Non repeating numbers using C++

Author: raunakpandey1

Bit Manipulation/Non_Repeating_Numbers.cpp

@@ -0,0 +1,72 @@
+
+/*
+Given an array A containing 2*N+2 positive numbers, out of which 2*N numbers exist in pairs whereas the other two number occur exactly once and are distinct. Find the other two numbers.
+
+
+Example 1:
+
+Input: 
+N = 2
+arr[] = {1, 2, 3, 2, 1, 4}
+Output:
+3 4 
+Explanation:
+3 and 4 occur exactly once.*/
+#include<bits/stdc++.h>
+using namespace std;
+ 
+
+class Solution
+{
+public:
+    vector<int> singleNumber(vector<int> nums) 
+    {
+        // Code here.
+        int res=0;
+        vector <int> vec,v;
+        for(int i=0;i<nums.size();i++)
+        {
+            res^=nums[i];
+            
+        }
+        int set_bit_no = res & ~(res-1);
+        int temp=res;
+            for(int i=0;i<nums.size();i++)
+            {
+                if((nums[i]& set_bit_no))
+                {
+                    vec.push_back(nums[i]);
+                }
+            }
+            for(int i=0;i<vec.size();i++)
+            {
+               temp^=vec[i]; 
+            }
+            v.push_back(temp);
+            res^=temp;
+            v.push_back(res);
+            sort(v.begin(),v.end());
+            return v;
+            
+    }
+};
+
+// { Driver Code Starts.
+int main(){
+    int T;
+    cin >> T;
+    while(T--)
+    {
+    	int n; 
+    	cin >> n;
+    	vector<int> v(2 * n + 2);
+    	for(int i = 0; i < 2 * n + 2; i++)
+    		cin >> v[i];
+    	Solution ob;
+    	vector<int > ans = ob.singleNumber(v);
+    	for(auto i: ans)
+    		cout << i << " ";
+    	cout << "\n";
+    }
+	return 0;
+}  // } Driver Code Ends