# 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 <bits/stdc++.h>
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[0]);
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
```

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.

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