Find missing number in another array which is shuffled copy
Last Updated :
11 Jul, 2022
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++
#include <bits/stdc++.h>
using namespace std;
int missingNumber( int arr1[], int arr2[],
int n)
{
int mnum = 0;
for ( int i = 0; i < n; i++)
mnum = mnum ^ arr1[i];
for ( int i = 0; i < n - 1; i++)
mnum = mnum ^ arr2[i];
return mnum;
}
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
class GFG
{
static int missingNumber( int arr1[],
int arr2[],
int n)
{
int mnum = 0 ;
for ( int i = 0 ; i < n; i++)
mnum = mnum ^ arr1[i];
for ( int i = 0 ; i < n - 1 ; i++)
mnum = mnum ^ arr2[i];
return mnum;
}
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
def missingNumber(arr1, arr2, n):
mnum = 0
for i in range (n):
mnum = mnum ^ arr1[i]
for i in range (n - 1 ):
mnum = mnum ^ arr2[i]
return mnum
arr1 = [ 4 , 8 , 1 , 3 , 7 ]
arr2 = [ 7 , 4 , 3 , 1 ]
n = len (arr1)
print ( "Missing number = " ,
missingNumber(arr1, arr2, n))
|
C#
using System;
class GFG
{
static int missingNumber( int []arr1,
int []arr2,
int n)
{
int mnum = 0;
for ( int i = 0; i < n; i++)
mnum = mnum ^ arr1[i];
for ( int i = 0; i < n - 1; i++)
mnum = mnum ^ arr2[i];
return mnum;
}
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));
}
}
|
PHP
<?php
function missingNumber( $arr1 , $arr2 ,
$n )
{
$mnum = 0;
for ( $i = 0; $i < $n ; $i ++)
$mnum = $mnum ^ $arr1 [ $i ];
for ( $i = 0; $i < $n - 1; $i ++)
$mnum = $mnum ^ $arr2 [ $i ];
return $mnum ;
}
$arr1 = array (4, 8, 1, 3, 7);
$arr2 = array (7, 4, 3, 1);
$n = count ( $arr1 );
echo "Missing number = "
, missingNumber( $arr1 , $arr2 , $n );
?>
|
Javascript
<script>
function missingNumber(arr1, arr2, n)
{
let mnum = 0;
for (let i = 0; i < n; i++)
mnum = mnum ^ arr1[i];
for (let i = 0; i < n - 1; i++)
mnum = mnum ^ arr2[i];
return mnum;
}
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>
|
Output
Missing number = 8
Time Complexity: O(n).
Space Complexity: O(1).
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...