# NUM239 - Editorial

Practice

Contest

Author: Ivan Safonov

Editorialist: Bhuvnesh Jain

CAKEWALK

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.

O(N + T)

# Space Complexity

O(N)

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.

2 Likes

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?

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

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;
}


}

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