TAAND - Editorial

ad-hoc
bitwise-operatn
easy-medium
ltime14

#1

Problem link : contest practice

Difficulty : Simple

Pre-requisites : Basic Math, Recursion, Bit Operations

Problem : Given an array find maximum AND(&) possible between any two elements.

##Explanation

At first, let’s consider some partial solutions.

How to get 50 points

Here you can go through each possible pair and check what AND value is. And we can find what is the maximum AND possible. Complexity of this will be O(N*N).

n=input()
arr=[]
ans=-1
for i=1 to n:
	arr*=input()
for i=1 to n:
	for j=i+1 to n:
		ans = max( ans, arr*&arr[j] )

How to get 100 points

When we are ANDing two numbers, we would like to have a 1 at the most significant bit(MSB). So, first we’ll try to get a 1 at MSB. Now, suppose we denote A*=b_in,b_in-1…b_i0, where b_i’s are the bits that could be 1 or 0.

Let’s say S be the set of A* whose b_in is 1 and S’ be the set of A* whose b_in is 0. If size of S ≥ 2 we are sure that our answer will be maximum if we chose a pair of numbers from S, because n’th bit of their AND will be 1 for sure. So, we know our answer lies in S.

However, if size of S is less than 2, we can never have n’th bit 1 in our answer. So, we’ll have to continue with n’th bit as 0. Note that our answer will now be in S’.

Now, we know our answer is in S or S’ and we also know n’th bit of our answer. So, our new subproblem is to find (n-1)‘th bit of our answer using numbers in S or S’. We can write a recursive code for this.
What will be the complexity? For each of the n bits, we’ll traverse whole array to sort according to their bits. So O(nN). We will be keeping n=30, because A ≤ 109.

n=input()
arr=[]
flag[n]={}
for i=0 to n-1:
	arr*=input()
print foo(30)

def foo(level):  //this function will find the level'th bit
                //and accordingly return answer
	if level==-1:  // we already have solved for all bits. return 0 from here
		return 0
	Scount=0  // stores the size of set S
	ans=0 // the answer in decimal form
	for i=0 to n-1:
		if flag*==0:  // flag*=0 means arr* is available for us to use
			if (arr*&(1<<level)):	// level'th bit of arr* is 1
				Scount++
	if Scount>=2: // our answer's level'th bit will be 1
		ans += (1<<level)
		// but we also have to set the flag of unavailable numbers
		for i=0 to n-1:
			if (arr*&(1<<level))==0:  // level'th bit is 0, set the flag
				flag*=1
	return ans+foo(level-1);  // return the current answer + answer for the next bit

Solutions : setter tester


#2

Why cant you just sort an array and try updating max for every two consecutive elements?
It passed to me.


#3

I submitted this code in java.It gave NZEC.Can anybody tell why?

                                                                                                                                                                                                                               import java.io.BufferedReader;

                                                                                                                       import java.io.InputStreamReader;
 import java.io.PrintWriter;
import java.util.ArrayList;                                                                                                                                   import java.util.StringTokenizer;

 class AndOperation {

      long n=0;

      Integer arr[];  
                                                                                                                         public static void main(String[] args) throws Exception {
   
  
     AndOperation a=new AndOperation();

     a.solve();
   }

public void solve()throws Exception{
    long result=0;
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out=new PrintWriter(System.out);
    n=Integer.parseInt(br.readLine());
    arr=new Integer[(int)n];
 long ok=0;
    if(n>=2)
        {
      for(int i=0;i<n;i++)
         arr*=Integer.parseInt(br.readLine());
         }
  for(int i=0;i<n;i++){
       for(int j=i+1;j<n;j++){
            ok=arr*& arr[j];
               if(ok>result)
                 result=ok;
  
               }
          }
     System.out.println(result);

     br.close();
}

}


#4

I have used this logic but unable to get ACCEPTED:

Sort numbers according to the range of power of 2 i.e 0-2, 2-4, 4-8, 8-16…and so on…

Now the last range in which we have at least two numbers contains those two integers

So just And the biggest two numbers in this range to get the answer

What is wrong with this?


#5

There must be some error in test cases as I did a simple sorting of numbers and took & of consecutive numbers and got an AC.But it’s is wrong solution as ans for test case :
3
10
3
19
should be 3 but it will give 2.
19 = 10011
10 = 01010
3 = 00011
ANS = 3
but by sorting ANS = 2
And it’s an AC code. So I think that the solutions should be rejudged.
Link : http://www.codechef.com/viewsolution/4392454


#6

The question says, 1 <= u < v <= N


#7

How can we solve this problem using Trie ? I read the blog and understood the idea of finding 2 elements whose XOR is maximum using trie. But if we have to use that concept here then if a particular bit is zero in the number then either of the branch of the trie can yield optimal answer. So how to choose a specific branch ?


