Find lost element from a duplicated array

Given two arrays which 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.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

One simple solution is to iterate over arrays and check element by element and flag the missing element when an unmatch is found, but this solution requires linear time over size of array.

Another efficient solution is based on binary search approach. Algorithm steps are as follows:

Start binary search in bigger array and get mid as (lo + hi) / 2
If value from both array is same then missing element must be in right part so set lo as mid
Else set hi as mid because missing element must be in left part of bigger array if mid elements are not equal.
Special case are handled separately as for single element and zero element array, single element itself will be the missing element.
If first element itself is not equal then that element will be the missing element./li>

Below is the implementation of above steps

C++

// C++ program to find missing element from same 
// arrays (except one missing element) 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find missing element based on binary 
// search approach.  arr1[] is of larger size and 
// N is size of it.  arr1[] and arr2[] are assumed 
// to be in same order. 
int findMissingUtil(int arr1[], int arr2[], int N) 
{ 
    // special case, for only element which is 
    // missing in second array 
    if (N == 1) 
        return arr1[0]; 
  
    // special case, for first element missing 
    if (arr1[0] != arr2[0]) 
        return arr1[0]; 
  
    // Initialize current corner points 
    int lo = 0,  hi = N - 1; 
  
    // loop until lo < hi 
    while (lo < hi) 
    { 
        int mid = (lo + hi) / 2; 
  
        // If element at mid indices are equal 
        // then go to right subarray 
        if (arr1[mid] == arr2[mid]) 
            lo = mid; 
        else
            hi = mid; 
  
        // if lo, hi becomes contiguous,  break 
        if (lo == hi - 1) 
            break; 
    } 
  
    // missing element will be at hi index of 
    // bigger array 
    return arr1[hi]; 
} 
  
// This function mainly does basic error checking 
// and calls findMissingUtil 
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"; 
} 
  
// Driver Code 
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

// Java program to find missing element 
// from same arrays 
// (except one missing element) 
  
import java.io.*; 
class MissingNumber { 
  
    /* Function to find missing element based 
     on binary search approach. arr1[] is of 
     larger size and N is size of it.arr1[] and  
     arr2[] are assumed to be in same order. */
    int findMissingUtil(int arr1[], int arr2[], 
                        int N) 
    { 
        // special case, for only element 
        // which is missing in second array 
        if (N == 1) 
            return arr1[0]; 
  
        // special case, for first 
        // element missing 
        if (arr1[0] != arr2[0]) 
            return arr1[0]; 
  
        // Initialize current corner points 
        int lo = 0, hi = N - 1; 
  
        // loop until lo < hi 
        while (lo < hi) { 
            int mid = (lo + hi) / 2; 
  
            // If element at mid indices are 
            // equal then go to right subarray 
            if (arr1[mid] == arr2[mid]) 
                lo = mid; 
            else
                hi = mid; 
  
            // if lo, hi becomes  
            // contiguous, break 
            if (lo == hi - 1) 
                break; 
        } 
  
        // missing element will be at hi  
        // index of bigger array 
        return arr1[hi]; 
    } 
  
    // This function mainly does basic error 
    // checking and calls findMissingUtil 
    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"); 
    } 
  
    // Driver Code 
    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); 
    } 
} 
  
// This code is contributed by Anshika Goyal. 

Python3

# Python3 program to find missing 
# element from same arrays  
# (except one missing element) 
  
# Function to find missing element based 
# on binary search approach. arr1[] is  
# of larger size and N is size of it.  
# arr1[] and arr2[] are assumed 
# to be in same order. 
def findMissingUtil(arr1, arr2, N): 
  
