# Find index of an extra element present in one sorted array

Given two sorted arrays. There is only 1 difference between the arrays. The 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
Explanation: 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
Explanation: The first array has an extra element 9.
The extra element is present at index 3.```

Method 1: This includes the basic approach to solve this particular problem.

Approach: The basic method is to iterate through the whole second array and check element by element if they are different. As the array is sorted, checking the adjacent position of two arrays should be similar until and unless the missing element is found.

Algorithm:

1. Traverse through the array from start to end.
2. Check if the element at i’th element of the two arrays is similar or not.
3. If the elements are not similar then print the index and break

Implementation:

## C++

 `// C++ program to find an extra ``// element present in arr1[]``#include ``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

 ``

## Javascript

 ``

Output
`6`

Complexity Analysis:

• Time complexity: O(n).
As one traversal through the array is needed, so the time complexity is linear.
• Space complexity: O(1).
Since no extra space is required, the time complexity is constant.

Method 2: This method is a better way to solve the above problem and uses the concept of binary search.

Approach:To find the index of the missing element in less than linear time, binary search can be used, the idea is all the indices greater than or equal to the index of the missing element will have different elements in both the arrays and all the indices less than that index will have the similar elements in both arrays.

Algorithm:

1. Create three variables, low = 0, high = n-1, mid, ans = n
2. Run a loop until low is less than or equal to high, i.e till our search range is less than zero.
3. If the mid element, i.e (low + high)/2, of both arrays is similar then update the search to second half of the search range, i.e low = mid + 1
4. Else update the search to the first half of the search range, i.e high = mid – 1, and update the answer to the current index, ans = mid
5. Print the index.

Implementation:

## C++

 `// C++ program to find an extra``// element present in arr1[]``#include ``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

 ``

## Javascript

 ``

Output
`6`

Complexity Analysis:

• Time complexity : O(log n).
The time complexity of binary search is O(log n)
• Space complexity : O(1).
As no extra space is required, so the time complexity is constant.

Method 3: This method solves the given problem using the predefined function.

Approach: To find the element which is different, find the sum of each array and subtract the sums and find the absolute value. Search the larger array and check if the absolute is equal to an index and return that index. If an element is missing and all the other elements are the same, then the difference of sums will be equal to missing element.

Algorithm:

1. Create a function to calculate the sum of two arrays.
2. Find the absolute difference between the sum of two arrays (value).
3. Traverse the larger array from start too end
4. If the element at any index is equal to value, then print the index and break the loop.

Implementation:

## C++

 `// C++ code for above approach``#include``using` `namespace` `std;` `// function return sum of array elements``int` `sum(``int` `arr[], ``int` `n)``{``    ``int` `summ = 0;``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``summ += arr[i];``    ``}``    ``return` `summ;``}` `// function return index of given element``int` `indexOf(``int` `arr[], ``int` `element, ``int` `n)``{``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(arr[i] == element)``        ``{``            ``return` `i;``        ``}``    ``}``    ``return` `-1;``}` `// Function to find Index``int` `find_extra_element_index(``int` `arrA[], ``                             ``int` `arrB[], ``                             ``int` `n, ``int` `m)``{` `    ``// Calculating extra element``    ``int` `extra_element = sum(arrA, n) - ``                        ``sum(arrB, m);``    ` `    ``// returns index of extra element``    ``return` `indexOf(arrA, extra_element, n);``}` `// Driver Code``int` `main()``{``    ``int` `arrA[] = {2, 4, 6, 8, 10, 12, 13};``    ``int` `arrB[] = {2, 4, 6, 8, 10, 12};``    ``int` `n = ``sizeof``(arrA) / ``sizeof``(arrA[0]);``    ``int` `m = ``sizeof``(arrB) / ``sizeof``(arrB[0]);``    ``cout << find_extra_element_index(arrA, arrB, n, m);``}` `// This code is contributed by mohit kumar`

## Java

 `// Java code for above approach``class` `GFG ``{` `    ``// Function to find Index ``    ``static` `int` `find_extra_element_index(``int``[] arrA, ``                                        ``int``[] arrB) ``    ``{` `        ``// Calculating extra element``        ``int` `extra_element = sum(arrA) - sum(arrB);``        ` `        ``// returns index of extra element``        ``return` `indexOf(arrA, extra_element);``    ``}``    ` `    ``// function return sum of array elements``    ``static` `int` `sum(``int``[] arr)``    ``{``        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < arr.length; i++)``        ``{``            ``sum += arr[i];``        ``}``        ``return` `sum;``    ``}``    ` `    ``// function return index of given element``    ``static` `int` `indexOf(``int``[] arr, ``int` `element) ``    ``{``        ``for` `(``int` `i = ``0``; i < arr.length; i++)``        ``{``            ``if` `(arr[i] == element)``            ``{``                ``return` `i;``            ``}``        ``}``        ``return` `-``1``;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args) ``    ``{``        ``int``[] arrA = {``2``, ``4``, ``6``, ``8``, ``10``, ``12``, ``13``};``        ``int``[] arrB = {``2``, ``4``, ``6``, ``8``, ``10``, ``12``};``        ``System.out.println(find_extra_element_index(arrA, arrB));``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 code for above approach` `# Function to find Index ``def` `find_extra_element_index(arrA, arrB):``    ` `    ``# Calculating extra element``    ``extra_element ``=` `sum``(arrA) ``-` `sum``(arrB)``    ` `    ``# returns index of extra element``    ``return` `arrA.index(extra_element)` `# Driver Code``arrA ``=` `[``2``, ``4``, ``6``, ``8``, ``10``, ``12``, ``13``]``arrB ``=` `[``2``, ``4``, ``6``, ``8``, ``10``, ``12``]``print``(find_extra_element_index(arrA,arrB))` `# This code is contributed by Dravid`

## C#

 `// C# code for above approach``using` `System;` `class` `GFG ``{` `    ``// Function to find Index ``    ``static` `int` `find_extra_element_index(``int``[] arrA, ``                                        ``int``[] arrB) ``    ``{` `        ``// Calculating extra element``        ``int` `extra_element = sum(arrA) - sum(arrB);``        ` `        ``// returns index of extra element``        ``return` `indexOf(arrA, extra_element);``    ``}``    ` `    ``// function return sum of array elements``    ``static` `int` `sum(``int``[] arr)``    ``{``        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < arr.Length; i++)``        ``{``            ``sum += arr[i];``        ``}``        ``return` `sum;``    ``}``    ` `    ``// function return index of given element``    ``static` `int` `indexOf(``int``[] arr, ``int` `element) ``    ``{``        ``for` `(``int` `i = 0; i < arr.Length; i++)``        ``{``            ``if` `(arr[i] == element)``            ``{``                ``return` `i;``            ``}``        ``}``        ``return` `-1;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args) ``    ``{``        ``int``[] arrA = {2, 4, 6, 8, 10, 12, 13};``        ``int``[] arrB = {2, 4, 6, 8, 10, 12};``        ``Console.WriteLine(find_extra_element_index(arrA, arrB));``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Javascript

 ``

Output
`6`

Complexity Analysis:

• Time Complexity: O(n).
Since only three traversals through the array is needed, So the time complexity is linear.
• Space Complexity: O(1).
As no extra space is required, so the time complexity is constant.

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