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

Jumping on the Clouds: Revisited

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  • devilpolten
     + 0 comments

    Jumping on Clouds Function below in python def jumpingOnClouds(c, k): n = len(c) energy = 100 # start with full energy i = 0 # start at the first cloud

    while True:
    # Move i to the next cloud considering the circular array
    i = (i + k) % n
    # Decrease energy by 1 for the jump
    energy -= 1  
    # If it's a thundercloud, lose an additional 2 energy
    if c[i] == 1:
        energy -= 2
    # If we're back at the start, break out of the loop
    if i == 0:
        break

    return energy

    This function iterates through the clouds, adjusts the index based on the jump size k, and updates the energy according to the cloud type.

    In a way, this challenge reminds me of editing a complex video sequence using VN Video Editor, where you often loop through clips and apply different effects strategically to enhance the final output.

  • g_azabache1
     + 0 comments

    This is my solution in C# 😉

        public static int jumpingOnClouds(int[] c, int k)
    {
        int index = 0;
        int points = 100;
        bool isExit = false;

        while (true)
        {
            index += k;
            if (index > c.Length - 1)
            {
                index = index % c.Length;
                isExit = true;
            }
            int value = c[index];

            if (value == 1)
                points -= 2;

            points--;
            if (index == 0 && isExit) break;
        }
        return points;
    }

  • vuhuutuan0
     + 0 comments

    Answer in javascript

    function jumpingOnClouds(c, k) {
  let energy = 100;
  let cloud = 0;

  const jump = () => {
    cloud = (cloud + k) % c.length;
    energy -= 1;
    if (c[cloud]) energy -= 2;
  };

  do {
    jump();
  } while (cloud != 0);

  return energy;
}

  • mohammed_ashraf7
     + 0 comments
    # jumpingOnClouds function below in python.
def jumpingOnClouds(c, k):
   
    energy = 100
    
    # got to next cloud by k steps
    i = 0 + k
    
    while energy > 0:
        '''The ternary operators are to check
        if its a thundercloud(c[i] == 1):
            then we decrement energy by 3(source to destination)
        else if its a clumbus (c[i] == 0):
            then we decrement energy by 1
        '''
        # prevent array out of index
        if i >= len(c):
            i -= len(c)
            
        if i == 0:
            # reached cloud zero, game stops
            energy = energy - 1 if c[i] == 0 else energy - 3
            break
        else:
            energy = energy - 1 if c[i] == 0 else energy - 3
            i += k
            
    return energy

  • khandanish5
     + 0 comments
        static int jumpingOnClouds(int[] c, int k) {
        int e = 100;
        int i = 0;
        do {
            if (c[i] == 0) {
                e -= 1;
            } else if (c[i] == 1) {
                e -= 3;
            }
            i = (i+k) % c.length;
        } while (i != 0);
        return e;
    }