# Find the element that appears once

• Difficulty Level : Hard
• Last Updated : 04 Feb, 2022

Given an array where every element occurs three times, except one element which occurs only once. Find the element that occurs once. The 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:
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 an odd number of times here. The idea is taken from here.
Run a loop for all elements in the array. At the end of every iteration, maintain the 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 the array, find out the common set bits in the new element and the previous value of ‘ones’. These common set bits are actually the bits that should be added to ‘twos’. So do bitwise XOR of the common set bits with ‘twos’. ‘twos’ also gets some extra bits that appear the third time. These extra bits are removed later.
Update ‘ones’ by doing XOR of new element with the previous value of ‘ones’. There may be some bits that 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 the above approach:

## C++

 `// C++ program to find the element` `// that occur only once` `#include ` `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 iterations 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);` `    ``cout << ``"The element with single occurrence is  "` `         ``<< getSingle(arr, n);` `    ``return` `0;` `}`   `// This code is contributed by rathbhupendra`

## C

 `// C program to find the element` `// that occur only once` `#include `   `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 iterations 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);` `    ``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 XOR` `        ``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 XOR` `        ``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"`

## C#

 `// C# code to find the element` `// that occur only once` `using` `System;` `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 code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { 3, 3, 2, 3 };` `        ``int` `n = arr.Length;` `        ``Console.WriteLine(``"The element with single"` `                          ``+ ``"occurrence is "` `+ getSingle(arr, n));` `    ``}` `}`   `// This Code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

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 the 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)%3 = 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

Note: this approach won’t work for negative numbers

Below is the implementation of the above approach:

## C++

 `// C++ program to find the element` `// that occur only once` `#include ` `using` `namespace` `std;` `#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) != 0)` `            ``result |= x;` `    ``}`   `    ``return` `result;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << ``"The element with single occurrence is "` `<< getSingle(arr, n);` `    ``return` `0;` `}`   `// This code is contributed by rathbhupendra`

## C

 `// C program to find the element` `// that occur only once` `#include ` `#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) != 0)` `            ``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);` `    ``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`

## Python3

 `# Python3 code to find the element ` `# that occur only once` `INT_SIZE ``=` `32`   `def` `getSingle(arr, n) :` `    `  `    ``# Initialize result` `    ``result ``=` `0` `    `  `    ``# Iterate through every bit` `    ``for` `i ``in` `range``(``0``, INT_SIZE) :` `        `  `        ``# Find sum of set bits ` `        ``# at ith position in all ` `        ``# array elements` `        ``sm ``=` `0` `        ``x ``=` `(``1` `<< i)` `        ``for` `j ``in` `range``(``0``, n) :` `            ``if` `(arr[j] & x) :` `                ``sm ``=` `sm ``+` `1` `                `  `        ``# The bits with sum not ` `        ``# multiple of 3, are the` `        ``# bits of element with ` `        ``# single occurrence.` `        ``if` `((sm ``%` `3``)!``=` `0``) :` `            ``result ``=` `result | x` `    `  `    ``return` `result` `    `  `# Driver program` `arr ``=` `[``12``, ``1``, ``12``, ``3``, ``12``, ``1``, ``1``, ``2``, ``3``, ``2``, ``2``, ``3``, ``7``]` `n ``=` `len``(arr)` `print``(``"The element with single occurrence is "``, getSingle(arr, n))`     `# This code is contributed ` `# by Nikita Tiwari.`

## C#

 `// C# code to find the element` `// that occur only once` `using` `System;`   `class` `GFG {` `    ``static` `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 Code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { 12, 1, 12, 3, 12, 1,` `                      ``1, 2, 3, 2, 2, 3, 7 };` `        ``int` `n = arr.Length;` `        ``Console.WriteLine(``"The element with single "` `                          ``+ ``"occurrence is "` `+ getSingle(arr, n));` `    ``}` `}`   `// This code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output

`The element with single occurrence is 7`

Time Complexity: O(n)

Auxiliary Space: O(1)

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

As we know that set does not contain any duplicate elements,
But, std::set is commonly implemented as a red-black binary search tree. Insertion on this data structure has a worst-case of O(log(n)) complexity, as the tree is kept balanced. we will be using set here.

