# Find the only repeating element in a sorted array of size n

Given a sorted array of n elements containing elements in range from 1 to n-1 i.e. one element occurs twice, the task is to find the repeating element in an array.

Examples :

```Input :  arr[] = { 1, 2 , 3 , 4 , 4}
Output :  4

Input :  arr[] = { 1 , 1 , 2 , 3 , 4}
Output :  1
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A naive approach is to scan the whole array and check if an element occurs twice, then return. This approach takes O(n) time.

An efficient method is to use Binary Search .
1- Check if the middle element is the repeating one.
2- If not then check if the middle element is at proper position if yes then start searching repeating element in right.
3- Otherwise the repeating one will be in left.

## C/C++

 `// C++ program to find the only repeating element in an ` `// array of size n and elements from range 1 to n-1. ` `#include ` `using` `namespace` `std; ` ` `  `// Returns index of second appearance of a repeating element ` `// The function assumes that array elements are in range from ` `// 1 to n-1. ` `int` `findRepeatingElement(``int` `arr[], ``int` `low, ``int` `high) ` `{ ` `    ``// low = 0 , high = n-1; ` `    ``if` `(low > high) ` `        ``return` `-1; ` ` `  `    ``int` `mid = (low + high) / 2; ` ` `  `    ``// Check if the mid element is the repeating one ` `    ``if` `(arr[mid] != mid + 1) ` `    ``{ ` `        ``if` `(mid > 0 && arr[mid]==arr[mid-1]) ` `            ``return` `mid; ` ` `  `        ``// If mid element is not at its position that means ` `        ``// the repeated element is in left ` `        ``return`  `findRepeatingElement(arr, low, mid-1); ` `    ``} ` ` `  `    ``// If mid is at proper position then repeated one is in ` `    ``// right. ` `    ``return` `findRepeatingElement(arr, mid+1, high); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int`  `arr[] = {1, 2, 3, 3, 4, 5}; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `index = findRepeatingElement(arr, 0, n-1); ` `    ``if` `(index != -1) ` `        ``cout << arr[index]; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the only repeating element in an ` `// array of size n and elements from range 1 to n-1. ` ` `  `class` `Test ` `{ ` `    ``// Returns index of second appearance of a repeating element ` `    ``// The function assumes that array elements are in range from ` `    ``// 1 to n-1. ` `    ``static` `int` `findRepeatingElement(``int` `arr[], ``int` `low, ``int` `high) ` `    ``{ ` `        ``// low = 0 , high = n-1; ` `        ``if` `(low > high) ` `            ``return` `-``1``; ` `      `  `        ``int` `mid = (low + high) / ``2``; ` `      `  `        ``// Check if the mid element is the repeating one ` `        ``if` `(arr[mid] != mid + ``1``) ` `        ``{ ` `            ``if` `(mid > ``0` `&& arr[mid]==arr[mid-``1``]) ` `                ``return` `mid; ` `      `  `            ``// If mid element is not at its position that means ` `            ``// the repeated element is in left ` `            ``return`  `findRepeatingElement(arr, low, mid-``1``); ` `        ``} ` `      `  `        ``// If mid is at proper position then repeated one is in ` `        ``// right. ` `        ``return` `findRepeatingElement(arr, mid+``1``, high); ` `    ``} ` `     `  `    ``// Driver method ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int`  `arr[] = {``1``, ``2``, ``3``, ``3``, ``4``, ``5``}; ` `        ``int` `index = findRepeatingElement(arr, ``0``, arr.length-``1``); ` `        ``if` `(index != -``1``) ` `            ``System.out.println(arr[index]); ` `    ``} ` `} `

