Given an array of integers. All numbers occur twice except one number which occurs once. Find the number in O(n) time & constant extra space.
Example :
Input: ar[] = {7, 3, 5, 4, 5, 3, 4} Output: 7
One solution is to check every element if it appears once or not. Once an element with a single occurrence is found, return it. Time complexity of this solution is O(n2).
A better solution is to use hashing.
- Traverse all elements and put them in a hash table. Element is used as key and the count of occurrences is used as the value in the hash table.
- Traverse the array again and print the element with count 1 in the hash table.
This solution works in O(n) time but requires extra space.
The best solution is to use XOR. XOR of all array elements gives us the number with a single occurrence. The idea is based on the following two facts.
a) XOR of a number with itself is 0.
b) XOR of a number with 0 is number itself.
Let us consider the above example. Let ^ be xor operator as in C and C++. res = 7 ^ 3 ^ 5 ^ 4 ^ 5 ^ 3 ^ 4 Since XOR is associative and commutative, above expression can be written as: res = 7 ^ (3 ^ 3) ^ (4 ^ 4) ^ (5 ^ 5) = 7 ^ 0 ^ 0 ^ 0 = 7 ^ 0 = 7
Below are implementations of above algorithm.
<?php
// PHP program to find the array
// element that appears only once
function findSingle( $ar , $ar_size )
{
// Do XOR of all
// elements and return
$res = $ar [0];
for ( $i = 1; $i < $ar_size ; $i ++)
$res = $res ^ $ar [ $i ];
return $res ;
}
// Driver code
$ar = array (2, 3, 5, 4, 5, 3, 4);
$n = count ( $ar );
echo "Element occurring once is "
, findSingle( $ar , $n );
?>
Output:
Element occurring once is 2
The time complexity of this solution is O(n) and it requires O(1) extra space.