Skip to main content

Jumping on the Clouds HackerRank Java Solution

Jumping on the Clouds HackerRank Java Solution | Programming Blog

Jumping on the Clouds HackerRank in Java
 

Emma is playing a new mobile game that starts with consecutively numbered clouds. Some of the clouds are thunderheads and others are cumulus. She can jump on any cumulus cloud having a number that is equal to the number of the current cloud plus 1 or 2. She must avoid the thunderheads. Determine the minimum number of jumps it will take Emma to jump from her starting postion to the last cloud. It is always possible to win the game. 

For each game, Emma will get an array of clouds numbered 0 if they are safe or 1 if they must be avoided. For example, 

c = [0, 1, 0 , 0 ,0, 1, 0] indexed from 0...6. The number on each cloud is its index in the list so she must avoid the clouds at indexes 1 and 5. 

She could follow the following two paths:
0 -> 2 -> 4 -> 6 or 0 -> 2 -> 3 -> 4 -> 6. The first path takes 3 jumps while the second takes 4.

Here is the link for Hacker Rank problem :

Lets see solution of this problem

Solution 1 :- Using For Loop

import java.util.Scanner;

public class JumpingOnTheClouds {

    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter the total number of cloud");
        int totalPath = sc.nextInt();
        int[] array = new int[totalPath];
        
        System.out.println("Enter Values of clouds");
        for (int i = 0; i < array.length; i++) {
            array[i] = sc.nextInt();
        }
        
        int jumps = 0;
        
        // Loop until arrays length
        for (int i = 0; i < array.length - 1; i++) {
            
            // Check i+1 cloud, if its exist and value is 0 then
            // jump into that
            if ((i+2) < array.length && array[i+2] == 0) {
                ++i;
                jumps++;
            
            // Check i+1 cloud, if it exist then jump into that
            } else if ((i+1) < array.length) {
                if (array[i+1] == 0) {
                    jumps++;
                }
            }
        }

        System.out.println("Minimum jumps required to win game : " +jumps);
    }

}
 

Output :-
Enter the total number of cloud
7
Enter Values of clouds
0 0 1 0 0 1 0
Minimum jumps required to win game : 4
__________________________________

Enter the total number of cloud
6
Enter Values of clouds
0 0 0 0 1 0
Minimum jumps required to win game : 3

Solution 2 :- Using While Loop

import java.util.Scanner;

public class JumpingOnTheClouds {

    public static void main(String[] args) {
       
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter the total number of cloud");
        int totalPath = sc.nextInt();
        int[] array = new int[totalPath];
       
        System.out.println("Enter Values of clouds");
        for (int i = 0; i < array.length; i++) {
            array[i] = sc.nextInt();
        }
       
        int jumps = 0;
        int i = 0;
        while(i < totalPath-1) {
            if (i < totalPath-2 && array[i+2]==0) {
                i++;
            }
            if(i != totalPath - 1) {
                jumps++;
            }
            i++;
        }
        System.out.println("Minimum jumps required to win game : " + jumps);
    }

}

Output :-
Enter the total number of cloud
7
Enter Values of clouds
0 0 1 0 0 1 0
Minimum jumps required to win game : 4
__________________________________

Enter the total number of cloud
6
Enter Values of clouds
0 0 0 0 1 0
Minimum jumps required to win game : 3

If you have any query related to above problem then comment down. I am happy to resolve it.

Check out other HackerRank problem's solution in Java :-

 

                                                                    Thank You

Comments

Post a Comment

Popular posts from this blog

Flipping the Matrix HackerRank Solution in Java with Explanation

Java Solution for Flipping the Matrix | Find Highest Sum of Upper-Left Quadrant of Matrix Problem Description : Sean invented a game involving a 2n * 2n matrix where each cell of the matrix contains an integer. He can reverse any of its rows or columns any number of times. The goal of the game is to maximize the sum of the elements in the n *n submatrix located in the upper-left quadrant of the matrix. Given the initial configurations for q matrices, help Sean reverse the rows and columns of each matrix in the best possible way so that the sum of the elements in the matrix's upper-left quadrant is maximal.  Input : matrix = [[1, 2], [3, 4]] Output : 4 Input : matrix = [[112, 42, 83, 119], [56, 125, 56, 49], [15, 78, 101, 43], [62, 98, 114, 108]] Output : 119 + 114 + 56 + 125 = 414 Full Problem Description : Flipping the Matrix Problem Description   Here we can find solution using following pattern, So simply we have to find Max of same number of box like (1,1,1,1). And last

Plus Minus HackerRank Solution in Java | Programming Blog

Java Solution for HackerRank Plus Minus Problem Given an array of integers, calculate the ratios of its elements that are positive , negative , and zero . Print the decimal value of each fraction on a new line with 6 places after the decimal. Example 1 : array = [1, 1, 0, -1, -1] There are N = 5 elements, two positive, two negative and one zero. Their ratios are 2/5 = 0.400000, 2/5 = 0.400000 and 1/5 = 0.200000. Results are printed as:  0.400000 0.400000 0.200000 proportion of positive values proportion of negative values proportion of zeros Example 2 : array = [-4, 3, -9, 0, 4, 1]  There are 3 positive numbers, 2 negative numbers, and 1 zero in array. Following is answer : 3/6 = 0.500000 2/6 = 0.333333 1/6 = 0.166667 Lets see solution Solution 1 import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.function.*; import java.util.regex.*; import java.util.stream.*; import static java.util.st