Last updated on August 5th, 2024 at 01:03 am

Here, We see ** Single Number problem Solution**. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

*List of all LeetCode Solution*

*List of all LeetCode Solution*

## Topics

Bit-Manipulation, Hash Table

## Companies

Airbnb, Palantir

## Level of Question

Easy

**Single Number LeetCode Solution**

## Table of Contents

**Problem Statement**

Given a **non-empty** array of integers ** nums**, every element appears

*twice*except for one. Find that single one.

Example 1:Input: nums = [2,2,1]Output: 1Example 2:Input: nums = [4,1,2,1,2]Output: 4Example 3:Input: nums = [1]Output: 1

**1. Single Number Leetcode Solution C++**

class Solution { public: int singleNumber(vector<int>& nums) { sort(nums.begin(),nums.end()); for(int i=1;i<nums.size();i+=2) { if(nums[i]!=nums[i-1]) return nums[i-1]; } return nums[nums.size()-1]; } };

#### 1.1 Explanation

**Sort the Array:**The array**nums**is sorted in ascending order.**Iterate Through Sorted Array:**Iterate through the sorted array in steps of 2. For each pair of elements, check if they are equal. If a pair of elements is not equal, return the first element of that pair (the single number).**Return the Last Element:**If no single element is found in the loop, the single number is the last element in the array.

#### 1.2 Time Complexity

**O(n log n)**.

#### 1.3 Space Complexity

**O(1)**.

**2. Single Number Leetcode Solution Java**

class Solution { public int singleNumber(int[] nums) { int res = 0; for (int i = 0; i < nums.length; i++) { res = res^nums[i]; } return res; } }

#### 2.1 Explanation

**Initialize Result:**Initialize a variable**res**to 0.- XOR Operation: Iterate through each element in the array and perform the XOR operation with
**res**. - Return Result: The result will be the single number because the XOR of two identical numbers is 0, and the XOR of a number with 0 is the number itself.

#### 2.2 Time Complexity

**O(n)**.

#### 2.3 Space Complexity

**O(1)**.

**3. Single Number Leetcode Solution JavaScript**

var singleNumber = function(nums) { let uniqNum = 0; for (let idx = 0; idx < nums.length; idx++) { uniqNum = uniqNum ^ nums[idx]; } return uniqNum; };

#### 3.1 Explanation

**Initialize Unique Number:**Initialize a variable**uniqNum**to 0.**XOR Operation**: Iterate through each element in the array and perform the XOR operation with**uniqNum**.**Return Unique Numbe**r: The result will be the single number for the same reason as explained in the Java code.

#### 3.2 Time Complexity

**O(n)**.

#### 3.3 Space Complexity

**O(1)**.

**4. Single Number Solution Python**

class Solution(object): def singleNumber(self, nums): return reduce(operator.xor, nums)

#### 4.1 Explanation

**Reduce Function with XOR:**

- The
**reduce**function applies the XOR operation cumulatively to the elements of the array**nums**. **operator.xor**is a function that performs the XOR operation between two numbers.

#### 4.2 Time Complexity

**O(n)**.

#### 4.3 Space Complexity

**O(1)**.