Given two sorted arrays. There is only 1 difference between the arrays. First array has one element extra added in between. Find the index of the extra element.

Examples:

Input : {2, 4, 6, 8, 9, 10, 12}; {2, 4, 6, 8, 10, 12}; Output : 4 The first array has an extra element 9. The extra element is present at index 4. Input : {3, 5, 7, 9, 11, 13} {3, 5, 7, 11, 13} Output : 3

**Method 1 (Basic)**

The basic method is to iterate through the whole second array and check element by element if they are different.

## C++

// C++ program to find an extra element present // in arr1[] #include <iostream> using namespace std; // Returns index of extra element in arr1[]. // n is size of arr2[]. Size of arr1[] is // n-1. int findExtra(int arr1[], int arr2[], int n) { for (int i=0; i<n; i++) if (arr1[i] != arr2[i]) return i; return n; } // Driver code int main() { int arr1[] = {2, 4, 6, 8, 10, 12, 13}; int arr2[] = {2, 4, 6, 8, 10, 12}; int n = sizeof(arr2) / sizeof(arr2[0]); // Solve is passed both arrays cout << findExtra(arr, arr2, n); return 0; }

## Java

// Java program to find an extra // element present in arr1[] class Main { // Returns index of extra element in arr1[]. // n is size of arr2[]. Size of arr1[] is // n-1. static int findExtra(int arr1[], int arr2[], int n) { for (int i=0; i<n; i++) if (arr1[i] != arr2[i]) return i; return n; } public static void main (String[] args) { int arr1[] = {2, 4, 6, 8, 10, 12, 13}; int arr2[] = {2, 4, 6, 8, 10, 12}; int n = arr2.length; // Solve is passed both arrays System.out.println(findExtra(arr1, arr2, n)); } } /* This code is contributed by Harsh Agarwal */

Output:

6

**Time complexity :** O(n)

**Method 2 (Using Binary search)**

We use binary search to check whether the same indices elements are different & reduce our search by a factor of 2 in each step.

## C++

// C++ program to find an extra element pre.sent // in arr1[] #include <iostream> using namespace std; // Returns index of extra element in arr1[]. // n is size of arr2[]. Size of arr1[] is // n-1. int findExtra(int arr1[], int arr2[], int n) { int index = 0; // Initialize result // left and right are end points denoting // the current range. int left = 0, right = n - 1; while (left <= right) { int mid = (left+right) / 2; // If middle element is same of both // arrays, it means that extra element // is after mid so we update left to mid+1 if (arr2[mid] == arr1[mid]) left = mid + 1; // If middle element is different of the // arrays, it means that the index we are // searching for is either mid, or before // mid. Hence we update right to mid-1. else { index = mid; right = mid - 1; } } // when right is greater than left our // search is complete. return index; } // Driver code int main() { int arr1[] = {2, 4, 6, 8, 10, 12, 13}; int arr2[] = {2, 4, 6, 8, 10, 12}; int n = sizeof(arr2) / sizeof(arr2[0]); // Solve is passed both arrays cout << findExtra(arr1, arr2, n); return 0; }

## Java

// Java program to find an extra // element present in arr1[] class Main { // Returns index of extra element in arr1[]. // n is size of arr2[]. Size of arr1[] is // n-1. static int findExtra(int arr1[], int arr2[], int n) { int index = 0; // Initialize result // left and right are end points denoting // the current range. int left = 0, right = n - 1; while (left <= right) { int mid = (left+right) / 2; // If middle element is same of both // arrays, it means that extra element // is after mid so we update left to mid+1 if (arr2[mid] == arr1[mid]) left = mid + 1; // If middle element is different of the // arrays, it means that the index we are // searching for is either mid, or before // mid. Hence we update right to mid-1. else { index = mid; right = mid - 1; } } // when right is greater than left our // search is complete. return index; } public static void main (String[] args) { int arr1[] = {2, 4, 6, 8, 10, 12,13}; int arr2[] = {2, 4, 6, 8, 10, 12}; int n = arr2.length; // Solve is passed both arrays System.out.println(findExtra(arr1, arr2, n)); } } /* This code is contributed by Harsh Agarwal */

Output:

6

**Time complexity :** O(log n)

### Asked in: Amazon

**Asked in ** Amazon

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