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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Jumping on the Clouds
  5. Discussions

Jumping on the Clouds

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3254 Discussions

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  • gcysne
     + 0 comments
    def jumpingOnClouds(c):
    idx = 0
    jumps = 0
    while idx != len(c)-1:
        # Greedy, if can jump 2, then do it
        idx = idx+2 if idx+2 < len(c) and c[idx+2] == 0 else idx+1
        jumps += 1
    return jumps

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/GNof9B-9CN0

    int jumpingOnClouds(vector<int> c) {
    int result = 0, index = 0;
    while(index < c.size() - 1){
        if(c[index + 2] == 1) index++;
        else index+=2;
        result++;
    }
    return result;
}

  • pathan_mazhar123
     + 1 comment

    In C#

    public static int jumpingOnClouds(List<int> c)
    {
        int jumps = 0;
        int position = 0;

        while (position < c.Count - 1)
        {
            if (position + 2 < c.Count && c[position + 2] == 0)
            {
                position += 2;
            }
            else
            {
                position += 1;
            }
            jumps++;
        }

        return jumps;

}

    }

  • pt023150
     + 0 comments

    java 8 

    public static int jumpingOnClouds(List<Integer> c) {
    // Write your code here
    int jump = 0;
    int start = 0;

    while (start < c.size() - 1) {
        if (start == c.size() - 2) {
            jump++;
            break;

        }
        int idx = (c.get(start + 2) != 0) ? 1 : 2;
        start += idx;
        jump++;

    }
    return jump;

  • kgouraw
     + 1 comment

    Problem is not very clear, what if there are consecutive thunderheads clouds. Example c=[0, 1, 1, 1, 0] , what will be the jumping rule?