# Leetcode.Solved.SingleNumber

``````/*
Approach
Using Bit Manipulation -

As we know XOR operation with 0 gives the same number
i.e, a XOR 0 = a
eg, for decimal no. 2=> 2 XOR 0 = 2
in binary, 010 XOR 000 = 010

Also we know that , XOR operation with same number gives 0
i.e, a XOR a = 0
eg, 2 XOR 2 = 0
in binary, 010 XOR 010 = 000

XOR is associative (like sum)
i.e, (2 XOR 3) XOR 4 = 2 XOR (3 XOR 4), So the order doesn't matter in performing XOR operation.
eg, 2^3^4^6 = 3^2^6^4 = 4^2^6^3 ......

So, using these three properties of XOR , we will solve the question. we will take ans variable with 0 as initial value. And then for each element i in array, we will perform the XOR operation of the element with 0, ans will become 0 if the same number is found (as a XOR a = 0) and so after the completion of the loop, only element with no duplicate number will remain and will be returned as ans.
*/

// https://leetcode.com/problems/single-number/description/
public class Solution {
public int SingleNumber(int[] nums) {
int ans=0; //since XOR with 0 returns same number
for(int i=0; i < nums.Length; i++){
ans ^= nums[i];  // ans = (ans) XOR (array element at i)
}
return ans;
}
}``````
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