Below is the implementation of above approach:

## C++

 `// C++ program to find the element` `// that occur only once`   `#include ` `using` `namespace` `std;`   `// function which find number` `int` `singleNumber(``int` `a[], ``int` `n)` `{` `    ``unordered_set<``int``> s(a, a + n);`   `    ``int` `arr_sum = accumulate(a, a + n, 0); ``// sum of array`   `    ``int` `set_sum = accumulate(s.begin(), s.end(), 0); ``// sum of set`   `    ``// applying the formula.` `    ``return` `(3 * set_sum - arr_sum) / 2;` `}`   `// driver function` `int` `main()` `{` `    ``int` `a[] = { 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 };`   `    ``int` `n = ``sizeof``(a) / ``sizeof``(a);`   `    ``cout << ``"The element with single occurrence is "` `<< singleNumber(a, n);` `}`   `// This code is contributed by Mohit Kumar 29 (IIIT gwalior)`

## Java

 `// Java program to find the element` `// that occur only once` `import` `java.util.*;`   `class` `GFG {`   `    ``// function which find number` `    ``static` `int` `singleNumber(``int` `a[], ``int` `n)` `    ``{` `        ``HashSet s = ``new` `HashSet();` `        ``for` `(``int` `i : a) {` `            ``s.add(i);` `        ``}`   `        ``int` `arr_sum = ``0``; ``// sum of array` `        ``for` `(``int` `i : a) {` `            ``arr_sum += i;` `        ``}`   `        ``int` `set_sum = ``0``; ``// sum of set` `        ``for` `(``int` `i : s) {` `            ``set_sum += i;` `        ``}`   `        ``// applying the formula.` `        ``return` `(``3` `* set_sum - arr_sum) / ``2``;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `a[] = { ``12``, ``1``, ``12``, ``3``, ``12``, ``1``, ``1``, ``2``, ``3``, ``2``, ``2``, ``3``, ``7` `};` `        ``int` `n = a.length;` `        ``System.out.println(``"The element with single "` `                           ``+ ``"occurrence is "` `+ singleNumber(a, n));` `    ``}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 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)))`

## C#

 `// C# program to find the element` `// that occur only once` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG {`   `    ``// function which find number` `    ``static` `int` `singleNumber(``int``[] a, ``int` `n)` `    ``{` `        ``HashSet<``int``> s = ``new` `HashSet<``int``>();` `        ``foreach``(``int` `i ``in` `a)` `        ``{` `            ``s.Add(i);` `        ``}`   `        ``int` `arr_sum = 0; ``// sum of array` `        ``foreach``(``int` `i ``in` `a)` `        ``{` `            ``arr_sum += i;` `        ``}`   `        ``int` `set_sum = 0; ``// sum of set` `        ``foreach``(``int` `i ``in` `s)` `        ``{` `            ``set_sum += i;` `        ``}`   `        ``// applying the formula.` `        ``return` `(3 * set_sum - arr_sum) / 2;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``int``[] a = { 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 };` `        ``int` `n = a.Length;` `        ``Console.WriteLine(``"The element with single "` `                          ``+ ``"occurrence is "` `+ singleNumber(a, n));` `    ``}` `}`   `// This code is contributed by PrinciRaj1992`

## PHP

 ``

## Javascript

 ``

Output

`The element with single occurrence is 7`

Time Complexity: O(Nlog(N))
Auxiliary Space: O(N)

### Method #4:Using Counter() function

• Calculate the frequency of array using Counter function
• Traverse in this Counter dictionary and check if any key has value 1
• If the value of any key is 1 return the key

Below is the implementation:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// function which find number` `int` `singlenumber(``int` `a[],``int` `N)` `{` `    ``// umap for finding frequency` `    ``unordered_map<``int``,``int``>fmap;` `  `  `    ``// traverse the array for frequency` `    ``for``(``int` `i = 0; i < N;i++)` `    ``{` `        ``fmap[a[i]]++;` `    ``}` `  `  `    ``// iterate over the map` `    ``for``(``auto` `it:fmap)` `    ``{` `      `  `        ``// check frequency whether it is one or not.` `        ``if``(it.second == 1)` `        ``{` `          `  `            ``// return it as we got the answer` `            ``return` `it.first;` `        ``}` `    ``}` `}`   `// Driver code` `int` `main()` `{` `  `  `    ``// given array` `    ``int` `a[]={12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7};` `  `  `    ``// size of the array` `    ``int` `N=``sizeof``(a)/``sizeof``(a);` `  `  `    ``// printing the returned value` `    ``cout << singlenumber(a,N);` `    ``return` `0;` `}`   `// This Code is contributed by ` `// Murarishetty Santhosh Charan`

