Given an array where every element occurs three times, except one element which occurs only once. Find the element that occurs once. Expected time complexity is O(n) and O(1) extra space.

**Examples:**

Input:arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 3}

Output:2

In the given array all element appear three times except 2 which appears once.

Input:arr[] = {10, 20, 10, 30, 10, 30, 30}

Output:20

In the given array all element appear three times except 20 which appears once.

We can use sorting to do it in O(nLogn) time. We can also use hashing, it has the worst case time complexity of O(n), but requires extra space.

The idea is to use bitwise operators for a solution that is O(n) time and uses O(1) extra space. The solution is not easy like other XOR based solutions, because all elements appear odd number of times here. Run a loop for all elements in array. At the end of every iteration, maintain following two values.

ones: The bits that have appeared 1st time or 4th time or 7th time … etc.

twos: The bits that have appeared 2nd time or 5th time or 8th time … etc.

Finally, we return the value of ‘ones’

How to maintain the values of ‘ones’ and ‘twos’?

‘ones’ and ‘twos’ are initialized as 0. For every new element in array, find out the common set bits in the new element and previous value of ‘ones’. These common set bits are actually the bits that should be added to ‘twos’. So do bitwise OR of the common set bits with ‘twos’. ‘twos’ also gets some extra bits that appear third time. These extra bits are removed later.

Update ‘ones’ by doing XOR of new element with previous value of ‘ones’. There may be some bits which appear 3rd time. These extra bits are also removed later.

Both ‘ones’ and ‘twos’ contain those extra bits which appear 3rd time. Remove these extra bits by finding out common set bits in ‘ones’ and ‘twos’.

Below is the implementation of above approach:

`// C++ program to find the element`

`// that occur only once`

`#include <bits/stdc++.h>`

`using`

`namespace`

`std;`

`int`

`getSingle(`

`int`

`arr[], `

`int`

`n)`

`{`

`` `int`

`ones = 0, twos = 0;`

`` `int`

`common_bit_mask;`

`` `// Let us take the example of`

`` `// {3, 3, 2, 3} to understand`

`` `// this`

`` `for`

`(`

`int`

`i = 0; i < n; i++) {`

``

`` `/* The expression "one & arr[i]" gives the bits that`

`` `are there in both 'ones' and new element from arr[].`

`` `We add these bits to 'twos' using bitwise OR`

`` `Value of 'twos' will be set as 0, 3, 3 and 1 after`

`` `1st, 2nd, 3rd and 4th iterations respectively */`

`` `twos = twos | (ones & arr[i]);`

`` `/* XOR the new bits with previous 'ones' to get all`

`` `bits appearing odd number of times`

`` `Value of 'ones' will be set as 3, 0, 2 and 3 after`

`` `1st, 2nd, 3rd and 4th iterations respectively */`

`` `ones = ones ^ arr[i];`

`` `/* The common bits are those bits which appear third`

`` `time So these bits should not be there in both`

`` `'ones' and 'twos'. common_bit_mask contains all`

`` `these bits as 0, so that the bits can be removed`

`` `from 'ones' and 'twos'`

`` `Value of 'common_bit_mask' will be set as 00, 00, 01`

`` `and 10 after 1st, 2nd, 3rd and 4th iterations`

`` `respectively */`

`` `common_bit_mask = ~(ones & twos);`

`` `/* Remove common bits (the bits that appear third`

`` `time) from 'ones'`

`` `Value of 'ones' will be set as 3, 0, 0 and 2 after`

`` `1st, 2nd, 3rd and 4th iterations respectively */`

`` `ones &= common_bit_mask;`

`` `/* Remove common bits (the bits that appear third`

`` `time) from 'twos'`

`` `Value of 'twos' will be set as 0, 3, 1 and 0 after`

`` `1st, 2nd, 3rd and 4th itearations respectively */`

`` `twos &= common_bit_mask;`

`` `// uncomment this code to see intermediate values`

`` `// printf (" %d %d n", ones, twos);`

`` `}`

`` `return`

`ones;`

`}`

`// Driver code`

`int`

`main()`

`{`

`` `int`

`arr[] = { 3, 3, 2, 3 };`

`` `int`

`n = `

`sizeof`

`(arr) / `

`sizeof`

`(arr[0]);`

`` `cout << `

`"The element with single occurrence is "`

`` `<< getSingle(arr, n);`

`` `return`

`0;`

`}`

**Output**

The element with single occurrence is 2

**Time Complexity :** O(n)

**Auxiliary Space :** O(1)