#8

my ans uses a different approach .
I just sorted the array and took and of the numbers and maintained that a&b <=max(a,b);
Is my approach correct?
my solution is link http://www.codechef.com/viewsolution/4386414


#9

I was bored to write a code on the formal algorithm I had developed i.e to check ranges of 2^n - where in you have to find the max range with at least two elements in that range and then compute AND operation on the closest two numbers. So instead, I checked the highest 5 elements and their AND operations with all elements and stored the max- giving AC(100) with O(10*n). Not very efficient and accurate but can be helpful in terms of saving time in contests.

http://www.codechef.com/viewsolution/4387562


#10

my solution is easy one :slight_smile:

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int main(){
int n;
cin>>n;
int a[n],b[n];
for(int i=0;i<n;i++){
cin>>a*;
b*=log2(a*);
// cout<<b*;
}
int mx=0,temp;
std::sort(a,a+n);
std::sort(b,b+n);
 
for(int i=n-1;i>=0;i--){
if(b*==b[i-1]){
mx=max(mx,a*&a[i-1]);
}
}
cout<<mx<<endl;
return 0;
}

#11

push back every result into vector and sort.print the last one.
it accepted.


#12

JUST STICK TO AND DEF.(it gives the minterm i.e. & of 2 number is smaller then small of 2 numbers, so… )
#include<stdio.h>
main(void)
{
int n,i,j,max=0;
scanf("%d",&n);
int a[n];
for(i=0;i<n;i++)
scanf("%d",&a*);

for(i=0;i<n;i++)
if (a*>max)
for(j=i+1;j<n;j++)
if((a*&a[j])>max) max=a*&a[j];
 
printf("%d",max);
return 0;
}

#13

what is the maths behind simply sorting the array and printing max of sum(by bitwise and) of consecutive elements. Why is there no need to test sum of non-consecutive elements?


#14

Not so much to do! simply sort the numbers and run a loop to find the maximum if AND of consecutive numbers And that gives you the answer.Here is the code for it:
#include<bits/stdc++.h>
int main()
{ios_base::sync_with_stdio(false);
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++)
{
cin>>a*;
}
sort(a,a+n);
int max=-1;
for(int i=0;i<n-1;i++)
{
int t=a*&a[i+1];
if(t>max)
max=t;
}
cout<<max;

return 0;

}

It is 100 points submission…now the logic:
Rather than running an O(n^2) algo you have many things to keep in mind.
1.Bigger numbers when ANDed will provide big results(generally).
2.A big and a small number will provide small results(most of the times)
3.So better not to find all nC2 pairs but compute from the consecutive pairs and get the solution.
HOPE IT HELPS!


#15

So i used the logic, for every number calculate the position of set bit and for that bit keep the record of two max numbers, for eg: in 2, 8, 10, a record will be kept for 8 and 10 for bit position 2. Keep a record of the max bit position where more than two elements have set bit for that particular bit position. Now print (arr[maxn][0] & arrmaxn) where maxn is highest bit power and 0 and two are the highest elements for that particular bit.

My solution passed for subtask 1 when i used set, but it doesnt pass anymore when i use array with the same logic.
Please review my code

http://pastebin.com/WvqeeZWA


#16

Can anyone tell me what’s wrong with this code.I understand that I am making redundant vectors. But why wrong answer.
Thanks!!!

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<algorithm>
    #include<vector>
    #include<cmath>
    int ans;
    using namespace std;
    int recursion(vector<int> v,int n)
    {
    	if(n==-1)
    		return 0;
    	vector<int> v1;
    	vector<int> v2;
    	int size1,size2;
    	size1=size2=0;
    	for(int i=0;i<v.size();i++)
    	{
    		if((1<<n)&v*)
    		{
    			v1.push_back(v*);
    			size1++;
    		}
    		else
    		{
    			v2.push_back(v*);
    			size2++;
    		}
    	}
    	if(size1>=2)
    	{
    		//cout<<" calling1"<<endl;
    		return (1<<n) + recursion(v1,n-1);
    	}
    	else
    	{
    		//cout<<" calling2"<<endl;
    		return recursion(v2,n-1);
    	}
    }
    int main()
    {
    	/*int n=4;
    	for(int i=0;i<18;i++)
    	{
    		cout<<((1<<n)&i) <<" ";
    	}*/
    	int n;
    	scanf("%d",&n);
    	vector<int> v(n);
    	for(int i=0;i<n;i++)
    		cin>>v*;
    	ans=0;
    	printf("%d

",recursion(v,32));
}


#17

very Weak test cases. @o007’s solution not giving right ans but got AC.


#18

the set of test cases for this question is not exhaustive, if u see my algorithm, u can get 100 points for the question but my algorithm fails in this test case -

3
85
42
21

my algorithm -
store all elements in array . sort the array. then traverse in reverse direction and compute the maximum of array*,array[i-1]
the max will be the ans


#19

I am getting RE(SIGSEGV) in the code - I haven’t allocated a very large amount of memory and I cannot detect any illegal memory reference made by me. I have used an iterative solution of the bit-representation logic.

Solution

I would be thankful for any help.


#20

It should be fail on testcase like 4 2 4 8 18 as pointed by abhisheklfs in problem’s comment , really weak test even my bruteforce got accepted.