Find the element that appears once | GeeksforGeeks

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.

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’.

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;

}

Following is another O(n) time complexity and O(1) extra space method suggested by aj. We can sum the bits in same positions for all the numbers and take modulo with 3. The bits for which sum is not multiple of 3, are the bits of number with single occurrence.

int getSingle(int arr[], int n)

{

    // Initialize result

    int result = 0;

    int x, sum;

    // Iterate through every bit

    for (int i = 0; i < INT_SIZE; i++)

    {

      // Find sum of set bits at ith position in all

      // array elements

      sum = 0;

      x = (1 << i);

      for (int j=0; j< n; j++ )

      {

          if (arr[j] & x)

            sum++;

      }

      // The bits with sum not multiple of 3, are the

      // bits of element with single occurrence.

      if (sum % 3)

        result |= x;

    }

    return result;

}

From http://www.careercup.com/question?id=7902674
ones - At any point of time, this variable holds XOR of all the elements which have 
appeared "only" once. 
twos - At any point of time, this variable holds XOR of all the elements which have 
appeared "only" twice. 

So if at any point time, 
1. A new number appears - It gets XOR'd to the variable "ones". 
2. A number gets repeated(appears twice) - It is removed from "ones" and XOR'd to the 
variable "twice". 
3. A number appears for the third time - It gets removed from both "ones" and "twice". 

The final answer we want is the value present in "ones" - coz, it holds the unique element. 

So if we explain how steps 1 to 3 happens in the code, we are done. 
Before explaining above 3 steps, lets look at last three lines of the code, 

not_threes = ~(ones & twos) 
ones & = not_threes 
twos & = not_threes 

All it does is, common 1's between "ones" and "twos" are converted to zero. 

For simplicity, in all the below explanations - consider we have got only 4 elements in the array (one unique element and 3 repeated elements - in any order). 

Explanation for step 1 
------------------------ 
Lets say a new element(x) appears. 
CURRENT SITUATION - Both variables - "ones" and "twos" has not recorded "x". 

Observe the statement "twos| = ones & x". 
Since bit representation of "x" is not present in "ones", AND condition yields nothing. So "twos" does not get bit representation of "x". 
But, in next step "ones ^= x" - "ones" ends up adding bits of "x". Thus new element gets recorded in "ones" but not in "twos". 

The last 3 lines of code as explained already, converts common 1's b/w "ones" and "twos" to zeros. 
Since as of now, only "ones" has "x" and not "twos" - last 3 lines does nothing. 

Explanation for step 2. 
------------------------ 
Lets say an element(x) appears twice. 
CURRENT SITUATION - "ones" has recorded "x" but not "twos". 

Now due to the statement, "twos| = ones & x" - "twos" ends up getting bits of x. 
But due to the statement, "ones ^ = x" - "ones" removes "x" from its binary representation. 

Again, last 3 lines of code does nothing. 
So ultimately, "twos" ends up getting bits of "x" and "ones" ends up losing bits of "x". 

Explanation for step 3. 
------------------------- 
Lets say an element(x) appears for the third time. 
CURRENT SITUATION - "ones" does not have bit representation of "x" but "twos" has. 

Though "ones & x" does not yield nothing .. "twos" by itself has bit representation of "x". So after this statement, "two" has bit representation of "x". 
Due to "ones^=x", after this step, "one" also ends up getting bit representation of "x". 

Now last 3 lines of code removes common 1's of "ones" and "twos" - which is the bit representation of "x".
Thus both "ones" and "twos" ends up losing bit representation of "x"
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