BINTREE - Editorial

PROBLEM LINK:

Practice
Contest

Author: Lalit Kundu
Tester: Shiplu Hawlader and Mahbubul Hasan
Editorialist: Lalit Kundu

DIFFICULTY:

EASY

PREREQUISITES:

Binary Tree
Lowest Common Ancestor

PROBLEM:

Given infinite full binary tree (each node has two children except leaf nodes), for queries of form (i,j) [i,j <= 10^9] print the length of shortest path between nodes labelled i and j.
Root is labelled 1. For each node labelled v, it’s left child is labelled 2*v and right child, 2*v+1.

QUICK EXPLANATION:

We convert i,j to base 2 (without leading zeroes)
Let i in base 2 be = a1a2…an
Let j in base 2 be = b1b2…bm
If ap=bp for all p<=k, then our answer is (n+m-2*k).

EXPLANATION:

If we are at a node labelled v, if we move left we get 2*v ie. append a 0 to binary representation of v. If we move right we get 2*v+1 ie. append a 1 to binary representation of v. Thus from binary value of a node v, we completely know the path taken from the root.
image1
For example, Node 10 in binary is 1010, here first 1 is root node, next is 0, means a left turn, next 1 means are right child, next 0 means a left child.
We convert i,j to base 2 (without leading zeroes)
Let i in base 2 be = a1a2…an
Let j in base 2 be = b1b2…bm
If ap=bp for all p<=k, means their Lowest Common Ancestor(LCA) in binary is a0a1…ak. So the distance between i and j is dist(i,LCA(i,j))+dist(j,LCA(i,j)).
Since i in base 2 has n digits, the distance between i and LCA(i,j) will be (n-k). Since those are the extra steps taken from LCA moving towards node.
Therefore our answer is (n-k)+(m-k).
For example, i=10, j=13.
i in base2 = 1010
j in base2 = 1101
So, k=1 and our answer is 4-1+4-1=6.

Complexity for each query= log2(i)+log2(j).

AUTHOR’S AND TESTER’S SOLUTIONS:

To be updated soon.

14 Likes

@pandu_49: i(10) in binary is 1010 and j(13) in binary is 1101, only for p<=1(k) is ap=bp true.
Another example, i(8) in binary is 1000 and j(2) in binary is 10, only for p<=2(k) is ap=bp true.
One more, i(10) in binary is 1010 and j(11) in binary is 1011, only for p<=3(k) is ap=bp true.

1 Like

#include <stdio.h>
#include<math.h>
long int poww(long int to)
{
if(to==0)
return 1;

return 2*poww(to-1);

}
long int other_path(long int h,long int l)
{
long int path=0;
while(h!=l){
path+=2;
h/=2;l/=2;
}
return path;
}

long int level(long int i)
{return (long int)(log10((double)i)/log10((double)2));}

int main(void) {

long int t;
scanf("%ld",&t);
long int i,j,l_i,l_j,l_df,ans;

while(t-- > 0)
{
	scanf("%ld %ld",&i,&j);
	if(i>j)
		{i=i+j; j=i-j; i=i-j;} // assigning max(j,i) to j and other i is always smaller one

    l_i=level(i);
	l_j=level(j);
			
	l_df=l_j-l_i;
	ans = l_df + other_path(j/poww(l_df),i); 
    /*bring them to same level and then decrease them to the common parent*/

        printf("%ld

",ans);
}

return 0;

}

Please tell why this code is wrong ?

1 Like

A simple yet an elegant solution on the basis of a simple observation. Helped to relate the binary tree structure to its binary representation and actually how the the LCA method inherently works on this principle! Cheers!
Btw this also made me realize the effect of using cin/cout compared to scanf/printf!!
http://www.codechef.com/status/BINTREE,tukku26
10 sec difference! :stuck_out_tongue:

Why is my code incorrect?

http://www.codechef.com/viewsolution/3705831

I have followed the method in the editorial. But I am getting WA.
what could be wrong?
http://www.codechef.com/viewsolution/7022849

1 Like

I tried all edge cases I could imagine, and have no idea where my logic is going wrong. I didn’t assume the numbers as binary numbers (as you did here though).

Could anyone give some test data?

1 Like

Plz write in a bit detail .
U have not explained what is k in here
when u said
ap=bp,p<=k
What is K ???

I don’t know why it is coming wa…I have tested it for all possible cases
link : https://www.codechef.com/viewsolution/10129222

As I understood , here k is value of the shared common ancestor of i and j. Or to understand it better follow this link

simplest solution that i came up with:
#include

using namespace std;

int main()
{
int t;
cin>>t;
while(t–)
{
long long int i,j,n=0;
cin>>i>>j;

    while(i!=j)
    {
        if(i>j)
        {
            i=i/2;
            n++;

        }
        if(j>i)
        {
            j=j/2;
            n++;
        }
    }

    cout<<n<<endl;


}
return 0;

}

5 Likes

ll -> long int or long long int

ll calc(ll a, ll b){

if(a == b)
    return 0;
else if(a > b && a != 1){
    if(a%2 == 0)
        return 1+calc(a/2, b);
    else 
        return 1+calc((a-1)/2, b);
}
else if(b>a && b != 1){
    if(b%2 == 0)
        return 1+calc(a, b/2);
    else 
        return 1+calc(a, (b-1)/2);
}

}

Guys please explain me in the above explanation how ‘k’ value could 1.
Please explain me.

http://code.hackerearth.com/72c7f6j

here is link to run this code on hackerearth.com and please tell how this code is incorrect .

ThankYou

use long instead of int. As given in the question, 1 ≤ i,j ≤ 109.

try using


scanf( " %ld " , &valueToBeInserted );

Best editorial!! THANKS

https://www.codechef.com/viewsolution/24299424

https://www.codechef.com/viewsolution/24299478

here, 24299424 is getting TLE and 24299478 is getting AC but, all I have done is taken a block of code and made it a function. Why is this happening?