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

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; }

## 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)

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