How to Solve Lonely Integer by Using Dictionary or Bit Manipulation

SOLUTION IN C# USING DICTIONARY

    public static int lonelyinteger(List<int> a)

    {

      Dictionary<int,int> mydict = new Dictionary<int,int>();

      int element = 0;

      foreach(int i in a){

          if(mydict.ContainsKey(i)){

              mydict[i] += 1;

          }

          else{

              mydict.Add(i, 1);

          }

          

      }

      foreach(KeyValuePair<int,int> k in mydict){

          if(mydict[k.Key] == 1){

              element = k.Key;

          }

      }

      return element;

SOLUTION IN C# USING BIT MANIPULATION : the best option 

public static int lonelyinteger(List<int> a)

    {

      int result=0;

      foreach(int i in a){

          result ^=i;

      }

      return result;

    }

 PROBLEM:

Given an array of integers, where all elements but one occur twice, find the unique element.

Example

The unique element is .

Function Description

Complete the lonelyinteger function in the editor below.

lonelyinteger has the following parameter(s):

• int a[n]: an array of integers

Returns

• int: the element that occurs only once

Input Format

The first line contains a single integer, , the number of integers in the array.
The second line contains  space-separated integers that describe the values in .

Constraints

• It is guaranteed that  is an odd number and that there is one unique element.
• , where .

Sample Input 0

1
1

Sample Output 0

1

Explanation 0

There is only one element in the array, thus it is unique.

Sample Input 1

3
1 1 2

Sample Output 1

2

Explanation 1

We have two 's, and  is unique.

Sample Input 2

5
0 0 1 2 1

Sample Output 2

2

Explanation 2

We have two 's, two 's, and one .  is unique.