HackerRank Jumping on the Clouds: Revisited Solution

Hello Programmers, In this post, you will know how to solve the HackerRank Jumping on the Clouds: Revisited Solution. This problem is a part of the HackerRank Algorithms Series.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

HackerRank Jumping on the Clouds: Revisited Solution

Task

A child is playing a cloud hopping game. In this game, there are sequentially numbered clouds that can be thunderheads or cumulus clouds. The character must jump from cloud to cloud until it reaches the start again.

There is an array of clouds, c and an energy level e = 100. The character starts from c[0] and uses 1 unit of energy to make a jump of size k to cloud c[(i + k) % n]. If it lands on a thundercloud, c[i] = 1, its energy (e) decreases by 2 additional units. The game ends when the character lands back on cloud 0.

Given the values of n, k, and the configuration of the clouds as an array c, determine the final value of e after the game ends.

Example. c = [0, 0, 1, 0]
k = 2

The indices of the path are 0 –> 2 -> 0. The energy level reduces by 1 for each jump to 98. The character landed on one thunderhead at an additional cost of 2 energy units. The final energy level is 96.

Note: Recall that % refers to the modulo operation. In this case, it serves to make the route circular. If the character is at c[n – 1] and jumps 1, it will arrive at c[0]

Function Description

Complete the jumpingOnClouds function in the editor below.

jumpingOnClouds has the following parameter(s):

int c[n]: the cloud types along the path
int k: the length of one jump

Returns

int: the energy level remaining.

Input Format

The first line contains two space-separated integers, n and k, the number of clouds and the jump distance.
The second line contains n space–separated integers c[i] where 0 <= i < n. Each cloud is described as follows:

If c[i] = 0, then cloud i is a cumulus cloud.
If c[i] = 1, then cloud i is a thunderhead

Constraints

2 <= n <= 25
1 <= k <= n
n % k = 0
c[i] = {0, 1}

Sample Input

STDIN Function
—– ——–
8 2 n = 8, k = 2
0 0 1 0 0 1 1 0 c = [0, 0, 1, 0, 0, 1, 1, 0]

Sample Output

Explanation

In the diagram below, red clouds are thunderheads and purple clouds are cumulus clouds:

Observe that our thunderheads are the clouds numbered 2, 5, and 6. The character makes the following sequence of moves:

Move: 0 -> 2, Energy: e = 100 – 1 – 2 = 97.
Move: 2 -> 4, Energy: e = 97 – 1 = 96.
Move: 4 -> 6, Energy: e = 96 – 1 – 2 = 93.
Move: 6 -> 0, Energy: e = 93 – 1 = 92.

HackerRank Jumping on the Clouds: Revisited Solution

Jumping on the Clouds: Revisited Solution in C

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
    int n,i,e=100; 
    int k; 
    scanf("%d %d",&n,&k);
    int *c = malloc(sizeof(int) * n);
    for(int c_i = 0; c_i < n; c_i++){
       scanf("%d",&c[c_i]);
    }
    do
        {
        i=(i+k)%n;
        if(c[i]==1)
            e=e-2;
        e--;
    }while(i!=0);
    printf("%d",e);
        
    return 0;
}

Jumping on the Clouds: Revisited Solution in Cpp

#include "bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
static const int INF = 0x3f3f3f3f; static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
template<typename T, typename U> static void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> static void amax(T &x, U y) { if(x < y) x = y; }
int main() {
	int n; int k;
	while(~scanf("%d%d", &n, &k)) {
		vector<int> c(n);
		for(int i = 0; i < n; ++ i)
			scanf("%d", &c[i]);
		int i = 0, E = 100;
		do {
			-- E;
			(i += k) %= n;
			if(c[i] == 1)
				E -= 2;
		} while(i != 0);
		printf("%d\n", E);
	}
	return 0;
}

Jumping on the Clouds: Revisited Solution in Java

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        int c[] = new int[n];
        for(int c_i=0; c_i < n; c_i++){
            c[c_i] = in.nextInt();
        }
        
        int curr = 0;
        int e = 100;
        curr = (curr+k)%n;
        e -= 1+c[curr]*2;
        while (curr != 0) {
            curr = (curr+k)%n;
            e -= 1+c[curr]*2;
        }
        System.out.println(e);
    }
}

Jumping on the Clouds: Revisited Solution in Python

#!/bin/python
import sys
n,k = raw_input().strip().split(' ')
n,k = [int(n),int(k)]
c = map(int,raw_input().strip().split(' '))
cur = 0
e = 100
while True:
    if c[cur]==1:
        e-=2
    e-=1
    cur = (cur+k)%n
    if cur==0:
        break
print e

Jumping on the Clouds: Revisited Solution using JavaScript

process.stdin.resume();
process.stdin.setEncoding('ascii');
var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;
process.stdin.on('data', function (data) {
    input_stdin += data;
});
process.stdin.on('end', function () {
    input_stdin_array = input_stdin.split("\n");
    main();    
});
function readLine() {
    return input_stdin_array[input_currentline++];
}
/////////////// ignore above this line ////////////////////
function main() {
    var n_temp = readLine().split(' ');
    var n = parseInt(n_temp[0]);
    var k = parseInt(n_temp[1]);
    c = readLine().split(' ');
    c = c.map(Number);
    
    var pos = 0;
    var E = 100;
    do {
        pos = (pos + k) % n;
        E -= 1;
        if (c[pos] === 1) {
            E -= 2;
        }
    } while (pos != 0);
    process.stdout.write("" + E);
}

Jumping on the Clouds: Revisited Solution in Scala

import scala.io.StdIn
object Solution {
  def main(args: Array[String]) {
    val Array(n, k) = StdIn.readLine().split("\\s+").map(_.toInt)
    val c = StdIn.readLine().split("\\s+").map(_.toInt)
    var e = 100
    var i = k % n
    e -= 1
    if (c(i) == 1) {
      e -= 2
    }
    while (i != 0) {
      i = (i + k) % n
      e -= 1
      if (c(i) == 1) {
        e -= 2
      }
    }
    println(e)
  }
}

Jumping on the Clouds: Revisited Solution in Pascal

var f:text; n,e,i,c,k:longint;
begin
  assign(f,''); reset(f); read(f,n,k); e:=100;
  for i:=0 to n-1 do
  begin
    read(f,c);
    if i mod k=0 then
    begin
      dec(e);
      if c=1 then dec(e,2)
    end
  end;
  close(f);
  assign(f,''); rewrite(f); write(f,e); close(f)
end.

Disclaimer: This problem (Jumping on the Clouds) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

Next: HackerRank Non Divisible Subset Solution

HackerRank Jumping on the Clouds: Revisited Solution

ByBrokenprogrammers

HackerRank Jumping on the Clouds: Revisited Solution

Task

HackerRank Jumping on the Clouds: Revisited Solution

Jumping on the Clouds: Revisited Solution in C

Jumping on the Clouds: Revisited Solution in Cpp

Jumping on the Clouds: Revisited Solution in Java

Jumping on the Clouds: Revisited Solution in Python

Jumping on the Clouds: Revisited Solution using JavaScript

Jumping on the Clouds: Revisited Solution in Scala

Jumping on the Clouds: Revisited Solution in Pascal

By Brokenprogrammers

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