Find lost element from a duplicated array
Given two arrays that are duplicates of each other except one element, that is one element from one of the array is missing, we need to find that missing element.
Examples:
Input: arr1[] = {1, 4, 5, 7, 9}
arr2[] = {4, 5, 7, 9}
Output: 1
1 is missing from second array.
Input: arr1[] = {2, 3, 4, 5}
arr2[] = {2, 3, 4, 5, 6}
Output: 6
6 is missing from first array.
One simple solution is to iterate over arrays and check element by element and flag the missing element when an unmatched element is found, but this solution requires linear time oversize of the array.
Approach:
We will iterate over each element of the first array and check if it exists in the second array.
If it doesn’t exist, that means it’s the missing element, and we’ll return it.
If we don’t find any missing element, we’ll return -1.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int find_Missing( int arr1[], int arr2[], int n1, int n2) {
int i, j;
bool found;
for (i = 0; i < n1; i++) {
found = false ;
for (j = 0; j < n2; j++) {
if (arr1[i] == arr2[j]) {
found = true ;
break ;
}
}
if (!found) {
return arr1[i];
}
}
return -1;
}
int main() {
int arr1[] = {1, 4, 5, 7, 9};
int arr2[] = {4, 5, 7, 9};
int n1 = sizeof (arr1) / sizeof (arr1[0]);
int n2 = sizeof (arr2) / sizeof (arr2[0]);
int missing = find_Missing(arr1, arr2, n1, n2);
if (missing == -1) {
cout << "No missing element" << endl;
}
else {
cout<< missing << endl;
}
return 0;
}
|
Java
import java.util.Arrays;
public class GFG {
static int findMissing( int [] arr1, int [] arr2, int n1, int n2) {
int i, j;
boolean found;
for (i = 0 ; i < n1; i++) {
found = false ;
for (j = 0 ; j < n2; j++) {
if (arr1[i] == arr2[j]) {
found = true ;
break ;
}
}
if (!found) {
return arr1[i];
}
}
return - 1 ;
}
public static void main(String[] args) {
int [] arr1 = { 1 , 4 , 5 , 7 , 9 };
int [] arr2 = { 4 , 5 , 7 , 9 };
int n1 = arr1.length;
int n2 = arr2.length;
int missing = findMissing(arr1, arr2, n1, n2);
if (missing == - 1 ) {
System.out.println( "No missing element" );
} else {
System.out.println(missing);
}
}
}
|
Python3
def find_Missing(arr1, arr2):
n1 = len (arr1)
n2 = len (arr2)
for i in range (n1):
found = False
for j in range (n2):
if arr1[i] = = arr2[j]:
found = True
break
if not found:
return arr1[i]
return - 1
arr1 = [ 1 , 4 , 5 , 7 , 9 ]
arr2 = [ 4 , 5 , 7 , 9 ]
missing = find_Missing(arr1, arr2)
if missing = = - 1 :
print ( "No missing element" )
else :
print (missing)
|
C#
using System;
class Program
{
static int FindMissing( int [] arr1, int [] arr2)
{
bool found;
for ( int i = 0; i < arr1.Length; i++)
{
found = false ;
for ( int j = 0; j < arr2.Length; j++)
{
if (arr1[i] == arr2[j])
{
found = true ;
break ;
}
}
if (!found)
{
return arr1[i];
}
}
return -1;
}
static void Main()
{
int [] arr1 = { 1, 4, 5, 7, 9 };
int [] arr2 = { 4, 5, 7, 9 };
int n1 = arr1.Length;
int n2 = arr2.Length;
int missing = FindMissing(arr1, arr2);
if (missing == -1)
{
Console.WriteLine( "No missing element" );
}
else
{
Console.WriteLine(missing);
}
}
}
|
Javascript
function find_Missing(arr1, arr2, n1, n2) {
for (let i = 0; i < n1; i++) {
let found = false ;
for (let j = 0; j < n2; j++) {
if (arr1[i] === arr2[j]) {
found = true ;
break ;
}
}
if (!found) {
return arr1[i];
}
}
return -1;
}
const arr1 = [1, 4, 5, 7, 9];
const arr2 = [4, 5, 7, 9];
const n1 = arr1.length;
const n2 = arr2.length;
const missing = find_Missing(arr1, arr2, n1, n2);
if (missing === -1) {
console.log( "No missing element" );
} else {
console.log(missing);
}
|
Time Complexity: O(M * N), where M and N represents the size of the given two arrays.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Another efficient solution is based on a binary search approach. Algorithm steps are as follows:
- Start a binary search in a bigger array and get mid as (lo + hi) / 2
- If the value from both arrays is the same then the missing element must be in the right part so set lo as mid
- Else set hi as mid because the missing element must be in the left part of the bigger array if mid-elements are not equal.