## Python

 `# Python program to find the only repeating element in an ` `# array of size n and elements from range 1 to n-1 ` ` `  `# Returns index of second appearance of a repeating element ` `# The function assumes that array elements are in range from ` `# 1 to n-1. ` `def` `findRepeatingElement(arr, low, high): ` ` `  `    ``# low = 0 , high = n-1 ` `    ``if` `low > high: ` `        ``return` `-``1` ` `  `    ``mid ``=` `(low ``+` `high) ``/` `2` ` `  `    ``# Check if the mid element is the repeating one ` `    ``if` `(arr[mid] !``=` `mid ``+` `1``): ` `     `  `        ``if` `(mid > ``0` `and` `arr[mid]``=``=``arr[mid``-``1``]): ` `            ``return` `mid ` ` `  `        ``# If mid element is not at its position that means ` `        ``# the repeated element is in left ` `        ``return`  `findRepeatingElement(arr, low, mid``-``1``) ` ` `  `    ``# If mid is at proper position then repeated one is in ` `    ``# right. ` `    ``return` `findRepeatingElement(arr, mid``+``1``, high) ` ` `  `# Driver code ` `arr ``=` `[``1``, ``2``, ``3``, ``3``, ``4``, ``5``] ` `n ``=` `len``(arr) ` `index ``=` `findRepeatingElement(arr, ``0``, n``-``1``) ` `if` `(index ``is` `not` `-``1``): ` `    ``print` `arr[index] ` ` `  `#This code is contributed by Afzal Ansari `

## C#

 `// C# program to find the only repeating ` `// element in an array of size n and  ` `// elements from range 1 to n-1. ` `using` `System; ` ` `  `class` `Test ` `{ ` `    ``// Returns index of second appearance of a ` `    ``// repeating element. The function assumes that ` `    ``// array elements are in range from 1 to n-1. ` `    ``static` `int` `findRepeatingElement(``int` `[]arr, ``int` `low, ` `                                              ``int` `high) ` `    ``{ ` `        ``// low = 0 , high = n-1; ` `        ``if` `(low > high) ` `            ``return` `-1; ` `     `  `        ``int` `mid = (low + high) / 2; ` `     `  `        ``// Check if the mid element  ` `        ``// is the repeating one ` `        ``if` `(arr[mid] != mid + 1) ` `        ``{ ` `            ``if` `(mid > 0 && arr[mid]==arr[mid-1]) ` `                ``return` `mid; ` `     `  `            ``// If mid element is not at its position ` `            ``// that means the repeated element is in left ` `            ``return` `findRepeatingElement(arr, low, mid-1); ` `        ``} ` `     `  `        ``// If mid is at proper position  ` `        ``// then repeated one is in right. ` `        ``return` `findRepeatingElement(arr, mid+1, high); ` `    ``} ` `     `  `    ``// Driver method ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `[]arr = {1, 2, 3, 3, 4, 5}; ` `        ``int` `index = findRepeatingElement(arr, 0, arr.Length-1); ` `        ``if` `(index != -1) ` `        ``Console.Write(arr[index]); ` `    ``} ` `} ` ` `  `// This code is contributed by Nitin Mittal. `

## PHP

 ` ``\$high``) ` `        ``return` `-1; ` ` `  `    ``\$mid` `= ``floor``((``\$low` `+ ``\$high``) / 2); ` ` `  `    ``// Check if the mid element ` `    ``// is the repeating one ` `    ``if` `(``\$arr``[``\$mid``] != ``\$mid` `+ 1) ` `    ``{ ` `        ``if` `(``\$mid` `> 0 && ``\$arr``[``\$mid``] ==  ` `                        ``\$arr``[``\$mid` `- 1]) ` `            ``return` `\$mid``; ` ` `  `        ``// If mid element is not at  ` `        ``// its position that means ` `        ``// the repeated element is in left ` `        ``return` `findRepeatingElement(``\$arr``, ``\$low``,  ` `                                    ``\$mid` `- 1); ` `    ``} ` ` `  `    ``// If mid is at proper position ` `    ``// then repeated one is in right. ` `    ``return` `findRepeatingElement(``\$arr``, ``\$mid` `+ 1,  ` `                                        ``\$high``); ` `} ` ` `  `// Driver code ` `\$arr` `= ``array``(1, 2, 3, 3, 4, 5); ` `\$n` `= sizeof(``\$arr``); ` `\$index` `= findRepeatingElement(``\$arr``, 0,  ` `                              ``\$n` `- 1); ` `if` `(``\$index` `!= -1) ` `echo` `\$arr``[``\$index``]; ` ` `  `// This code is contributed ` `// by nitin mittal.  ` `?> `

Output :

```3
```

Time Complexity : O(log n)

This article is contributed by Sahil Chhabra (KILLER). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.