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CARVANS - Editorial







Simple Math


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.


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.


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:
        answer = answer + 1

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]:
        answer = answer + 1


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.


Can be found here


Can be found here.

This question is marked "community wiki".

asked 24 Sep '12, 00:50

fushar's gravatar image

3★fushar ♦♦
accept rate: 0%

edited 25 Dec '12, 13:34

admin's gravatar image

0★admin ♦♦

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.

(24 Sep '12, 01:22) kunalkrishna852★

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

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

(24 Sep '12, 02:33) anton_lunyov ♦♦6★

Yes. All three of us missed that :P Will fix it for the problems in the practice section.

(24 Sep '12, 09:54) gamabunta ♦♦1★

12next »

we don't even need arrays!!


answered 24 Sep '12, 00:54

abhinav1592's gravatar image

accept rate: 0%

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
    then answer := answer + 1
  speed = max(speed, maxSpeed)


answered 24 Sep '12, 00:57

anton_lunyov's gravatar image

6★anton_lunyov ♦♦
accept rate: 12%

why this shows time limit exceed?

(24 Sep '12, 02:26) ak11111110★

Use scanf, printf instead of cin, cout.

(24 Sep '12, 02:31) anton_lunyov ♦♦6★

if W==1 you still need to input speed of the only car of this test. Otherwise you consider its speed as a number of cars for the next test. Also while(--N) as well as while(--W) is wrong. By this you miss the last test case, last car respectively. Use while(N--) and while(W--) instead.

(24 Sep '12, 02:43) anton_lunyov ♦♦6★

@anton_lunyov're right, corrected those.

Got a correct one ! thanks

(24 Sep '12, 02:53) ak11111110★

Did you even run the program on local machine and test? its full of errors!

printf("1\n");    // this was using front slash
while(--W)        // --W and not W-- this was causing TLE
printf("%d\n",counter); // reference was passed here!
(24 Sep '12, 03:05) kriateive5★

Sorry. This was my advise. Second while really should have --W. I didn't see the scanf before :)

(24 Sep '12, 03:09) anton_lunyov ♦♦6★
showing 5 of 6 show all

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 read abhinav1592's solution and it gives the answer as 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 cars.. as cars cannot overtake.. Please help me understand the problem..


answered 03 Oct '12, 01:02

mancoolgunda's gravatar image

accept rate: 0%

edited 03 Oct '12, 01:06

The third car also moves with speed 4 since it can not overtake the second car. So yes, we should compare only consecutive cars but the speed of each car may change.

(03 Oct '12, 17:43) anton_adm ♦♦2★

solved - but have a question.

Can anyone point out (line 29 - commented out) why do we get "Runtime Error" 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.


answered 26 Sep '12, 04:02

ujjalcal's gravatar image

accept rate: 0%

Why am i getting a WA here ?


answered 25 Jul '14, 12:18

ak795's gravatar image

accept rate: 0% please figure out where i am wrong....


answered 07 Oct '14, 19:10

vickycod's gravatar image

accept rate: 6%

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


int main(){ int t,n,count; long long int a[10000]; scanf("%d",&t); while(t--){ 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); } }


answered 16 Jan '15, 12:19

nischitpra's gravatar image

accept rate: 0%

Even this problem can be solved without the help of any array : in o(n-1)time sol is:


answered 06 Mar '15, 01:17

ritwikenator's gravatar image

accept rate: 16%


using namespace std; int main() { int t,n; cin>>t; while(t--) { 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); }

This answer is marked "community wiki".

answered 08 Jul '16, 14:14

ny826's gravatar image

accept rate: 0%

what is wrong in this question

(08 Jul '16, 14:14) ny8261★

please help me

(08 Jul '16, 14:15) ny8261★


using namespace std;

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

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


This answer is marked "community wiki".

answered 04 Nov '16, 02:56

rrishabh's gravatar image

accept rate: 0%

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Asked: 24 Sep '12, 00:50

Seen: 7,341 times

Last updated: 19 May, 14:27