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Question

It is a problem from hackerrank.

https://www.hackerrank.com/challenges/jumping-on-the-clouds/problem?h_l=interview&isFullScreen=false&playlist_slugs%5B%5D%5B%5D%5B%5D=interview-preparation-kit&playlist_slugs%5B%5D%5B%5D%5B%5D=warmup

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int n;
    cin>>n;
    int jump=0;
    int a[101];
    a[0]=0;
    for (int i = 1; i < n; i++) {
        cin>> a[i];
    }
    for(int j=1;j<n;j++)
    {
        if(a[j]==1)
        {
            jump++;
        }
        if(a[j]==0)
        {
            jump++;
            if(a[j-1]==0 && a[j-1]==1)
                jump--;
        }
    }
    cout<<jump;
    return 0;
}

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. So, we have to print the minimum number of jumps needed to win the game.

Here's the input/output:

7
0 0 1 0 0 1 0

Expected output is: 4

My wrong output: 6

Answer 1

The first obvious problem I see is that the second loop will never end, because j is never incremented. Furthermore the condition

if(a[j-1]==0 && a[j-1]==1)

makes no sense. a[j-1] can only be either 0 or 1, not both unless you are running this on a quantum computer. Fix these issues and work from there.

Answer 2

Loop through every position in the consecutive cloud List advancing your position 2 jumps forward if possible. Otherwise, you just jump once. Regardless, you'll count the number of iterations taken until you've reached the end. Also, because each iteration is a jump, you don't need to jump off the last cloud, so go to size() minus one.

Time-complexity: O(n) [n = number of clouds]

Space-complexity: O(1)

Java 8

public static int jumpingOnClouds(List<Integer> c) 
{
    //Track number of jumps
    int jumpCount = 0;
    
    //Keep jumping until you can't anymore
    for(int i = 0; i < c.size()-1; jumpCount++)
    {
        //If possible, long jump (2 clouds forward)
        if(i < c.size() - 2 && c.get(i+2) != 1)
            i+=2;
        else //Otherwise, short jump (1 cloud forward)
            i++;
    }
    
    //Return count of jumps taken
    return jumpCount;
}