C Program for Find the element that appears once

Last Updated : 22 Dec, 2021

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.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C/C++

#include <stdio.h> 
  
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; 
} 
  
int main() 
{ 
    int arr[] = {3, 3, 2, 3}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    printf("The element with single occurrence is %d ", 
            getSingle(arr, n)); 
    return 0; 
} 