Add solution to day 22

05- May/22- Top K Frequent Elements/22- Top K Frequent Elements (Ahmed Gamal).cpp

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+// Author: Ahmed Gamal
+
+// for this problem we will use bucket sort to sort the elements according to their frequency (we will use counting sort to count the frequency of each element)
+// we can use counting sort because the elements are in a range from -10^4 to 10^4 so we can use an array of size 2*10^4+1 to count the frequency of each element
+
+// after counting the frequency of each element we will put each element in a bucket according to its frequency
+// then we will start from the end of the buckets (the bucket with the highest frequency) and we will put the elements in the answer vector until we reach the k elements
+
+class Solution {
+public:
+    vector<int> topKFrequent(vector<int>& nums, int k) {
+        // sf: shift factor (to shift the elements to the right to make them non-negative)
+        // freq: array to count the frequency of each element
+        const int sf = 1e4;
+        vector<int> freq(sf << 1 | 1);
+
+        // counting frequency of each element
+        for(auto& i : nums) {
+            freq[i + sf]++;
+        }
+
+        // c: buckets to put the elements in according to their frequency
+        vector<vector<int>> c(sf + 1);
+
+        // putting the elements in the buckets according to their frequency
+        // remember that the elements are shifted to the right by sf so we need to shift them back to the left
+        for(int i = 0; i < freq.size(); i++) {
+            c[freq[i]].emplace_back(i - sf);
+        }
+
+        // ans: the answer vector
+        // i: the index of the bucket we are currently in
+        vector<int> ans;
+        int i = c.size() - 1;
+
+        // putting the elements in the answer vector until we reach the k elements
+        while(ans.size() < k) {
+            // if the bucket is empty we will move to the previous bucket, otherwise we will put the elements in the answer vector
+            while(not c[i].empty() and ans.size() < k) {
+                ans.emplace_back(c[i].back());
+                c[i].pop_back();
+            }
+            if(c[i].empty()) {
+                --i;
+            }
+        }
+
+        return ans;
+    }
+};