Updates code to Swift 4. Refactored a bit of the readme.

Author: kelvinlauKL

Egg Drop Problem/EggDrop.playground/Contents.swift

@@ -1,9 +1,12 @@
- //: Playground - noun: a place where people can play
+#if swift(>=4.0)
+  print("Hello, Swift 4!")
+#endif
 
-import Cocoa
+drop(numberOfEggs: 2, numberOfFloors: 2) //expected answer: 2
+drop(numberOfEggs: 2, numberOfFloors: 4) //expected answer: 3
+drop(numberOfEggs: 2, numberOfFloors: 6) //expected answer: 3
+drop(numberOfEggs: 5, numberOfFloors: 5) //expected answer: 2
+drop(numberOfEggs: 5, numberOfFloors: 20) //expected answer: 4
 
-eggDrop(numberOfEggs: 2, numberOfFloors: 2) //expected answer: 2
-eggDrop(numberOfEggs: 2, numberOfFloors: 4) //expected answer: 3
-eggDrop(numberOfEggs: 2, numberOfFloors: 6) //expected answer: 3
-eggDrop(numberOfEggs: 5, numberOfFloors: 5) //expected answer: 2
-eggDrop(numberOfEggs: 5, numberOfFloors: 20) //expected answer: 4
+drop(numberOfEggs: 2, numberOfFloors: 36)
+drop(numberOfEggs: 2, numberOfFloors: 10)

Egg Drop Problem/EggDrop.playground/Sources/EggDrop.swift

@@ -1,39 +1,27 @@
-public func eggDrop(numberOfEggs: Int, numberOfFloors: Int) -> Int {
-    if numberOfEggs == 0 || numberOfFloors == 0{ //edge case: When either number of eggs or number of floors is 0, answer is 0
-        return 0
-    }
-    if numberOfEggs == 1 || numberOfFloors == 1{ //edge case: When either number of eggs or number of floors is 1, answer is 1
-        return 1
-    }
-    
-    var eggFloor = [[Int]](repeating: [Int](repeating: 0, count: numberOfFloors+1), count: numberOfEggs+1) //egg(rows) floor(cols) array to store the solutions
-    var attempts: Int = 0
-    
-    for var floorNumber in (0..<(numberOfFloors+1)){
-        eggFloor[1][floorNumber] = floorNumber      //base case: if there's only one egg, it takes 'numberOfFloors' attempts
-    }
-    eggFloor[2][1] = 1 //base case: if there are two eggs and one floor, it takes one attempt
-    
-    for var eggNumber in (2..<(numberOfEggs+1)){
-        for var floorNumber in (2..<(numberOfFloors+1)){
-            eggFloor[eggNumber][floorNumber] = Int.max   //setting the final result a high number to find out minimum
-            for var visitingFloor in (1..<(floorNumber+1)){
-                //there are two cases
-                //case 1: egg breaks. meaning we'll have one less egg, and we'll have to go downstairs -> visitingFloor-1
-                //case 2: egg doesn't break. meaning we'll still have 'eggs' number of eggs, and we'll go upstairs -> floorNumber-visitingFloor
-                attempts = 1 + max(eggFloor[eggNumber-1][visitingFloor-1], eggFloor[eggNumber][floorNumber-visitingFloor])//we add one taking into account the attempt we're taking at the moment
-                
-                if attempts < eggFloor[eggNumber][floorNumber]{ //finding the min
-                    eggFloor[eggNumber][floorNumber] = attempts;
-                }
-            }
+public func drop(numberOfEggs: Int, numberOfFloors: Int) -> Int {
+  guard numberOfEggs != 0 && numberOfFloors != 0 else { return 0 }
+  guard numberOfEggs != 1 && numberOfFloors != 1 else { return 1 }
+  
+  var eggFloor: [[Int]] = .init(repeating: .init(repeating: 0, count: numberOfFloors + 1), count: numberOfEggs + 1)
+  var attempts = 0
+  
+  for floorNumber in stride(from: 0, through: numberOfFloors, by: 1) {
+    eggFloor[1][floorNumber] = floorNumber
+  }
+  eggFloor[2][1] = 1
+  
+  for eggNumber in stride(from: 2, through: numberOfEggs, by: 1) {
+    for floorNumber in stride(from: 2, through: numberOfFloors, by: 1) {
+      eggFloor[eggNumber][floorNumber] = Int.max
+      for visitingFloor in stride(from: 1, through: floorNumber, by: 1) {
+        attempts = 1 + max(eggFloor[eggNumber - 1][visitingFloor - 1], eggFloor[eggNumber][floorNumber - visitingFloor])
+        
+        if attempts < eggFloor[eggNumber][floorNumber] {
+          eggFloor[eggNumber][floorNumber] = attempts
         }
+      }
     }
-    
-    return eggFloor[numberOfEggs][numberOfFloors]
-}
-
-//Helper function to find max of two integers
-public func max(_ x1: Int, _ x2: Int) -> Int{
-    return x1 > x2 ? x1 : x2
+  }
+  
+  return eggFloor[numberOfEggs][numberOfFloors]
 }

Egg Drop Problem/EggDrop.playground/playground.xcworkspace/contents.xcworkspacedata

@@ -12,7 +12,7 @@ If the object is incredibly resilient, and you may need to do the testing on the
 
 ## Description
 
-You're in a building with **m** floors and you are given **n** eggs. What is the minimum number attempts it will take to find out the lowest floor that breaks the egg?
+You're in a building with **m** floors and you are given **n** eggs. What is the maximum number of attempts it will take to find out the floor that breaks the egg?
 
 For convenience, here are a few rules to keep in mind:
 
@@ -22,9 +22,8 @@ For convenience, here are a few rules to keep in mind:
 - If an egg breaks, then it would break if dropped from a higher floor.
 - If an egg survives, then it would survive a shorter fall.
 
-The problem is not actually to find the critical floor, but merely to decide floors from which eggs should be dropped so that total number of trials are minimized.
-
 ## Solution
+
 - eggNumber -> Number of eggs at the moment
 - floorNumber -> Floor number at the moment
 - visitingFloor -> Floor being visited at the moment