Find the element that appears once in a sorted array

Given a sorted array in which all elements appear twice (one after one) and one element appears only once. Find that element in O(log n) complexity.

Example:

Input:   arr[] = {1, 1, 3, 3, 4, 5, 5, 7, 7, 8, 8}
Output:  4

Input:   arr[] = {1, 1, 3, 3, 4, 4, 5, 5, 7, 7, 8}
Output:  8

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A Simple Solution is to traverse the array from left to right. Since the array is sorted, we can easily figure out the required element.

An Efficient Solution can find the required element in O(Log n) time. The idea is to use Binary Search. Below is an observation in input array.
All elements before the required have first occurrence at even index (0, 2, ..) and next occurrence at odd index (1, 3, …). And all elements after the required element have first occurrence at odd index and next occurrence at even index.

1) Find the middle index, say ‘mid’.

2) If ‘mid’ is even, then compare arr[mid] and arr[mid + 1]. If both are same, then the required element after ‘mid’ else before mid.

3) If ‘mid’ is odd, then compare arr[mid] and arr[mid – 1]. If both are same, then the required element after ‘mid’ else before mid.

Below is C implementation based on above idea.

// C program to find the element that appears only once
#include<stdio.h>

// A Binary Search based function to find the element
// that appears only once
void search(int *arr, int low, int high)
{
     // Base cases
    if (low > high)
       return;

    if (low==high)
    {
        printf("The required element is %d ", arr[low]);
        return;
    }

    // Find the middle point
    int mid = (low + high) / 2;

    // If mid is even and element next to mid is
    // same as mid, then output element lies on
    // right side, else on left side
    if (mid%2 == 0)
    {
        if (arr[mid] == arr[mid+1])
            search(arr, mid+2, high);
        else
            search(arr, low, mid);
    }
    else  // If mid is odd
    {
        if (arr[mid] == arr[mid-1])
            search(arr, mid-2, high);
        else
            search(arr, low, mid-1);
    }
}

// Driver program
int main()
{
    int arr[] = {1, 1, 2, 4, 4, 5, 5, 6, 6};
    int len = sizeof(arr)/sizeof(arr[0]);
    search(arr, 0, len-1);
    return 0;
}

Output:

The required element is 2

Time Complexity: O(Log n)