 3, 7}, 
        arr2[] = {7, 4, 3, 1}
Output : 8

Input : arr1[] = {12, 10, 15, 23, 11, 30}, 
        arr2[] = {15, 12, 23, 11, 30}
Output : 10

A simple solution is to one by one consider every element of first array and search in second array. As soon as we find a missing element, we return. Time complexity of this solution is O(n²)

An efficient solution is based on XOR. The combined occurrence of each element is twice, one in 'arr1' and other in 'arr2', except one element which only has a single occurrence in 'arr1'. We know that (a Xor a) = 0. So, simply XOR the elements of both the arrays. The result will be the missing number. 

Implementation:

// C++ implementation to find the
// missing number in shuffled array
// C++ implementation to find the
// missing number in shuffled array
#include <bits/stdc++.h>
using namespace std;

// Returns the missing number
// Size of arr2[] is n-1
int missingNumber(int arr1[], int arr2[], 
                                   int n)
{
    // Missing number 'mnum'
    int mnum = 0;

    // 1st array is of size 'n'
    for (int i = 0; i < n; i++)
        mnum = mnum ^ arr1[i];

    // 2nd array is of size 'n - 1'
    for (int i = 0; i < n - 1; i++)
        mnum = mnum ^ arr2[i];

    // Required missing number
    return mnum;
}

// Driver Code
int main()
{
    int arr1[] = {4, 8, 1, 3, 7};
    int arr2[] = {7, 4, 3, 1};
    int n = sizeof(arr1) / sizeof(arr1[0]);
    cout << "Missing number = "
        << missingNumber(arr1, arr2, n);
    return 0;
}

// Java implementation to find the
// missing number in shuffled array

class GFG 
{
    // Returns the missing number
    // Size of arr2[] is n-1
    static int missingNumber(int arr1[], 
                             int arr2[],
                                  int n)
    {
        // Missing number 'mnum'
        int mnum = 0;
    
        // 1st array is of size 'n'
        for (int i = 0; i < n; i++)
            mnum = mnum ^ arr1[i];
    
        // 2nd array is of size 'n - 1'
        for (int i = 0; i < n - 1; i++)
            mnum = mnum ^ arr2[i];
    
        // Required missing number
        return mnum;
    }
    
    // Driver Code
    public static void main (String[] args) 
    {
        int arr1[] = {4, 8, 1, 3, 7};
        int arr2[] = {7, 4, 3, 1};
        int n = arr1.length;
        
        System.out.println("Missing number = "
            + missingNumber(arr1, arr2, n));
    }
}

# Python3 implementation to find the
# missing number in shuffled array

# Returns the missing number
# Size of arr2[] is n - 1
def missingNumber(arr1, arr2, n):

    # missing number 'mnum'
    mnum = 0

    # 1st array is of size 'n'
    for i in range(n):
        mnum = mnum ^ arr1[i]

    # 2nd array is of size 'n - 1'
    for i in range(n - 1):
        mnum = mnum ^ arr2[i]

    # Required missing number
    return mnum

# Driver Code
arr1 = [4, 8, 1, 3, 7]
arr2= [7, 4, 3, 1]
n = len(arr1)
print("Missing number = ", 
    missingNumber(arr1, arr2, n))
        
# This code is contributed by Anant Agarwal.

 // C# implementation to find the
// missing number in shuffled array
using System;

class GFG
{
    // Returns the missing number
    // Size of arr2[] is n-1
    static int missingNumber(int []arr1, 
                             int []arr2, 
                                  int n)
    {
        // Missing number 'mnum'
        int mnum = 0;
    
        // 1st array is of size 'n'
        for (int i = 0; i < n; i++)
            mnum = mnum ^ arr1[i];
    
        // 2nd array is of size 'n - 1'
        for (int i = 0; i < n - 1; i++)
            mnum = mnum ^ arr2[i];
    
        // Required missing number
        return mnum;
    }
    
    // Driver Code
    public static void Main () 
    {
        int []arr1 = {4, 8, 1, 3, 7};
        int []arr2 = {7, 4, 3, 1};
        int n = arr1.Length;
        
    Console.Write("Missing number = "
            + missingNumber(arr1, arr2, n));
    }
}

// This code is contributed by nitin mittal.

<?php
// PHP implementation to find the
// missing number in shuffled array
// PHP implementation to find the
// missing number in shuffled array

// Returns the missing number
// Size of arr2[] is n-1
function missingNumber($arr1, $arr2, 
                                $n)
{
    
    // Missing number 'mnum'
    $mnum = 0;

    // 1st array is of size 'n'
    for ($i = 0; $i < $n; $i++)
        $mnum = $mnum ^ $arr1[$i];

    // 2nd array is of size 'n - 1'
    for ($i = 0; $i < $n - 1; $i++)
        $mnum = $mnum ^ $arr2[$i];

    // Required missing number
    return $mnum;
}

    // Driver Code
    $arr1 = array(4, 8, 1, 3, 7);
    $arr2 = array(7, 4, 3, 1);
    $n = count($arr1);
    echo "Missing number = "
        , missingNumber($arr1, $arr2, $n);
        
// This code is contributed by anuj_67. 
?>

<script>

// Javascript implementation to find the
// missing number in shuffled array
// Javascript implementation to find the
// missing number in shuffled array

// Returns the missing number
// Size of arr2[] is n-1
function missingNumber(arr1, arr2, n)
{
    // Missing number 'mnum'
    let mnum = 0;

    // 1st array is of size 'n'
    for (let i = 0; i < n; i++)
        mnum = mnum ^ arr1[i];

    // 2nd array is of size 'n - 1'
    for (let i = 0; i < n - 1; i++)
        mnum = mnum ^ arr2[i];

    // Required missing number
    return mnum;
}

// Driver Code
    let arr1 = [4, 8, 1, 3, 7];
    let arr2 = [7, 4, 3, 1];
    let n = arr1.length;
    document.write("Missing number = "
        + missingNumber(arr1, arr2, n));

</script>

Missing number = 8

Time Complexity:  O(n).
Space Complexity: O(1).