- A special case is handled separately as for single element and zero elements array, the single element itself will be the missing element.
If the first element itself is not equal then that element will be the missing element./li>
Below is the implementation of the above steps
C++
#include <bits/stdc++.h>
using namespace std;
int findMissingUtil( int arr1[], int arr2[], int N)
{
if (N == 1)
return arr1[0];
if (arr1[0] != arr2[0])
return arr1[0];
int lo = 0, hi = N - 1;
while (lo < hi)
{
int mid = (lo + hi) / 2;
if (arr1[mid] == arr2[mid])
lo = mid;
else
hi = mid;
if (lo == hi - 1)
break ;
}
return arr1[hi];
}
void findMissing( int arr1[], int arr2[], int M, int N)
{
if (N == M-1)
cout << "Missing Element is "
<< findMissingUtil(arr1, arr2, M) << endl;
else if (M == N-1)
cout << "Missing Element is "
<< findMissingUtil(arr2, arr1, N) << endl;
else
cout << "Invalid Input" ;
}
int main()
{
int arr1[] = {1, 4, 5, 7, 9};
int arr2[] = {4, 5, 7, 9};
int M = sizeof (arr1) / sizeof ( int );
int N = sizeof (arr2) / sizeof ( int );
findMissing(arr1, arr2, M, N);
return 0;
}
|
Java
import java.io.*;
class MissingNumber {
int findMissingUtil( int arr1[], int arr2[],
int N)
{
if (N == 1 )
return arr1[ 0 ];
if (arr1[ 0 ] != arr2[ 0 ])
return arr1[ 0 ];
int lo = 0 , hi = N - 1 ;
while (lo < hi) {
int mid = (lo + hi) / 2 ;
if (arr1[mid] == arr2[mid])
lo = mid;
else
hi = mid;
if (lo == hi - 1 )
break ;
}
return arr1[hi];
}
void findMissing( int arr1[], int arr2[],
int M, int N)
{
if (N == M - 1 )
System.out.println( "Missing Element is "
+ findMissingUtil(arr1, arr2, M) + "\n" );
else if (M == N - 1 )
System.out.println( "Missing Element is "
+ findMissingUtil(arr2, arr1, N) + "\n" );
else
System.out.println( "Invalid Input" );
}
public static void main(String args[])
{
MissingNumber obj = new MissingNumber();
int arr1[] = { 1 , 4 , 5 , 7 , 9 };
int arr2[] = { 4 , 5 , 7 , 9 };
int M = arr1.length;
int N = arr2.length;
obj.findMissing(arr1, arr2, M, N);
}
}
|
Python3
def findMissingUtil(arr1, arr2, N):
if N = = 1 :
return arr1[ 0 ];
if arr1[ 0 ] ! = arr2[ 0 ]:
return arr1[ 0 ]
lo = 0
hi = N - 1
while (lo < hi):
mid = (lo + hi) / 2
if arr1[mid] = = arr2[mid]:
lo = mid
else :
hi = mid
if lo = = hi - 1 :
break
return arr1[hi]
def findMissing(arr1, arr2, M, N):
if N = = M - 1 :
print ( "Missing Element is" ,
findMissingUtil(arr1, arr2, M))
elif M = = N - 1 :
print ( "Missing Element is" ,
findMissingUtil(arr2, arr1, N))
else :
print ( "Invalid Input" )
arr1 = [ 1 , 4 , 5 , 7 , 9 ]
arr2 = [ 4 , 5 , 7 , 9 ]
M = len (arr1)
N = len (arr2)
findMissing(arr1, arr2, M, N)
|
C#
using System;
class GFG {
