Find the odd appearing element in O(Log n) time
Last Updated :
15 Dec, 2022
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 Solution 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 appearing 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++
#include<stdio.h>
void search( int *arr, int low, int high)
{
if (low > high)
return ;
if (low==high)
{
printf ( "The required element is %d " , arr[low]);
return ;
}
int mid = (low+high)/2;
if (mid%2 == 0)
{
if (arr[mid] == arr[mid+1])
search(arr, mid+2, high);
else
search(arr, low, mid);
}
else
{
if (arr[mid] == arr[mid-1])
search(arr, mid+1, high);
else
search(arr, low, mid-1);
}
}
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
class GFG
{
static void search( int arr[], int low, int high)
{
if (low > high)
return ;
if (low == high)
{
System.out.printf( "The required element is %d "
, arr[low]);
return ;
}
int mid = (low + high)/ 2 ;
if (mid % 2 == 0 )
{
if (arr[mid] == arr[mid + 1 ])
search(arr, mid + 2 , high);
else
search(arr, low, mid);
}
else
{
if (arr[mid] == arr[mid - 1 ])
search(arr, mid + 1 , high);
else
search(arr, low, mid - 1 );
}
}
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 );
}
}
|
Python3
def search(arr, low, high):
if low > high:
return None
if low = = high:
return arr[low]
mid = (low + high) / / 2 ;
if mid % 2 = = 0 :
if arr[mid] = = arr[mid + 1 ]:
return search(arr, mid + 2 , high)
else :
return search(arr, low, mid)
else :
if arr[mid] = = arr[mid - 1 ]:
return search(arr, mid + 1 , high)
else :
return search(arr, low, mid - 1 )
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#
using System;
class GFG {
static void search( int []arr, int low, int high)
{
if (low > high)
return ;
if (low == high)
{
Console.WriteLine( "The required element is " +
arr[low]);
return ;
}
int mid = (low + high)/2;
if (mid % 2 == 0)
{
if (arr[mid] == arr[mid + 1])
search(arr, mid + 2, high);
else
search(arr, low, mid);
}
else
{
if (arr[mid] == arr[mid - 1])
search(arr, mid + 1, high);
else
search(arr, low, mid - 1);
}
}
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);
}
}
|
PHP
<?php
function search( $arr , $low , $high )
{
if ( $low > $high )
return ;
if ( $low == $high )
{
echo "The required element is " ,
$arr [ $low ];
return ;
}
$mid = ( $low + $high ) / 2;
if ( $mid % 2 == 0)
{
if ( $arr [ $mid ] == $arr [ $mid + 1])
search( $arr , $mid + 2, $high );
else
search( $arr , $low , $mid );
}
else
{
if ( $arr [ $mid ] == $arr [ $mid - 1])
search( $arr , $mid + 1, $high );
else
search( $arr , $low , $mid - 1);
}
}
$arr = array (1, 1, 2, 2, 1, 1, 2,
2, 13, 1, 1, 40, 40);
$len = count ( $arr );;
search( $arr , 0, $len - 1);
?>
|
Javascript
<script>
function search(arr, low, high)
{
if (low > high)
return ;
if (low == high)
{
document.write( "The required element is " +
arr[low]);
return ;
}
let mid = Math.floor((low + high) / 2);
if (mid % 2 == 0)
{
if (arr[mid] == arr[mid + 1])
search(arr, mid + 2, high);
else
search(arr, low, mid);
}
else
{
if (arr[mid] == arr[mid-1])
search(arr, mid + 1, high);
else
search(arr, low, mid - 1);
}
}
let arr = [ 1, 1, 2, 2, 1, 1, 2, 2, 13, 1, 1, 40, 40 ];
let len = arr.length;
search(arr, 0, len-1);
</script>
|
Output :
The required element is 13
Time Complexity: O(Log n)
Auxiliary Space: O(1), ignoring recursion stack and O(log n) considering recursion stack.
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