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(arr1, arr2, n); return 0; }

## Java

// Java program to find an extra // element present in arr1[] class GFG { // 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; } // Driver Code 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

## Python3

# Python 3 program to find an # extra element present in arr1[] # Returns index of extra . # element in arr1[] n is # size of arr2[]. Size of # arr1[] is n-1. def findExtra(arr1, arr2, n) : for i in range(0, n) : if (arr1[i] != arr2[i]) : return i return n # Driver code arr1 = [2, 4, 6, 8, 10, 12, 13] arr2 = [2, 4, 6, 8, 10, 12] n = len(arr2) # Solve is passed both arrays print(findExtra(arr1, arr2, n)) # This code is contributed # by Nikita Tiwari.

## C#

// C# program to find an extra // element present in arr1[] using System; class GfG { // 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; } // Driver code public static void Main () { 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 Console.Write(findExtra(arr1, arr2, n)); } } // This code is contributed by parashar.

## PHP

<?php // PHP program to find an extra // element present in arr1[] // Returns index of extra element // in arr1[]. n is size of arr2[]. // Size of arr1[] is n-1. function findExtra($arr1, $arr2, $n) { for ($i = 0; $i < $n; $i++) if ($arr1[$i] != $arr2[$i]) return $i; return $n; } // Driver code $arr1 = array (2, 4, 6, 8, 10, 12, 13); $arr2 = array(2, 4, 6, 8, 10, 12); $n = sizeof($arr2); // Solve is passed // both arrays echo findExtra($arr1, $arr2, $n); // This code is contributed by ajit ?>

**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 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) { // Initialize result int index = n; // 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 GFG { // 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) { // Initialize result int index = n; // 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 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

## Python3

# Python3 program to find an extra # element present in arr1[] # Returns index of extra element # in arr1[]. n is size of arr2[]. # Size of arr1[] is n-1. def findExtra(arr1, arr2, n) : index = n # Initialize result # left and right are end points # denoting the current range. left = 0 right = n - 1 while (left <= right) : mid = (int)((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 arr1 = [2, 4, 6, 8, 10, 12, 13] arr2 = [2, 4, 6, 8, 10, 12] n = len(arr2) # Solve is passed both arrays print(findExtra(arr1, arr2, n)) # This code is contributed by Nikita Tiwari.

## C#

// C# program to find an extra // element present in arr1[] using System; class GFG { // 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) { // Initialize result int index = n; // 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 public static void Main () { 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 Console.Write(findExtra(arr1, arr2, n)); } } // This code is contributed by nitin mittal.

## PHP

<?php // PHP program to find an extra // element present in arr1[] // Returns index of extra element // in arr1[]. n is size of arr2[]. // Size of arr1[] is n-1. function findExtra($arr1, $arr2, $n) { // Initialize result $index = $n; // left and right are // end points denoting // the current range. $left = 0; $right = $n - 1; while ($left <= $right) { $mid = ($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 { $arr1 = array(2, 4, 6, 8, 10, 12, 13); $arr2 = array(2, 4, 6, 8, 10, 12); $n = sizeof($arr2) / sizeof($arr2[0]); // Solve is passed both arrays echo findExtra($arr1, $arr2, $n); return 0; } // This code is contributed by nitin mittal ?>

**Output :**

6

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

### Asked in : Amazon

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