New Problem Solution - "1884. Egg Drop With 2 Eggs and N Floors"

README.md

@@ -12,6 +12,7 @@ LeetCode
 |1946|[Largest Number After Mutating Substring](https://leetcode.com/problems/largest-number-after-mutating-substring/) | [C++](./algorithms/cpp/largestNumberAfterMutatingSubstring/LargestNumberAfterMutatingSubstring.cpp)|Medium|
 |1945|[Sum of Digits of String After Convert](https://leetcode.com/problems/sum-of-digits-of-string-after-convert/) | [C++](./algorithms/cpp/leetcode/sumOfDigitsOfStringAfterConvert/SumOfDigitsOfStringAfterConvert.cpp)|Easy|
 |1935|[Maximum Number of Words You Can Type](https://leetcode.com/problems/maximum-number-of-words-you-can-type/) | [C++](./algorithms/cpp/maximumNumberOfWordsYouCanType/MaximumNumberOfWordsYouCanType.cpp)|Easy|
+|1884|[Egg Drop With 2 Eggs and N Floors](https://leetcode.com/problems/egg-drop-with-2-eggs-and-n-floors/) | [C++](./algorithms/cpp/eggDropWith2EggsAndNFloors/EggDropWith2EggsAndNFloors.cpp)|Medium|
 |1882|[Process Tasks Using Servers](https://leetcode.com/problems/process-tasks-using-servers/) | [C++](./algorithms/cpp/processTasksUsingServers/ProcessTasksUsingServers.cpp)|Medium|
 |1881|[Maximum Value after Insertion](https://leetcode.com/problems/maximum-value-after-insertion/) | [C++](./algorithms/cpp/maximumValueAfterInsertion/MaximumValueAfterInsertion.cpp)|Medium|
 |1880|[Check if Word Equals Summation of Two Words](https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/) | [C++](./algorithms/cpp/checkIfWordEqualsSummationOfTwoWords/CheckIfWordEqualsSummationOfTwoWords.cpp)|Easy|

algorithms/cpp/eggDropWith2EggsAndNFloors/EggDropWith2EggsAndNFloors.cpp

@@ -0,0 +1,89 @@
+// Source : https://leetcode.com/problems/egg-drop-with-2-eggs-and-n-floors/
+// Author : Hao Chen
+// Date   : 2021-11-15
+
+/***************************************************************************************************** 
+ *
+ * You are given two identical eggs and you have access to a building with n floors labeled from 1 to 
+ * n.
+ * 
+ * You know that there exists a floor f where 0 <= f <= n such that any egg dropped at a floor higher 
+ * than f will break, and any egg dropped at or below floor f will not break.
+ * 
+ * In each move, you may take an unbroken egg and drop it from any floor x (where 1 <= x <= n). If the 
+ * egg breaks, you can no longer use it. However, if the egg does not break, you may reuse it in 
+ * future moves.
+ * 
+ * Return the minimum number of moves that you need to determine with certainty what the value of f is.
+ * 
+ * Example 1:
+ * 
+ * Input: n = 2
+ * Output: 2
+ * Explanation: We can drop the first egg from floor 1 and the second egg from floor 2.
+ * If the first egg breaks, we know that f = 0.
+ * If the second egg breaks but the first egg didn't, we know that f = 1.
+ * Otherwise, if both eggs survive, we know that f = 2.
+ * 
+ * Example 2:
+ * 
+ * Input: n = 100
+ * Output: 14
+ * Explanation: One optimal strategy is:
+ * - Drop the 1st egg at floor 9. If it breaks, we know f is between 0 and 8. Drop the 2nd egg starting
+ *   from floor 1 and going up one at a time to find f within 7 more drops. Total drops is 1 + 7 = 8.
+ * - If the 1st egg does not break, drop the 1st egg again at floor 22. If it breaks, we know f is 
+ * between 9
+ *   and 21. Drop the 2nd egg starting from floor 10 and going up one at a time to find f within 12 
+ * more
+ *   drops. Total drops is 2 + 12 = 14.
+ * - If the 1st egg does not break again, follow a similar process dropping the 1st egg from floors 
+ * 34, 45,
+ *   55, 64, 72, 79, 85, 90, 94, 97, 99, and 100.
+ * Regardless of the outcome, it takes at most 14 drops to determine f.
+ * 
+ * Constraints:
+ * 
+ * 	1 <= n <= 1000
+ ******************************************************************************************************/
+
+/*
+ 1 floor  - 1 attemp
+ 2 floors - 2 attemps
+ 3 floors - 2 attemps
+ 4 floors - 3 attemps
+ 5 floors - 3 attemps
+ 6 floors - 3 attemps
+ 7 floors - 4 attemps
+ ...
+ 11 floors - 5 attemps
+ ...
+ 16 floors - 6 attemps
+ ...
+ 22 floors - 7 attemps
+ ...
+ 
+ we can see: 
+ floor(1) = 1
+ floor(2) = 2
+ floor(4) = 3
+ floor(7) = 4
+ floor(11) = 5
+ floor(16) = 6
+ floor(22) = 7
+ ...
+ 
+ */
+
+
+class Solution {
+public:
+    int twoEggDrop(int n) {
+        int c = 1, i=1;
+        while(c < n) {
+            c += i;
+            i++;
+        }
+        return c==n ? i : i-1;
+    }
+};