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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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1188 Discussions

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  • 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

    `

  • gandymovspc
     + 1 comment

    Solution in JavaScript

    function jumpingOnClouds(c, k) {
	let energy = 100;
	let position = 0;
	do{
		position = (position + k) % c.length;
		energy -= 1;
		if (c[position] === 1){
			energy -= 2;
		}
	} while (position !== 0);
	return energy;
}

  • raveen_re
     + 0 comments

    Solution in JS

    function jumpingOnClouds(c, k) {
    let n = c.length;
    
    let i = k >= n ? Math.abs(n - k) : k;
    let e = 100 - (c[i] === 0 ? 1 : 3);
    
    while(i != 0 || e <= 0){
        i = i + k;
        if(i >= n){
            i = Math.abs(n - i);
        }
        
        if(c[i] === 0){
            e--;
        }
        
        else{
           e = e - 3;
        }
    }
    
    return e >= 0 ? e : 0;
}

  • yashdeora98294
     + 0 comments

    Here is problem solution in Python, Java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-jumping-on-the-clouds-revisited-problem-solution.html

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/retgbr5jSsg

    int jumpingOnClouds(vector<int> c, int k) {
    int position = 0, energie = 100;
    do{
        position = (position + k) % c.size();
        energie -= 1 + c[position] * 2;
    }while(position != 0);
    return energie;
}