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

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. The idea is taken from 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’.

## 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; }

## Java

//JAVA code to find the element //that occur only once class GFG { // Method to find the element that occur only once static int getSingle(int arr[], int n) { int ones = 0, twos = 0; int common_bit_mask; for(int i=0; i<n; i++ ) { /*"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*/ twos = twos | (ones & arr[i]); /*"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*/ 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'*/ common_bit_mask = ~(ones & twos); /*Remove common bits (the bits that appear third time) from 'ones'*/ ones &= common_bit_mask; /*Remove common bits (the bits that appear third time) from 'twos'*/ twos &= common_bit_mask; } return ones; } // Driver method public static void main(String args[]) { int arr[] = {3, 3, 2, 3}; int n = arr.length; System.out.println("The element with single occurrence is " + getSingle(arr, n)); } } // Code contributed by Rishab Jain

## Python3

# Python3 code to find the element that # appears once def getSingle(arr, n): ones = 0 twos = 0 for i in range(n): # 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 twos = twos | (ones & arr[i]) # 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 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' common_bit_mask = ~(ones & twos) # Remove common bits (the bits that # appear third time) from 'ones' ones &= common_bit_mask # Remove common bits (the bits that # appear third time) from 'twos' twos &= common_bit_mask return ones # driver code arr = [3, 3, 2, 3] n = len(arr) print("The element with single occurrence is ", getSingle(arr, n)) # This code is contributed by "Abhishek Sharma 44"

Output:

The element with single occurrence is 2

Time Complexity: O(n)

Auxiliary Space: O(1)

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.

Let us consider the example array {5, 5, 5, 8}. The 101, 101, 101, 1000

Sum of first bits%3 = (1 + 1 + 1 + 0)%3 = 0;

Sum of second bits%3 = (0 + 0 + 0 + 0)%0 = 0;

Sum of third bits%3 = (1 + 1 + 1 + 0)%3 = 0;

Sum of fourth bits%3 = (1)%3 = 1;

Hence number which appears once is 1000

## C/C++

#include <stdio.h> #define INT_SIZE 32 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; } // Driver program to test above function int main() { int arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7}; int n = sizeof(arr) / sizeof(arr[0]); printf("The element with single occurrence is %d ", getSingle(arr, n)); return 0; }

## Java

// JAVA code to find the element // that occur only once class GFG { static final int INT_SIZE = 32; // Method to find the element that occur only once static int getSingle(int arr[], int n) { 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) == 0) sum++; } // The bits with sum not multiple of 3, are the // bits of element with single occurrence. if ((sum % 3) == 0) result |= x; } return result; } // Driver method public static void main(String args[]) { int arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7}; int n = arr.length; System.out.println("The element with single occurrence is " + getSingle(arr, n)); } } // Code contributed by Rishab Jain

The element with single occurrence is 7

Another approach suggested by *Abhishek Sharma 44*. Add each number once and multiply the sum by 3, we will get thrice the sum of each element of the array. Store it as thrice_sum. Subtract the sum of the whole array from the thrice_sum and divide the result by 2. The number we get is the required number (which appears once in the array).

Array [] : [a, a, a, b, b, b, c, c, c, d]

Mathematical Equation = ( 3*(a+b+c+d) – (a + a + a + b + b + b + c + c + c + d) ) / 2

In more simple words: ( 3*(sum_of_array_without_duplicates) – (sum_of_array) ) / 2

let arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 3} Required no = ( 3*(sum_of_array_without_duplicates) - (sum_of_array) ) / 2 = ( 3*(12 + 1 + 3 + 2) - (12 + 1 + 12 + 3 + 12 + 1 + 1 + 2 + 3 + 3))/2 = ( 3* 18 - 50) / 2 = (54 - 50) / 2 = 2 (required answer)

As we know that **set** does not contain any duplicate element, we will be using set here.

Below is the implementation of above approach:

# Python program to find the element # that occur only once # function which find number def singleNumber(nums): # applying the formula. return (3 * sum(set(nums)) - sum(nums)) / 2 # driver function. a = [12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7] print ("The element with single occurrence is ", int(singleNumber(a)))

Output:

The element with single occurrence is 7

This article is compiled by **Sumit Jain** and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.