added topKFrequentElements

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|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|
 |337|[House Robber III](https://leetcode.com/problems/house-robber-iii/) | [C++](./algorithms/cpp/houseRobber/houseRobberIII.cpp)|Medium|
 |334|[Increasing Triplet Subsequence](https://leetcode.com/problems/increasing-triplet-subsequence/) | [C++](./algorithms/cpp/increasingTripletSubsequence/increasingTripletSubsequence.cpp)|Medium|

algorithms/cpp/topKFrequentElements/topKFrequentElements.cpp

@@ -0,0 +1,65 @@
+// Source : https://leetcode.com/problems/top-k-frequent-elements/
+// Author : Calinescu Valentin
+// Date   : 2016-05-02
+
+/*************************************************************************************** 
+ *
+ * Given a non-empty array of integers, return the k most frequent elements.
+ * 
+ * For example,
+ * Given [1,1,1,2,2,3] and k = 2, return [1,2].
+ * 
+ * Note: 
+ * You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
+ * Your algorithm's time complexity must be better than O(n log n), where n is the 
+ * array's size.
+ * 
+ ***************************************************************************************/
+
+class Solution {
+public:
+    struct element//structure consisting of every distinct number in the vector,
+    //along with its frequency
+    {
+        int number, frequency;
+        bool operator < (const element arg) const
+        {
+            return frequency < arg.frequency;
+        }
+    };
+    priority_queue <element> sol;//we use a heap so we have all of the elements sorted
+    //by their frequency
+    vector <int> solution;
+
+    vector<int> topKFrequent(vector<int>& nums, int k) {
+        sort(nums.begin(), nums.end());
+        int i = 1;
+        for(; i < nums.size(); i++)
+        {
+            int freq = 1;
+            while(i < nums.size() && nums[i] == nums[i - 1])
+            {
+                i++;
+                freq++;
+            }
+            element el;
+            el.number = nums[i - 1];
+            el.frequency = freq;
+            sol.push(el);
+        }
+        if(i == nums.size())//if we have 1 distinct element as the last
+        {
+            element el;
+            el.number = nums[nums.size() - 1];
+            el.frequency = 1;
+            sol.push(el);
+        }
+        while(k)//we extract the first k elements from the heap
+        {
+            solution.push_back(sol.top().number);
+            sol.pop();
+            k--;
+        }
+        return solution;
+    }
+};