# CARVANS - Editorial

CAKEWALK

Simple Math

## PROBLEM

There are N cars on a narrow road that are moving in the same direction. Each car has maximum speed. No car can overtake other cars on this road. Each car will move at the maximum possible speed while avoiding any collisions. Count the number of cars which are moving at their maximum speeds.

## QUICK EXPLANATION

Intuitively, a car is moving at its maximum speed if and only if all cars in front of it are moving at greater speeds (otherwise it will overtake the slower car). Therefore, the answer is the number of such cars.

## EXPLANATION

Suppose that the cars are numbered 1 through N from front to back, and the maximum speed of the i-th car is maxSpeed[i]. From the intuitive observation above, we can directly come up with this naive solution:

```answer = 0
for i = 1; i <= N; i++:
allGreater = true
for j = 1; j <= i-1; j++:
if maxSpeed[j] < maxSpeed[i]:
allGreater = false
if allGreater:
```

Unfortunately, this solution runs in O(N^2) time, which will surely time out. We will need other observations.

Consider each car. From the problem statement, each car will:

• Avoid any collisions. Since the road is narrow, therefore, it will not move at greater speed than the car directly in front of it (if any).
• Move at the maximum possible speed. Therefore, it will move at speed of exactly min(the maximum speed of the car, the speed of the car directly in front of it).

From those observations, we can calculate the speed of each car in O(1) time. When calculating the speed of the i-th car, we have to know the speed of the (i-1)st car. Therefore, we must calculate the speeds in the right order (i.e., from the first car to the last car on the road). After that, we compare the speed of each car with its maximum speed.

A direct implementation of the above solution is as follows. This solution runs in O(N) time, which will pass the time limit.

```answer = 0

speed[1] = maxSpeed[1]
for i = 2; i <= N; i++:
speed[i] = min(maxSpeed[i], speed[i-1])

for i = 1; i <= N; i++:
if speed[i] == maxSpeed[i]:
```

Exercise

Try to solve this problem without creating the additional speed[]/maxSpeed[] array. Hint: we can always store only the speed of the last car we consider instead of storing all speeds in speed[] array.

## SETTERâ€™S SOLUTION

Can be found here

## TESTERâ€™S SOLUTION

Can be found here.

5 Likes

we donâ€™t even need arrays!! http://www.codechef.com/viewsolution/1364489

7 Likes

The problem can be solved even without creating an array maxSpeed[]. We can do all calculations during input stage:

```answer := 1
for i:=2 to N do
if speed >= maxSpeed
speed = max(speed, maxSpeed)
```
6 Likes

solved - http://www.codechef.com/viewsolution/1374650 but have a question.

Can anyone point out (line 29 - commented out) why do we get â€śRuntime Errorâ€ť if I do a br.close (closing the BufferedReader). I thought closing the BufferedReader was a good practice.

Once I commented it out, the submission was accepted.

I wanted to know what would be the output for the following test case
8 4 6 5
Would it be 3 or 2?
I think the answer should be 3 because in the problem it is clearly mentioned that cars cannot overtake as the track is not wide enough.
So the first car is travelling at max speed. The second one is also travelling at max speed since its speed is less than 8. The third one is not at max speed. The fourth one is travelling at max speed.
We just have to compare the speeds of consecutive cars right??? and not with all the carsâ€¦ as cars cannot overtakeâ€¦ Please help me understand the problemâ€¦

1 Like

Why am i getting a WA here ?
http://www.codechef.com/viewsolution/4375902

please figure out where i am wrongâ€¦

Sir, may i know why this is giving worng answer? thanks

#include<stdio.h>
int main(){
int t,n,count;
long long int a[10000];
scanf("%d",&t);
while(tâ€“){
count=1;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
if(a[i]<=a[i-1] &&i !=0){
count++;
}
}
printf("%d\n",count);
}
}

Even this problem can be solved without the help of any array :
in o(n-1)time
sol is:http://www.codechef.com/viewsolution/6400303

#include
using namespace std;
int main()
{
int t,n;
cin>>t;
while(tâ€“)
{
cin>>n;
int a[n],i,count=1;
for(i=0;i<n;i++)
cin>>a[i];
for(i=1;i<n;i++)
{
if(a[i]<a[i-1])
count++;
}
cout<<count<<"\n";
}
return(0);
}

#include
using namespace std;

int main()
{
int t;
cin>>t;
while(tâ€“){
int n;
cin>>n;
int a[n+1];
for(int i=1;i<=n;i++){
cin>>a[i];

``````    }
int ans=0;
a[0]=-1;
for(int i=1;i<=n;i++){
if(a[i]<a[i-1]) ans++;
}
cout<<ans<<endl;
}
return 0;
``````

}

â€śâ€ťâ€śThis approach works too ,and is faster than other appraoches mentioned aboveâ€ť""
â€śâ€ťâ€śAC in one goâ€ť""

``````test=int(raw_input())
for _ in range(test):
N=int(raw_input())
count=1
arr=list()
arr=[int(i) for i in raw_input().split()]
for i in range(len(arr)-1):
if(arr[i+1]>arr[i]):
arr[i+1]=arr[i]
else:
count+=1
print count``````

can some one find mistake in this code

import java.util.;
import java.lang.
;
import java.io.*;
class klf
{
public static void main(String[]args) throws Exception
{
Scanner cin=new Scanner(System.in);
Stackst;
int t=cin.nextInt();
while(tâ€“>0)
{
st=new Stack();
int n=cin.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++)
{
a[i]=cin.nextInt();
}
int count=0;
for(int i=0;i<n;i++)
{
if(st.empty())
{
st.push(a[i]);
count++;
}
else if(a[i]<st.peek())
{
st.push(a[i]);
count++;
}
else
{
st.push(a[i]);
}
}
while(!st.empty())
st.pop();
System.out.println(count);
}
}
}

strong text
Can anybody please let me know what is wrong with my codeâ€¦
strong text
#include<stdio.h>

int main(){
int rounds,cars,speed,count=0;
for(scanf("%d",&rounds);rounds;roundsâ€“){
int min=10000;
count=0;
for(scanf("%d",&cars);cars;carsâ€“){
scanf("%d",&speed);
if(speed<=min){
count++;
min=speed;
}
}
printf("%d\n",count);

}

``````return 0;
``````

}

Does Input to this problem is in file?
Am i need to take input from files?

Just have a look : https://www.codechef.com/viewsolution/22546307,
You will not need an array!

the obvious/funny thing is :since at anytime we are dealing with two consecutive cars only, problem can be solved without the need of an array.

why this shows time limit exceed?

http://www.codechef.com/viewsolution/1370908

Use scanf, printf instead of cin, cout.

2 Likes

By the way, why 4MB is 4,000,000,000 bytes

It is mentioned in three problems CARVANS, PPERM and COALSCAM.