## Java

 `// Java program for the above approach`   `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {` `    `  `    ``// function which find number` `    ``static` `int` `singlenumber(``int` `a[],``int` `N)` `    ``{` `        ``// umap for finding frequency` `        ``Map fmap` `            ``= ``new` `HashMap();` `            `  `        `  `        ``// traverse the array for frequency` `        ``for``(``int` `i = ``0``; i < N;i++)` `        ``{` `            ``if``(!fmap.containsKey(a[i]))` `                ``fmap.put(a[i],``0``);` `             `  `            ``fmap.put(a[i],fmap.get(a[i])+``1``);` `        ``}` `        `  `        ``// iterate over the map` `        ``for``(Map.Entry me : fmap.entrySet())` `        ``{` `            `  `            ``// check frequency whether it is one or not.` `            ``if``(me.getValue()==``1``)` `            ``{` `                `  `                ``// return it as we got the answer` `                ``return` `me.getKey();` `            ``}` `        ``}` `        ``return` `-``1``;` `        `  `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main (String[] args) {` `        `  `        `  `        ``// given array` `    ``int` `a[]={``12``, ``1``, ``12``, ``3``, ``12``, ``1``, ``1``, ``2``, ``3``, ``2``, ``2``, ``3``, ``7``};` `    `  `    ``// size of the array` `    ``int` `N= a.length;` `    `  `    `  `        ``// printing the returned value` `        ``System.out.println(``"The element with single occurrence is "``+singlenumber(a,N));` `    ``}` `}`   `// This code is contributed by avanitrachhadiya2155`

## Python3

 `from` `collections ``import` `Counter` `# Python3 program to find the element` `# that occur only once`   `# function which find number` `def` `singleNumber(nums):` `  `  `    ``# storing the frequencies using Counter` `    ``freq ``=` `Counter(nums)` `    `  `    ``# traversing the Counter dictionary` `    ``for` `i ``in` `freq:` `      `  `        ``# check if any value is 1` `        ``if``(freq[i] ``=``=` `1``):` `            ``return` `i`     `# 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)))` `# This code is contributed by vikkycirus`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `using` `System.Linq;`   `public` `class` `GFG {` `    `  `// function which find number` `static` `void` `singlenumber(``int``[] a, ``int` `N)` `{` `  `  `    ``// umap for finding frequency` `    ``Dictionary<``int``, ``int``> fmap = ``new` `Dictionary<``int``, ``int``>();`   `    ``// traverse the array for frequency` `     ``for` `(``int` `i = 0; i < N; i++)` `        ``{` `            ``if` `(fmap.ContainsKey(a[i]))` `                ``fmap[a[i]]++;` `            ``else` `                ``fmap.Add(a[i], 1);` `        ``}`   `  `  `    ``// iterate over the map` `    ``foreach` `(``int` `it ``in` `fmap.Keys.ToList())` `    ``{` `      `  `        ``// check frequency whether it is one or not.` `        ``if``(fmap[it] == 1)` `        ``{` `          `  `            ``// return it as we got the answer` `            ``Console.Write(``"The element with single occurrence is "` `+ it);` `        ``}` `    ``}` `}`   `// Driver Code` `public` `static` `void` `Main (``string``[] args) {` `    `  `     ``// given array` `    ``int``[] arr = {12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7};` `    `  `    ``// size of the array` `    ``int` `n= arr.Length;` `    `  `    `  `    ``// printing the returned value` `    ``singlenumber(arr, n);` `}` `}`   `// This code is contributed by splevel62.`

## Javascript

 ``

Output

`The element with single occurrence is  7`

Time Complexity: O(n)

Auxiliary Space: O(n)