Jumping on the Clouds


Problem Statement :


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

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



Function Description

Complete the jumpingOnClouds function in the editor below.

jumpingOnClouds has the following parameter(s):

int c[n]: an array of binary integers



Returns

int: the minimum number of jumps required



Input Format

The first line contains an integer n, the total number of clouds. The second line contains n space-separated binary integers describing clouds c[ i ] where 0  <=  i  < n.

Constraints

2  <=  n  <=   100


Output Format

Print the minimum number of jumps needed to win the game.


Solution :



title-img 


                            Solution in C :

In    C :





#include<stdio.h>
#include<math.h>
 
int main(int argc, char const *argv[])
{
	int n,i,count=0;
	scanf("%d",&n);
	int arr[n];

	for(i=0;i<n;i++)
	{
		scanf("%d",&arr[i]);
	}

	for(i=0;i<n-1;i++)
	{
		if(arr[i+2]==0)
		{
			count++;
			i=i+1;
		}

		else
			count++;
	}

	printf("%d\n",count);
	return 0;
}
                        


                        Solution in C++ :

In   C++  :






#include<bits/stdc++.h>

using namespace std;

int a[100], dp[100];

int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) scanf("%d", a + i);
	
	dp[0] = 0;
	for (int i = 1; i < n; i++) {
		dp[i] = dp[i - 1] + 1;
		if (i > 1) dp[i] = min(dp[i], dp[i - 2] + 1);
		
		if (a[i] == 1) dp[i] = n + 1;
	}
	
	printf("%d\n", dp[n - 1]);

	return 0;
}
                    


                        Solution in Java :

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 input = new Scanner(System.in);
        int rounds = input.nextInt();
        int[] ar = new int[rounds];
        int i = 0;
        for(i = 0; i < rounds; i++)
            ar[i] = input.nextInt();
        int count = 0;
        i = 0;
        while(i != rounds-1)
        {
            if(i != ar.length - 2 && ar[i+2] == 0)
                i+=2;
            else
                i++;
            count++;
        }    
        System.out.println(count);
    }
}
                    


                        Solution in Python : 
                            
In  Python3   :







#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the jumpingOnClouds function below.
def jumpingOnClouds(c):
    i = 0
    jumpCount = -1
    while i < len(c):
        jumpCount += 1
        if (i < len(c)-2) and (c[i+2] == 0):
            i+=1
        i += 1
    return jumpCount
            
            
        

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    n = int(input())

    c = list(map(int, input().rstrip().split()))

    result = jumpingOnClouds(c)

    fptr.write(str(result) + '\n')

    fptr.close()


View More Similar Problems

Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

View Solution →

Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

View Solution →

Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

View Solution →

Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

View Solution →

Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

View Solution →

Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty

View Solution →