    # special case, for only element  
    # which is missing in second array 
    if N == 1: 
        return arr1[0]; 
  
    # special case, for first 
    # element missing 
    if arr1[0] != arr2[0]: 
        return arr1[0] 
  
    # Initialize current corner points 
    lo = 0
    hi = N - 1
      
    # loop until lo < hi 
    while (lo < hi): 
      
        mid = (lo + hi) / 2
  
        # If element at mid indices 
        # are equal then go to  
        # right subarray 
        if arr1[mid] == arr2[mid]: 
            lo = mid 
        else: 
            hi = mid 
  
        # if lo, hi becomes  
        # contiguous, break 
        if lo == hi - 1: 
            break
      
    # missing element will be at 
    # hi index of bigger array 
    return arr1[hi] 
  
# This function mainly does basic 
# error checking and calls  
# findMissingUtil 
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") 
  
# Driver Code 
arr1 = [1, 4, 5, 7, 9] 
arr2 = [4, 5, 7, 9] 
M = len(arr1) 
N = len(arr2) 
findMissing(arr1, arr2, M, N) 
  
# This code is contributed by Smitha Dinesh Semwal 

C#

// C# program to find missing element from  
// same arrays (except one missing element) 
using System; 
  
class GFG { 
  
    /* Function to find missing element based 
    on binary search approach. arr1[] is of 
    larger size and N is size of it.arr1[] and  
    arr2[] are assumed to be in same order. */
    static int findMissingUtil(int []arr1,  
                            int []arr2, int N) 
    { 
          
        // special case, for only element 
        // which is missing in second array 
        if (N == 1) 
            return arr1[0]; 
  
        // special case, for first 
        // element missing 
        if (arr1[0] != arr2[0]) 
            return arr1[0]; 
  
        // Initialize current corner points 
        int lo = 0, hi = N - 1; 
  
        // loop until lo < hi 
        while (lo < hi) { 
            int mid = (lo + hi) / 2; 
  
            // If element at mid indices are 
            // equal then go to right subarray 
            if (arr1[mid] == arr2[mid]) 
                lo = mid; 
            else
                hi = mid; 
  
            // if lo, hi becomes  
            // contiguous, break 
            if (lo == hi - 1) 
                break; 
        } 
  
        // missing element will be at hi  
        // index of bigger array 
        return arr1[hi]; 
    } 
  
    // This function mainly does basic error 
    // checking and calls findMissingUtil 
    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"); 
    } 
  
    // Driver Code 
    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); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP program to find missing  
// element from same arrays  
// (except one missing element) 
  
// Function to find missing  
// element based on binary 
// search approach. arr1[]  
// is of larger size and 
// N is size of it. arr1[] 
// and arr2[] are assumed 
// to be in same order. 
function findMissingUtil($arr1, $arr2, $N) 
{ 
      
    // special case, for only  
    // element which is 
    // missing in second array 
    if ($N == 1) 
        return $arr1[0]; 
  
    // special case, for first 
    // element missing 
    if ($arr1[0] != $arr2[0]) 
        return $arr1[0]; 
  
    // Initialize current  
    // corner points 
    $lo = 0; 
    $hi = $N - 1; 
  
    // loop until lo < hi 
    while ($lo < $hi) 
    { 
        $mid = ($lo + $hi) / 2; 
  
        // If element at mid indices are  
        // equal then go to right subarray 
        if ($arr1[$mid] == $arr2[$mid]) 
            $lo = $mid; 
        else
            $hi = $mid; 
  
        // if lo, hi becomes  
        // contiguous, break 
        if ($lo == $hi - 1) 
            break; 
    } 
  
    // missing element will be  
    // at hi index of 
    // bigger array 
    return $arr1[$hi]; 
} 
  
// This function mainly  
// does basic error checking 
// and calls findMissingUtil 
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"; 
} 
  
    // Driver Code 
    $arr1 = array(1, 4, 5, 7, 9); 
    $arr2 = array(4, 5, 7, 9); 
    $M = count($arr1); 
    $N = count($arr2); 
    findMissing($arr1, $arr2, $M, $N); 
  
// This code is contributed by anuj_67. 
?> 

Output :

Missing Element is 1

What if input arrays are not in same order?
In this case, missing element is simply XOR of all elements of both arrays. Thanks to Yolo Song for suggesting this.

