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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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  • toolulopeolagun1
     + 0 comments

    Guys, this is a math problem. The solution lies in understanding that it's just walks and jumps. The solution can be defined as: floor((n+walks)/2), where n is the total number of clouds. A walk in this case represents a forced walk (when there is an even number of clouds between two consecutive thunderclouds)

    My solution in python:

    def jumpingOnClouds(c):
    # Write your code here
    n = len(c)
    walks = 0
    last_thunderhead = -1
    for i in range(n):
        if c[i]:
            if (i - last_thunderhead)%2:
                walks += 1
            last_thunderhead = i
    return math.floor((n + walks)/2)

  • highsoft_soi
     + 0 comments

    C++

    int jumpingOnClouds(vector<int> &c) {
    int cnt = 0;

    for (int j = 0; j < c.size() - 1; j = j + 2) {
        if (c[j + 2] == 1) {
            j--;
            cnt++;
        } else { cnt++; }
    }

    return cnt;
}

  • mirianquispe1505
     + 0 comments
    function jumpingOnClouds(c: number[]): number {
    let jumps = 0
    for(let i=0; i < c.length - 1; i++){
        const nextDoubleCloud = c[i + 2]
        if(nextDoubleCloud == 0) {
            i++
        }
        jumps++
    }
    return jumps
}

  • wifivop476
     + 0 comments

    Jumping on the Clouds sounds like such a fun and creative concept! Whether it's about tackling challenges or dreaming big, it's a reminder to aim high. Just like in life, though, it’s always smart to stay grounded while pursuing those big dreams. What’s the story behind this idea?

  • vishwasai160599
     + 0 comments

    def jumpingOnClouds(c): # Write your code here cnt=0 i=0 while i+2