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.

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; // Funtion 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 { /* Funtion 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) # Funtion 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

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

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