CPP

// C++ program to find missing element from one array 
// such that it has all elements of other array except 
// one.  Elements in two arrays can be in any order. 
#include <bits/stdc++.h> 
using namespace std; 
  
// This function mainly does XOR of all elements 
// of arr1[] and arr2[] 
void findMissing(int arr1[], int arr2[], int M, 
                 int N) 
{ 
    if (M != N-1 && N != M-1) 
    { 
        cout << "Invalid Input"; 
        return; 
    } 
  
    // Do XOR of all element 
    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; 
} 
  
// Driver Code 
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

// Java program to find missing element 
// from one array such that it has all  
// elements of other array except one. 
// Elements in two arrays can be in any order. 
  
import java.io.*; 
class Missing { 
      
    // This function mainly does XOR of  
    // all elements of arr1[] and arr2[] 
    void findMissing(int arr1[], int arr2[],  
                              int M, int N) 
    { 
        if (M != N - 1 && N != M - 1) { 
        System.out.println("Invalid Input"); 
            return; 
        } 
  
        // Do XOR of all element 
        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); 
    } 
  
    // Driver Code 
    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); 
    } 
} 
  
// This code is contributed by Anshika Goyal. 

Python3

# Python 3 program to find 
# missing element from one array 
# such that it has all elements 
# of other array except 
# one. Elements in two arrays 
# can be in any order. 
  
# This function mainly does XOR of all elements 
# of arr1[] and arr2[] 
def findMissing(arr1,arr2, M, N): 
    if (M != N-1 and N != M-1): 
      
        print("Invalid Input") 
        return
      
  
    # Do XOR of all element 
    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) 
  
# Driver Code 
arr1 = [4, 1, 5, 9, 7] 
arr2 = [7, 5, 9, 4] 
M = len(arr1)  
N = len(arr2) 
findMissing(arr1, arr2, M, N) 
  
# This code is contributed 
# by Smitha Dinesh Semwal 

C#

// C# program to find missing element 
// from one array such that it has all  
// elements of other array except one. 
// Elements in two arrays can be in 
// any order. 
using System; 
class GFG { 
      
    // This function mainly does XOR of  
    // all elements of arr1[] and arr2[] 
    static void findMissing(int []arr1, 
                            int []arr2,  
                            int M, int N) 
    { 
        if (M != N - 1 && N != M - 1) 
        { 
            Console.WriteLine("Invalid Input"); 
            return; 
        } 
  
        // Do XOR of all element 
        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); 
    } 
  
    // Driver Code 
    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); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP program to find missing 
// element from one array 
// such that it has all elements 
// of other array except 
// one. Elements in two arrays  
// can be in any order. 
  
// This function mainly does 
// XOR of all elements 
// of arr1[] and arr2[] 
function findMissing($arr1, $arr2,  
                           $M, $N) 
{ 
    if ($M != $N - 1 && $N != $M - 1) 
    { 
        echo "Invalid Input"; 
        return; 
    } 
  
    // Do XOR of all element 
    $res = 0; 
    for ($i = 0; $i < $M; $i++) 
        $res = $res ^ $arr1[$i]; 
          
    for ($i = 0; $i < $N; $i++) 
    $res = $res ^ $arr2[$i]; 
    echo "Missing element is ", $res; 
} 
  
    // Driver Code 
    $arr1 = array(4, 1, 5, 9, 7); 
    $arr2 = array(7, 5, 9, 4); 
    $M = sizeof($arr1); 
    $N = sizeof($arr2); 
  
    findMissing($arr1, $arr2, $M, $N); 
  
// This code is contributed by aj_36 
?> 

Output :

Missing Element is 1

This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

CPP

Java

Python3

C#

PHP

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