Find lost element from a duplicated array

30 Jun, 2022
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.
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) / 2If the value from both arrays is the same then the missing element must be in the right part so set lo as midElse 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> 
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 
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// 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 findMissingUtilvoid 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 Codeint 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 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 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# findMissingUtildef 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 Codearr1 = [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# 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 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 findMissingUtilfunction 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.?>
<script>    // Javascript 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        let lo = 0, hi = N - 1;          // loop until lo < hi        while (lo < hi) {            let mid = parseInt((lo + hi) / 2, 10);              // 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)            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>
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
Java
Python3
C#
PHP
Javascript
// 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 Codeint 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 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.
# 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 Codearr1 = [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# 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 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?>
<script>     // Javascript 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)        {            document.write("Invalid Input");            return;        }         // Do XOR of all element        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>
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.
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.
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