diff --git a/Week2/349_Intersection_of_2_arrays.cpp b/Week2/349_Intersection_of_2_arrays.cpp
new file mode 100644
index 0000000..06cdd35
--- /dev/null
+++ b/Week2/349_Intersection_of_2_arrays.cpp
@@ -0,0 +1,47 @@
+class Solution {
+public:
+    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
+        //if (nums1.size() > nums2.size()) return intersection(nums2, nums1);
+
+	// The idea here is to use some hash table data structure (set or map).
+	// This table shows the presense of element in first array.	
+	// Then while traversing 2nd array we update answer if given element is present in table(first array),
+	// update table by removing current element. 
+
+        vector<int> res;
+        unordered_map<int, bool> exist;
+        for (int i = 0; i < nums1.size(); ++i) {
+            exist[nums1[i]] = true;
+        }
+        for (int i = 0; i < nums2.size(); ++i) {
+            if (exist.find(nums2[i]) != exist.end() && exist[nums2[i]]) {
+                res.push_back(nums2[i]);
+                exist[nums2[i]] = false;
+            }
+        }
+        return res;
+    }
+};
+
+/*
+    nums1.size() = m, u - number of unique elements in nums1
+    nums2.size() = n
+    Time complexity T=O(max(m, n)) in average, worst case is T=O(m * max(m, n)).
+    Memory complexity M = O(u)
+
+*/
+
+/*
+    	[]
+	[]
+
+    	[1,2,3]
+	[]
+
+	[4]
+	[1,2,3]
+	
+	[1,2,2,2,2,3,3,3,1]
+	[1,2,5]	
+
+*/