Given an array where all elements appear even number of times except one. All repeating occurrences of elements appear in pairs and these pairs are not adjacent (there cannot be more than two consecutive occurrences of any element). Find the element that appears odd number of times.

Note that input like {2, 2, 1, 2, 2, 1, 1} is valid as all repeating occurrences occur in pairs and these pairs are not adjacent. Input like {2, 1, 2} is invalid as repeating elements don’t appear in pairs. Also, input like {1, 2, 2, 2, 2} is invalid as two pairs of 2 are adjacent. Input like {2, 2, 2, 1} is also invalid as there are three consecutive occurrences of 2.

**Example :**

Input: arr[] = {1, 1, 2, 2, 1, 1, 2, 2, 13, 1, 1, 40, 40, 13, 13} Output: 13 Input: arr[] = {1, 1, 2, 2, 3, 3, 4, 4, 3, 600, 600, 4, 4} Output: 3

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A **Simple Solutio**n is to sort the array and then traverse the array from left to right. Since the array is sorted, we can easily figure out the required element. Time complexity of this solution is O(n Log n)

A **Better Solution** is to do XOR of all elements, result of XOR would give the odd appearing element. Time complexity of this solution is O(n). See XOR based solution for add apearing for more details.

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.

Since the element appears odd number of times, there must be a single occurrence of the element. For example, in {2, 1, 1, 2, 2), the first 2 is the odd occurrence. So the idea is to find this odd occurrence using Binary Search.

All elements before the odd occurrence have first occurrence at even index (0, 2, ..) and next occurrence at odd index (1, 3, …). And all elements afterhave 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 there is an odd occurrence of the element after ‘mid’ else before mid.

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

Below is the implementation based on above idea.

## C/C++

// C program to find the element that appears odd number of time #include<stdio.h> // A Binary Search based function to find the element // that appears odd times 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+1, high); else search(arr, low, mid-1); } } // Driver program int main() { int arr[] = {1, 1, 2, 2, 1, 1, 2, 2, 13, 1, 1, 40, 40}; int len = sizeof(arr)/sizeof(arr[0]); search(arr, 0, len-1); return 0; }

## Java

// Java program to find the element // that appears odd number of time class GFG { // A Binary Search based function to find // the element that appears odd times static void search(int arr[], int low, int high) { // Base cases if (low > high) return; if (low == high) { System.out.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); } // If mid is odd else { if (arr[mid] == arr[mid - 1]) search(arr, mid + 1, high); else search(arr, low, mid - 1); } } // Driver program public static void main(String[] args) { int arr[] = {1, 1, 2, 2, 1, 1, 2, 2, 13, 1, 1, 40, 40}; int len = arr.length; search(arr, 0, len-1); } } // This code is contributed by // Smitha DInesh Semwal

## Python

# Python program to find the element that appears odd number of times # O(log n) approach # Binary search based function # Returns the element that appears odd number of times def search(arr, low, high): # Base case if low > high: return None if low == high: return arr[low] # Find the middle point mid = (low + high)/2; # If mid is even if mid%2 == 0: # If the element next to mid is same as mid, # then output element lies on right side, # else on left side if arr[mid] == arr[mid+1]: return search(arr, mid+2, high) else: return search(arr, low, mid) else: # else if mid is odd if arr[mid] == arr[mid-1]: return search(arr, mid+1, high) else: # (mid-1) because target element can only exist at even place return search(arr, low, mid-1) # Test array arr = [ 1, 1, 2, 2, 1, 1, 2, 2, 13, 1, 1, 40, 40 ] result = search(arr, 0, len(arr)-1 ) if result is not None: print "The required element is %d " % result else: print "Invalid array"

## C#

// C# program to find the element // that appears odd number of time using System; class GFG { // A Binary Search based function to find // the element that appears odd times static void search(int []arr, int low, int high) { // Base cases if (low > high) return; if (low == high) { Console.WriteLine("The required element is "+ 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); } // If mid is odd else { if (arr[mid] == arr[mid - 1]) search(arr, mid + 1, high); else search(arr, low, mid - 1); } } // Driver program public static void Main() { int []arr = {1, 1, 2, 2, 1, 1, 2, 2, 13, 1, 1, 40, 40}; int len = arr.Length; search(arr, 0, len-1); } } // This code is contributed by Sam007

## PHP

<?php // PHP program to find the // element that appears odd // number of times // A Binary Search based // function to find the // element that appears odd times function search($arr, $low, $high) { // Base cases if ($low > $high) return; if ($low == $high) { echo "The required element is ", $arr[$low]; return; } // Find the middle point $mid = ($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, $mid); } // If mid is odd else { if ($arr[$mid] == $arr[$mid - 1]) search($arr, $mid + 1, $high); else search($arr, $low, $mid - 1); } } // Driver Code $arr = array(1, 1, 2, 2, 1, 1, 2, 2, 13, 1, 1, 40, 40); $len = count($arr);; search($arr, 0, $len - 1); // This code is contributed by anuj_67. ?>

**Output :**

The required element is 13

**
Time Complexity:** O(Log n)

This article is contributed by Mehboob Elahi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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