Hello Programmers, In this post, you will learn how to solve HackerRank Jumping on the Clouds: Revisited Solution. This problem is a part of the HackerRank Algorithms Series.
One more thing to add, don’t straight away look for the solutions, first try to solve the problems by yourself. If you find any difficulty after trying several times, then look for the solutions. We are going to solve the HackerRank Algorithms problems using C, CPP, JAVA, PYTHON, JavaScript & SCALA Programming Languages.
HackerRank Jumping on the Clouds: Revisited
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
92
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
HackerRank 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;
}HackerRank 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;
}HackerRank 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);
}
}HackerRank 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 eHackerRank 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);
}HackerRank 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)
}
}HackerRank 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 Chase2learn. This tutorial is only for Educational and Learning purposes.