added intersectionOfTwoArraysII

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|350|[Intersection of Two Arrays II](https://leetcode.com/problems/intersection-of-two-arrays-ii/) | [C++](./algorithms/cpp/intersectionOfTwoArraysII/intersectionOfTwoArraysII.cpp)|Easy|
 |349|[Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays/) | [C++](./algorithms/cpp/intersectionOfTwoArrays/intersectionOfTwoArrays.cpp)|Easy|
 |347|[Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/) | [C++](./algorithms/cpp/topKFrequentElements/topKFrequentElements.cpp)|Medium|
 |345|[Reverse Vowels of a String](https://leetcode.com/problems/reverse-vowels-of-a-string/) | [C++](./algorithms/cpp/reverseVowelsOfAString/reverseVowelsOfAString.cpp)|Easy|

algorithms/cpp/intersectionOfTwoArraysII/intersectionOfTwoArraysII.cpp

@@ -0,0 +1,61 @@
+// Source : https://leetcode.com/problems/intersection-of-two-arrays-ii/
+// Author : Calinescu Valentin
+// Date   : 2016-05-22
+
+/*************************************************************************************** 
+ *
+ * Given two arrays, write a function to compute their intersection.
+ * 
+ * Example:
+ * Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].
+ * 
+ * Note:
+ * Each element in the result should appear as many times as it shows in both arrays.
+ * The result can be in any order.
+ * 
+ * Follow up:
+ * What if the given array is already sorted? How would you optimize your algorithm?
+ * What if nums1's size is small compared to num2's size? Which algorithm is better?
+ * What if elements of nums2 are stored on disk, and the memory is limited such that you
+ * cannot load all elements into the memory at once?
+ * 
+ ***************************************************************************************/
+
+ /* Solution
+  * --------
+  *
+  * Follow up:
+  * 
+  * 1)If the given array is already sorted we can skip the sorting.
+  * 
+  * 2)If nums1 is significantly smaller than nums2 we can only sort nums1 and then binary
+  * search every element of nums2 in nums1 with a total complexity of (MlogN) or if nums2
+  * is already sorted we can search every element of nums1 in nums2 in O(NlogM)
+  *
+  * 3)Just like 2), we can search for every element in nums2, thus having an online
+  * algorithm.
+  */
+
+class Solution { // O(NlogN + MlogM)
+public:
+    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
+        sort(nums1.begin(), nums1.end());//we sort both vectors in order to intersect
+        sort(nums2.begin(), nums2.end());//them later in O(N + M), where N = nums1.size()
+        vector <int> solution;                                         //M = nums2.size() 
+        int index = 0;
+        bool finished = false;
+        for(int i = 0; i < nums1.size() && !finished; i++)
+        {
+            while(index < nums2.size() && nums1[i] > nums2[index])//we skip over the
+                index++;//smaller elements in nums2
+            if(index == nums2.size())//we have reached the end of nums2 so we have no more
+                finished = true;//elements to add to the intersection
+            else if(nums1[i] == nums2[index])//we found a common element
+            {
+                solution.push_back(nums1[i]);
+                index++;
+            }
+        }
+        return solution;
+    }
+};