Imperative and functional-programming solutions for Jumping on the Cloud challenge

Question

Problem Statement

Emma is playing a new mobile game that starts with consecutively numbered clouds. Some of the clouds are thunderheads and others are cumulus. She can jump on any cumulus cloud having a number that is equal to the number of the current cloud plus 1 or 2. She must avoid the thunderheads. Determine the minimum number of jumps it will take Emma to jump from her starting postion to the last cloud. It is always possible to win the game.

For each game, Emma will get an array of clouds numbered 0 if they are safe or 1 if they must be avoided. For example, c=[0,1,0,0,0,1,0] indexed from 0...6. The number on each cloud is its index in the list so she must avoid the clouds at indexes 1 and 5 . She could follow the following two paths:
0 -> 2 -> 4 -> 6 or 0 -> 2 -> 3 -> 3 -> 4 -> 6 . The first path takes 3 jumps while the second takes 4.

Function Description

Complete the jumpingOnClouds function in the editor below. It should return the minimum number of jumps required, as an integer.

jumpingOnClouds has the following parameter(s):

• c: an array of binary integers

Input Format

The first line contains an integer n, the total number of clouds. The second line contains n space-separated binary integers describing clouds c[i] where 0 <= i < n.

Output Format

Print the minimum number of jumps needed to win the game.

Sample Input 0

7
0 0 1 0 0 1 0

Sample Output 0

4

Explanation 0: Emma must avoid c[2] and c[5]. She can win the game with a minimum of 4 jumps. Sample Input 1

6
0 0 0 0 1 0

Sample Output 1

3

Explanation 1: The only thundercloud to avoid is c[4]. Emma can win the game in 3 jumps.

Imperative style Solution:

// Complete the jumpingOnClouds function below.
def jumpingOnClouds(c: Array[Int]): Int = {
    var i=0 
    var length = c.length
    var jumps = 0 
    while(i < length -1) { 
        if(i < length-2 && c(i+2) == 0 ) i+=2
        else i+=1
        jumps +=1
    }
    jumps 
}

Functional-programming style Solution using Recursion:

 def jumpingOnClouds(c: Array[Int]): Int = {
     val limit = c.length -2 
     def rec(jumps: Int, index: Int): Int = {
         if (index > limit) jumps 
         else { 
             val jumpingOffset: Int = if(index < limit && c(index + 2) == 0 ) 2 else 1
             rec(jumps+1, index + jumpingOffset)
         }
     }
     rec(0,0) 
 }

Answer 1

The thing about these "story" problems is that the story is often distracting, if not outright misleading, from the underlying problem to be solved.

In this case your code solves the challenge by calculating and counting the number of jumps it takes to get from the beginning to the end. More or less the way the story is layed out. But it's worth noting:

1. The minimum number of jumps for a sequence of N zeros is always N/2.
2. At the end of every sequence of zeros is a 1 to be jumped over, except at the very end.

def jumpingOnClouds(c :Array[Int]) :Int =
  c.mkString.split("1").foldLeft(0)(_ + _.length/2 + 1) - 1

This isn't the most efficient solution possible but since the input Array is limited to no more than 100 elements I decided to go for brevity over efficiency.