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# PROBLEM LINK:

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

EASY

# 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.

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.

This question is marked "community wiki".

asked 14 Apr '14, 15:07

3.0k93164187
accept rate: 12%

4★kunal361
6.0k133272

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

(14 Apr '14, 21:07) 2★

Best editorial!! THANKS

(20 Sep '16, 16:21) 2★

11 Answers:

simplest solution that i came up with:

# include<iostream>

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;

}

link
This answer is marked "community wiki".

answered 27 Oct '16, 23:00

11
accept rate: 0%

 1 @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. answered 14 Apr '14, 21:49 2★azix6 10●2 accept rate: 0%

# 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\n",ans);
}

return 0;

}

Please tell why this code is wrong ?

answered 19 Apr '14, 13:52

2113
accept rate: 0%

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

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

ThankYou

(19 Apr '14, 14:08)
 1 I have followed the method in the editorial. But I am getting WA. what could be wrong? http://www.codechef.com/viewsolution/7022849 answered 25 May '15, 19:37 11●1 accept rate: 0%
 1 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? answered 16 Jun '15, 22:07 0★suryak 11●1 accept rate: 0%
 0 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! :p answered 24 Apr '14, 19:03 4★tukku26 86●1●3 accept rate: 12%
 0 Why is my code incorrect? http://www.codechef.com/viewsolution/3705831 answered 01 May '14, 18:18 1 accept rate: 0%
 0 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 ??? answered 25 Dec '15, 21:42 2★vchhabra 105●1●10 accept rate: 12%
 0 I don't know why it is coming wa...I have tested it for all possible cases link : https://www.codechef.com/viewsolution/10129222 answered 19 May '16, 16:22 1 accept rate: 0% use long instead of int. As given in the question, 1 ≤ i,j ≤ 109. try using scanf( " %ld " , &valueToBeInserted ); (08 Jul '16, 20:46)
 0 As I understood , here k is value of the shared common ancestor of i and j. Or to understand it better follow this link http://www.geeksforgeeks.org/lowest-common-ancestor-binary-tree-set-1/ answered 21 Jul '16, 17:05 1 accept rate: 0%
 0 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); } } answered yesterday 1●1 accept rate: 0%
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question asked: 14 Apr '14, 15:07

question was seen: 10,939 times

last updated: yesterday