static int findMissingUtil( int []arr1,
int []arr2, int N)
{
if (N == 1)
return arr1[0];
if (arr1[0] != arr2[0])
return arr1[0];
int lo = 0, hi = N - 1;
while (lo < hi) {
int mid = (lo + hi) / 2;
if (arr1[mid] == arr2[mid])
lo = mid;
else
hi = mid;
if (lo == hi - 1)
break ;
}
return arr1[hi];
}
static void findMissing( int []arr1, int []arr2,
int M, int N)
{
if (N == M - 1)
Console.WriteLine( "Missing Element is "
+ findMissingUtil(arr1, arr2, M) + "\n" );
else if (M == N - 1)
Console.WriteLine( "Missing Element is "
+ findMissingUtil(arr2, arr1, N) + "\n" );
else
Console.WriteLine( "Invalid Input" );
}
public static void Main()
{
int []arr1 = { 1, 4, 5, 7, 9 };
int []arr2 = { 4, 5, 7, 9 };
int M = arr1.Length;
int N = arr2.Length;
findMissing(arr1, arr2, M, N);
}
}
|
Javascript
<script>
function findMissingUtil(arr1, arr2, N)
{
if (N == 1)
return arr1[0];
if (arr1[0] != arr2[0])
return arr1[0];
let lo = 0, hi = N - 1;
while (lo < hi) {
let mid = parseInt((lo + hi) / 2, 10);
if (arr1[mid] == arr2[mid])
lo = mid;
else
hi = mid;
if (lo == hi - 1)
break ;
}
return arr1[hi];
}
function findMissing(arr1, arr2, M, N)
{
if (N == M - 1)
document.write( "Missing Element is "
+ findMissingUtil(arr1, arr2, M) + "</br>" );
else if (M == N - 1)
document.write( "Missing Element is "
+ findMissingUtil(arr2, arr1, N) + "</br>" );
else
document.write( "Invalid Input" + "</br>" );
}
let arr1 = [ 1, 4, 5, 7, 9 ];
let arr2 = [ 4, 5, 7, 9 ];
let M = arr1.length;
let N = arr2.length;
findMissing(arr1, arr2, M, N);
</script>
|
PHP
<?php
function findMissingUtil( $arr1 , $arr2 , $N )
{
if ( $N == 1)
return $arr1 [0];
if ( $arr1 [0] != $arr2 [0])
return $arr1 [0];
$lo = 0;
$hi = $N - 1;
while ( $lo < $hi )
{
$mid = ( $lo + $hi ) / 2;
if ( $arr1 [ $mid ] == $arr2 [ $mid ])
$lo = $mid ;
else
$hi = $mid ;
if ( $lo == $hi - 1)
break ;
}
return $arr1 [ $hi ];
}
function findMissing( $arr1 , $arr2 ,
$M , $N )
{
if ( $N == $M - 1)
echo "Missing Element is "
, findMissingUtil( $arr1 ,
$arr2 , $M ) ;
else if ( $M == $N - 1)
echo "Missing Element is "
, findMissingUtil( $arr2 ,
$arr1 , $N );
else
echo "Invalid Input" ;
}
$arr1 = array (1, 4, 5, 7, 9);
$arr2 = array (4, 5, 7, 9);
$M = count ( $arr1 );
$N = count ( $arr2 );
findMissing( $arr1 , $arr2 , $M , $N );
?>
|
Output
Missing Element is 1
Time Complexity: O(logM + logN), where M and N represents the size of the given two arrays.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
What if input arrays are not in the same order?
In this case, the missing element is simply XOR of all elements of both arrays. Thanks to Yolo Song for suggesting this.
