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()


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