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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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  • naufalfh19
     + 0 comments

    python3 Solution

    def jumpingOnClouds(c):
    pos = 0
    res = -1
    while pos < len(c):
        if pos + 2 < len(c) and c[pos+2] != 1:
            res += 1
            pos += 2
        else:
            res += 1
            pos += 1
        
    return res

  • leandes
     + 0 comments

    This max score is not fair compared with some other challenges way more intricate than this one...

    Java 8 solution:

    class Result {

    public static int jumpingOnClouds(List<Integer> c) {
        int i=0, jumps=0;
        while (i<c.size()) {
            i = i + ((i < c.size()-2 && c.get(i+2)==0) ? 2 : 1);
            jumps++;
        }
        return jumps-1;
    }

}

  • yashdeora98294
     + 0 comments

    Here is jumping on the CLOUD problem solution in python, java, c++, c and javascript - https://programmingoneonone.com/hackerrank-jumping-on-the-clouds-problem-solution.html

  • 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;
}