NUM239 - Editorial

Problem Link

Practice

Contest

Author: Ivan Safonov

Tester: Hasan Jaddouh

Editorialist: Bhuvnesh Jain

Difficulty

CAKEWALK

Prerequisites

Prefix sums

Problem

You are given to integers L and R. Find the number of integers which have the last digit as 2, 3 or 9 and lie in the range [L, R].

Explanation

As the constrainsts are small on L and R, we can pre-calculate all good numbers. Then each test case just asks to find out how many good numbers lie within the given range. This can be easily answered using prefix-sums. If you don’t know about prefix-sums, you can read it here. Basically the idea is the following:

\sum_{i=l}^{i=r} A_i = \sum_{i=1}^{i=r} A_i - \sum_{i=1}^{i=l-1} A_i
\sum_{i=l}^{i=r} A_i = \text{(Prefix-sum at i=r)} - \text{(Prefix-sum at i=l-1)}

Below is a pseudo-code for it:


	def init():
		for i in [1, 1000000]:
			last_digit = i % 10
			if last_digit == 2 or last_digit == 3 or last_digit == 9:
				good[i] = 1
			else:
				good[i] = 0

		# Build prefix sum
		for i in [1, 1000000]:
			good[i] += good[i-1]

	def solve(l, r):
		ans = good[r] - good[l-1]
		return ans

The time complexity of the above approach will be O(N + T), where N = 100000 is for pre-computation and T is for answering every test case in O(1) using prefix-sums. The space complexity of the above approach will be O(N) for storing the prefix array of good elements.

Bonus Problem

Solve the problem without any pre-computation and in O(1) memory.

Idea:

Click to view

Note that every consecutive number from “x…0” to “x…9” has 3 good numbers.

Feel free to share your approach, if it was somewhat different.

Time Complexity

O(N + T)

Space Complexity

O(N)

Solution Links

Setter’s solution

Tester’s solution

Editorialist solution

2 Likes

Since constraints were easy there was no need to define an array to hold good/pretty numbers.

a simple brute force from l to r and checking the last digit would yield the same result.

See my submission: https://www.codechef.com/viewsolution/19019309

Time Complexity: O(N)

Space Complexity: O(1)

My Solution:

Time Complexity : O(T)

Space Complexity : O(1)

It will be useful if no. of test case is quite large.

/* Written By : Anish Ray */

#include <bits/stdc++.h>

#define ll long long
#define v(k) vector
#define mod 1000000007
#define m(a,b) map <a,b>
#define ull unsigned long long
#define fi first
#define se second

using namespace std;

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);

int t;
cin>>t;

while(t--)
{
	int l,r;
	cin>>l>>r;
	
	int c=0;
	int f=0;
	
	for(int i=0;i<10;i++)
	{
		if(l==r)
		{
			if(l%10==3||l%10==9||l%10==2)
				c+=1;
				
			f=1;
			break;
		}
	
		if(l%10==2)
		{
			break;
		}
		else if(l%10==3||l%10==9)
			c++;
		
		l++;
	}
	
	if(f!=1)
	{
		for(int i=0;i<10;i++)
		{
			if(r==l)
			{
				if(l%10==3||l%10==9||l%10==2)
					c+=1;
				
				f=1;
				break;
			}
	
			if(r%10==1)
			{
				break;
			}
			else if(r%10==9||r%10==3||r%10==2)
				c++;
		
			r--;
		}
	}
	
	if(f!=1)
	{
		c+=((r-l+1)/10)*3;
	}
	
	cout<<c<<endl;
}

return 0;

}

Its a simple Brute Force application. Amazed to see such a simple problem in LTIME61B.
My code goes here…(in cpp).## Heading ##


#include <iostream>
using namespace std;

int main()
{
     int t,cnt,digit,l,r;
     cin>>t;
     while(t--)
    {
        cnt = 0;
        cin>>l>>r;
        for(int i=l;i<=r;i++)
        {
            digit = i % 10;
            if(digit == 2 || digit == 3 || digit == 9)
                cnt++;
        }
        cout<<cnt<<endl;
    }
    return 0;
}

3 Likes

Wtf… I wasted a lot of time on this question due to some connection issue and due to some problem with codechef IDE…
Didn’t knew that 1st question is that low in div2 (as I was in div2 before contest and I was in div1 before previous contest)…
This is completely brute Force…

I did O(1) space approach

@l_returns Pro tip #1 (#nooffence) Never use codechef ide in short contest for testing soln. It will ~1 min to give verdict. And sometimes it also shows “Submission Limit Reached” . Alternative offline ide(Best Option), CodeForces Ide (A bit slower but can be used) both are relatively faster. I personally use these options for testing purpose.

Time Complexity: O(T)
Space Complexity: O(1)
Java solution: https://www.codechef.com/viewsolution/26602770
Just check,

  1. how many of (2,3,9) ending numbers are present between L and next round figure (multiple of 10)
  2. how many of (2,3,9) ending numbers are present between R and previous round figure(multiple of 10)
  3. find the difference between these two round figures and divide by 10 and multiply by 3 (because there are 3 possibilites for each 0-10, 10-20, 20-30 etc…)
  4. sum of all the counts from above 3 steps gives the answer.

Eg: L=11 and R=33
1) count = 3, nextL = 20
2) count = 2, previousR = 30
3) count = (previousR-nextL)*3 = ((30-20)/10)*3 = 3
4) total count = 3+2+3 = 8

No need of any brute-force approach.

1 Like

Hi I am beginner. Why am I getting partially solved mark for this question? If my code is not right for all test cases where can I get those test cases to test on different IDE?
A little help please
Thanks in advance!

I did it in O(1) time complexity and O(1) space complexity
No Loops Bro…,I liked it

void solve(){
    int l,r;cin>>l>>r;l--;
    int a1=(r/10)*3;
    int a2=(l/10)*3;
    r=r%10;l=l%10;
    if(r>1)a1++;if(r>2)a1++;if(r==9)a1++;
    if(l>1)a2++;if(l>2)a2++;if(l==9)a2++;
    cout<<a1-a2<<endl;
 }

yes, i too do it with the same approach :smile:

Idk what is wrong with my solution.Can somebody help??

#include <bits/stdc++.h>
using namespace std;
int main()
{
int cases,i,n,a,b;
cin>>cases;
int temp1,temp,count;
while(cases–)
{
cin>>a>>b;
int temp=a%10;
if(temp<=2)
count=3;
else if(temp==3)
count=2;
else
if(temp<=9)
count=1;

	a=a+10-temp ;
while(a+10<b)
{
	a=a+10;
	count+=3;
}
while(b%10!=0)
{
	if(b%10==2 || b%10==3 || b%10==9)
		count++;
	b--;
}



	cout<<count<<endl;
}

}

@likecs @shivam_g1470 @anishray042 @hunterr_196 @l_returns @aryanc403

what is wrong with my code
https://www.codechef.com/viewsolution/36882644
my approach is for example l=12 and r =33
then l=l-1 mean l=11( because range is inclusive )
now lx =l%10 and rx =r%10
lx =1 and rx=3

now for get no of magical no l=( l/10*2 )and r =( r/10)*3
if lx>1 then add 1 if lx>2 and 1 and i lx>8 add 1
same for y
we get r=11 and l=3
finaly answer = r-l
11-3 =8 which is right
it is showing wa