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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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  • anarod_val25
     + 0 comments
    int jumpingOnClouds(vector<int> c, int k) {
    int e = 100;
    int cloud = 0;
    
    while (e>0) {
        cloud += k;  
        e--;
        
        if(cloud>=c.size()){
            cloud-=c.size();
        }
        
        if(c[cloud]==1){
            e-=2;
        }
        
        if(cloud==0){
            return e;
        }
    }
    
    return e;
}

  • aliraza4130633
     + 0 comments

    Door locks are essential for securing homes, offices, and commercial spaces, offering protection against unauthorized access and enhancing overall safety. Modern locks come in various types, from traditional key-operated to advanced digital and smart systems, catering to different security needs. Jumping on the clouds: revisited, just like exploring new heights requires preparation and awareness, choosing the right door locks ensures your property remains safe, reliable, and protected against potential threats.

  • greivin_lopez
     + 0 comments
    def jumpingOnClouds(c, k):
    t = n = len(c)
    e = 100
    while t > 0:
        t = (t + k) % n
        e -= 3 if c[t] == 1 else 1
    return e

  • bashersmmo20
     + 0 comments

    To be honest the discription is really bad.

  • antor_nahidul
     + 0 comments

    My python solution

    	```
i =k
	e = 100
	if c[0] == 1:
			e -= 3
	else:
			e -= 1
	if len(c)== k:
			return e
	while k != 0:
			if c[k] == 1:
					e -= 3
			else:
					e -= 1
			k += i
			if k >= len(c):
					k -= len(c)
			print(k, c[k], e)
	print(k, c[k], e)
	return e

    `