CPP
#include <bits/stdc++.h>
using namespace std;
void findMissing( int arr1[], int arr2[], int M,
int N)
{
if (M != N-1 && N != M-1)
{
cout << "Invalid Input" ;
return ;
}
int res = 0;
for ( int i=0; i<M; i++)
res = res^arr1[i];
for ( int i=0; i<N; i++)
res = res^arr2[i];
cout << "Missing element is " << res;
}
int main()
{
int arr1[] = {4, 1, 5, 9, 7};
int arr2[] = {7, 5, 9, 4};
int M = sizeof (arr1) / sizeof ( int );
int N = sizeof (arr2) / sizeof ( int );
findMissing(arr1, arr2, M, N);
return 0;
}
|
Java
import java.io.*;
class Missing {
void findMissing( int arr1[], int arr2[],
int M, int N)
{
if (M != N - 1 && N != M - 1 ) {
System.out.println( "Invalid Input" );
return ;
}
int res = 0 ;
for ( int i = 0 ; i < M; i++)
res = res ^ arr1[i];
for ( int i = 0 ; i < N; i++)
res = res ^ arr2[i];
System.out.println( "Missing element is "
+ res);
}
public static void main(String args[])
{
Missing obj = new Missing();
int arr1[] = { 4 , 1 , 5 , 9 , 7 };
int arr2[] = { 7 , 5 , 9 , 4 };
int M = arr1.length;
int N = arr2.length;
obj.findMissing(arr1, arr2, M, N);
}
}
|
Python3
def findMissing(arr1,arr2, M, N):
if (M ! = N - 1 and N ! = M - 1 ):
print ( "Invalid Input" )
return
res = 0
for i in range ( 0 ,M):
res = res^arr1[i];
for i in range ( 0 ,N):
res = res^arr2[i]
print ( "Missing element is" ,res)
arr1 = [ 4 , 1 , 5 , 9 , 7 ]
arr2 = [ 7 , 5 , 9 , 4 ]
M = len (arr1)
N = len (arr2)
findMissing(arr1, arr2, M, N)
|
C#
using System;
class GFG {
static void findMissing( int []arr1,
int []arr2,
int M, int N)
{
if (M != N - 1 && N != M - 1)
{
Console.WriteLine( "Invalid Input" );
return ;
}
int res = 0;
for ( int i = 0; i < M; i++)
res = res ^ arr1[i];
for ( int i = 0; i < N; i++)
res = res ^ arr2[i];
Console.WriteLine( "Missing element is "
+ res);
}
public static void Main()
{
int []arr1 = {4, 1, 5, 9, 7};
int []arr2 = {7, 5, 9, 4};
int M = arr1.Length;
int N = arr2.Length;
findMissing(arr1, arr2, M, N);
}
}
|
Javascript
<script>
function findMissing(arr1, arr2, M, N)
{
if (M != N-1 && N != M-1)
{
document.write( "Invalid Input" );
return ;
}
let res = 0;
for (let i=0; i<M; i++)
res = res^arr1[i];
for (let i=0; i<N; i++)
res = res^arr2[i];
document.write( "Missing element is " + res);
}
let arr1 = [4, 1, 5, 9, 7];
let arr2 = [7, 5, 9, 4];
let M = arr1.length;
let N = arr2.length;
findMissing(arr1, arr2, M, N);
</script>
|
PHP
<?php
function findMissing( $arr1 , $arr2 ,
$M , $N )
{
if ( $M != $N - 1 && $N != $M - 1)
{
echo "Invalid Input" ;
return ;
}
$res = 0;
for ( $i = 0; $i < $M ; $i ++)
$res = $res ^ $arr1 [ $i ];
for ( $i = 0; $i < $N ; $i ++)
$res = $res ^ $arr2 [ $i ];
echo "Missing element is " , $res ;
}
$arr1 = array (4, 1, 5, 9, 7);
$arr2 = array (7, 5, 9, 4);
$M = sizeof( $arr1 );
$N = sizeof( $arr2 );
findMissing( $arr1 , $arr2 , $M , $N );
?>
|
Output
Missing element is 1
Time Complexity: O(M + N), where M and N represents the size of the given two arrays.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Last Updated :
16 Oct, 2023
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