Find the element that appears once others appears THRICE.

Find the element that appears once others appears THRICE.

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Given an array of integers. All numbers occur thrice except one number which occurs once. Find the number in O(n) time & constant extra space.

Let us first understand what we want to achieve? what is the input and what will be the expected output?

Question:
Input: {17, 10, 17, 2, 17, 10, 10, 4, 4, 4}
Output: 2

Input: {17}
Output: 17

Input: {17, 17, 17, 2}


Output: 2

Algorithm

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This problem can be solved in multiple ways but solution might not fulfill time and space complexity constraints given in problem statement.

Time complexity to solve this problem should not be greater then O(n)
Space complexity to solve this problem should not be greater then O(1)

Bit Manipulation will be helpful in this situation.

Before proceeding further lets just see XOR operation and facts.

Lets see some XOR operation,
  1.   5 ^ 5     = 0101 ^ 0101 = 0000
  2.   1 ^ 1     = 0001 ^ 0001 = 0000
  3.   10 ^ 10 = 1010 ^ 1010 = 0000
From the above few operation, we can see that when a number is XOR with same number then result is 0.

Also XOR is Associative and Commutative, it means
Associative    = giving the same result regardless of grouping, i.e. so that a*(b*c)=(a*b)*c 
Commutative = independent of order; as in e.g. "a x b = b x a"

1.  1 ^ 1 = 0
2.  1 ^ 1 ^ 2 ^ 2 = 0
3.  1 ^ 2 ^ 1 ^ 2 = 0

Irrespective of order, when a same number will be XOR twice then result will be 0.

So, Now lets see our original problem statement.
Given an array of integers. All numbers occur thrice except one number which occurs once. Find the number.

If all the number appeared twice except one number and we need to find that number then it is easy,

Let's take a small example, say given array is,
1. {1, 1 , 2}
Answer is 2 here, XOR all the numbers. result will be the number that appeared only once.
From the fact we saw that when a number is XOR with same number then result is 0.
 = (1 ^ 1) ^ 2
 = (0) ^ 2
 =  2 
Number that appear only once is 2.

In our question, all numbers are appeared thrice, So how to solve that?
If we do XOR of 3 same number then result will be same number, So XOR all number will not helpful here.

So idea here is to take 2 variable, say var1 and var2

1. var1 will keep a note that variable appeared first time.


2. var2 will keep a note that same variable appeared second time.


3. Now, when the same variable appears third time, we have to discard the element as number appears 3 times which is fine and it is not the variable we are looking for. Also, if the number appears 3 times properly, then we have to initialize var1 and var2 to 0 to start looking for new element.So for all the numbers which appears thrice, var1 and var2 will become 0 but only for number which appears only once, var1 will be set with that value.

  Lets see what each line do in program.

secondTimeOccurence =  secondTimeOccurence | (firstTimeOccurence & arr[i]);
firstTimeOccurence  =  firstTimeOccurence ^ arr[i]; 
int neutraliser     =  ~(firstTimeOccurence & secondTimeOccurence);
firstTimeOccurence  =  firstTimeOccurence & neutraliser;
secondTimeOccurence =  secondTimeOccurence & neutraliser;

First line initialize the var2
Second line initialize var1
Third, Forth and Fifth line task is to check if the number appears thrice then reinitialize var1 and var2 to 0.

Lets take a simple example and see, given array is {1, 1, 1, 2}
  1. 1st iteration encounter 1 
      secondTimeOccurence = 0 
      firstTimeOccurence = 1 
  
  2. 2nd iteration encounter 1 
      secondTimeOccurence = 1  
      firstTimeOccurence = 0  

  3. 3rd iteration encounter 1 
      secondTimeOccurence = 0  
      firstTimeOccurence = 0  

  4. 4th iteration encounter 2 
      secondTimeOccurence = 0 
      firstTimeOccurence = 2  

Number that appear only once is present in variable firstTimeOccurence that is 2 here.

Java Program to Find the element that appears once.

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package javabypatel;

public class FindElementThatAppearOnceRestAllAppearThrice {

 public static void main(String[] args) {
  int arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 3};
  System.out.println(findElementThatAppearOnce(arr));
 }

 private static int findElementThatAppearOnce(int arr[]){

  int firstTimeOccurence = 0;
  int secondTimeOccurence = 0;

  for (int i=0; i < arr.length; i++){
   secondTimeOccurence =  secondTimeOccurence | (firstTimeOccurence & arr[i]);
   firstTimeOccurence = firstTimeOccurence ^ arr[i]; 
   int neutraliser = ~(firstTimeOccurence & secondTimeOccurence);
   firstTimeOccurence = firstTimeOccurence & neutraliser;
   secondTimeOccurence = secondTimeOccurence & neutraliser;
  }
  return firstTimeOccurence;
 }
}

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Enjoy !!!! 

If you find any issue in post or face any error while implementing, Please comment.