New Problem Solution "933. Number of Recent Calls"

Author: haoel

README.md

@@ -39,6 +39,7 @@ LeetCode
 |951|[Flip Equivalent Binary Trees](https://leetcode.com/problems/flip-equivalent-binary-trees/) | [Python](./algorithms/python/FlipEquivalentBinaryTrees/flipEquiv.py)|Medium|
 |950|[Reveal Cards In Increasing Order](https://leetcode.com/problems/reveal-cards-in-increasing-order/) | [Python](./algorithms/python/RevealCardsInIncreasingOrder/deckRevealedIncreasing.py)|Medium|
 |941|[Valid Mountain Array](https://leetcode.com/problems/valid-mountain-array/) | [Python](./algorithms/python/ValidMountainArray/validMountainArray.py)|Easy|
+|933|[Number of Recent Calls](https://leetcode.com/problems/number-of-recent-calls/) | [C++](./algorithms/cpp/numberOfRecentCalls/NumberOfRecentCalls.cpp)|Easy|
 |931|[Minimum Falling Path Sum](https://leetcode.com/problems/minimum-falling-path-sum/) | [C++](./algorithms/cpp/minimumFallingPathSum/MinimumFallingPathSum.cpp)|Medium|
 |922|[Sort Array By Parity II](https://leetcode.com/problems/sort-array-by-parity-ii/) | [C++](./algorithms/cpp/sortArrayByParity/SortArrayByParity.II.cpp)|Easy|
 |914|[X of a Kind in a Deck of Cards](https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/) | [Python](./algorithms/python/XOfAKindInADeckOfCards/hasGroupsSizeX.py)|Easy|

algorithms/cpp/numberOfRecentCalls/NumberOfRecentCalls.cpp

@@ -0,0 +1,59 @@
+// Source : https://leetcode.com/problems/number-of-recent-calls/
+// Author : Hao Chen
+// Date   : 2020-07-26
+
+/***************************************************************************************************** 
+ *
+ * Write a class RecentCounter to count recent requests.
+ * 
+ * It has only one method: ping(int t), where t represents some time in milliseconds.
+ * 
+ * Return the number of pings that have been made from 3000 milliseconds ago until now.
+ * 
+ * Any ping with time in [t - 3000, t] will count, including the current ping.
+ * 
+ * It is guaranteed that every call to ping uses a strictly larger value of t than before.
+ * 
+ * Example 1:
+ * 
+ * Input: inputs = ["RecentCounter","ping","ping","ping","ping"], inputs = [[],[1],[100],[3001],[3002]]
+ * Output: [null,1,2,3,3]
+ * 
+ * Note:
+ * 
+ * 	Each test case will have at most 10000 calls to ping.
+ * 	Each test case will call ping with strictly increasing values of t.
+ * 	Each call to ping will have 1 <= t <= 10^9.
+ * 
+ ******************************************************************************************************/
+
+class RecentCounter {
+public:
+    RecentCounter() {
+
+    }
+
+    int ping(int t) {
+        req.push_back(t);
+        return req.size() - binary_search(t-3000);
+    }
+private:
+    vector<int> req;
+    int binary_search(int x) {
+        int low=0, high=req.size()-1;
+        while(low < high) {
+            int mid = low + (high -low) / 2;
+            if ( req[mid] == x ) return mid;
+            if ( req[mid] < x ) low = mid + 1;
+            else high = mid - 1;
+        }
+        cout << "x="  << x << "\tlow=" << low << endl;
+        return x > req[low] ? low+1 : low ;
+    }
+};
+
+/**
+ * Your RecentCounter object will be instantiated and called as such:
+ * RecentCounter* obj = new RecentCounter();
+ * int param_1 = obj->ping(t);
+ */