Queries to find the maximum Xor value between X and the nodes of a given level of a perfect binary tree

Given a perfect binary tree of N nodes, with nodes having values from 1 to N as depicted in the image below and Q queries where every query consists of two integers L and X. The task is to find the maximum possible value of X XOR Y where Y can be any node at level L.

Examples:

Input: Q[] = {{2, 5}, {3, 15}}
Output:
7
11
1st Query: Level 2 has numbers 2 and 3.
Therefore, 2^5 = 7 and 3^5 = 6.
Hence, the answer is 7.
2nd Query: Level 3 has numbers 4, 5, 6 and 7
and 4^15 = 11 is the maximum possible.

Input: Q[] = {{ 1, 15 }, { 5, 23 }}
Output:
14
15

Approach: The numbers in level L contains L bits, e.g, in level 2 the numbers are 2 and 3 which can be represented using 2 bits in binary. Similarly, in level 3 the numbers are 4, 5, 6 and 7 can be represented with 3 bits.
So we have to find an L bit number which gives maximum xor with X. Store the bits of X in an array a[]. Now, fill an array b[] of size L with elements opposite to that in a[], i.e., if a[i] is equal to 0 then put b[i] equal to 1 and vice-versa.
Note that the L-1 index of b[]must always be 1. If size of a[] is less than b[] then fill the remaining indexes of b[] with 1. The array b[] is filled opposite to that of array a[] so that the number made from the bits of array b[] gives maximum xor value. Lastly, calculate the number made from b[] and return its xor with X as the anwser to the query.



Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAXN 60
  
// Function to solve queries of the maximum xor value
// between the nodes in a given level L of a
// perfect binary tree and a given value X
int solveQuery(int L, int X)
{
    // Initialize result
    int res;
  
    // Initialize array to store bits
    int a[MAXN], b[L];
  
    // Initialize a copy of X
    // and size of array
    int ref = X, size_a = 0;
  
    // Storing the bits of X
    // in the array a[]
    while (ref > 0) {
        a[size_a] = ref % 2;
        ref /= 2;
        size_a++;
    }
  
    // Filling the array b[]
    for (int i = 0; i < min(size_a, L); i++) {
        if (a[i] == 1)
            b[i] = 0;
        else
            b[i] = 1;
    }
  
    for (int i = min(size_a, L); i < L; i++)
        b[i] = 1;
  
    b[L - 1] = 1;
  
    // Initializing variable which gives
    // maximum xor
    int temp = 0, p = 1;
  
    for (int i = 0; i < L; i++) {
        temp += b[i] * p;
        p *= 2;
    }
  
    // Getting the maximum xor value
    res = temp ^ X;
  
    // Return the result
    return res;
}
  
// Driver code
int main()
{
    int queries[][2] = { { 2, 5 }, { 3, 15 } };
    int q = sizeof(queries) / sizeof(queries[0]);
  
    // Perform queries
    for (int i = 0; i < q; i++)
        cout << solveQuery(queries[i][0],
                           queries[i][1])
             << endl;
  
    return 0;
}
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// Java implementation of the approach
  
class GFG
{
      
    static int MAXN = 60
      
    // Function to solve queries of the maximum xor value 
    // between the nodes in a given level L of a 
    // perfect binary tree and a given value X 
    static int solveQuery(int L, int X) 
    
        // Initialize result 
        int res; 
      
        // Initialize array to store bits 
        int []a = new int [MAXN];
        int []b = new int[L]; 
      
        // Initialize a copy of X 
        // and size of array 
        int refer = X, size_a = 0
      
        // Storing the bits of X 
        // in the array a[] 
        while (refer > 0)
        
            a[size_a] = refer % 2
            refer /= 2
            size_a++; 
        
      
        // Filling the array b[] 
        for (int i = 0; i < Math.min(size_a, L); i++) 
        
            if (a[i] == 1
                b[i] = 0
            else
                b[i] = 1
        
      
        for (int i = Math.min(size_a, L); i < L; i++) 
            b[i] = 1
      
        b[L - 1] = 1
      
        // Initializing variable which gives 
        // maximum xor 
        int temp = 0, p = 1
      
        for (int i = 0; i < L; i++) 
        
            temp += b[i] * p; 
            p *= 2
        
      
        // Getting the maximum xor value 
        res = temp ^ X; 
      
        // Return the result 
        return res; 
    
      
    // Driver code 
    static public void main (String args[])
    
        int [][]queries= { { 2, 5 }, { 3, 15 } }; 
        int q = queries.length; 
      
        // Perform queries 
        for (int i = 0; i < q; i++) 
            System.out.println( solveQuery(queries[i][0], 
                            queries[i][1]) );
          
    
}
  
// This code is contributed by AnkitRai01
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# Python3 implementation of the approach
  
MAXN = 60
  
# Function to solve queries of the maximum xor value
# between the nodes in a given level L of a
# perfect binary tree and a given value X
def solveQuery(L, X):
      
    # Initialize result
    res = 0
  
    # Initialize array to store bits
    a = [0 for i in range(MAXN)]
    b = [0 for i in range(MAXN)]
  
    # Initialize a copy of X
    # and size of array
    ref = X
    size_a = 0
  
    # Storing the bits of X
    # in the array a[]
    while (ref > 0):
        a[size_a] = ref % 2
        ref //= 2
        size_a+=1
  
    # Filling the array b[]
    for i in range(min(size_a,L)):
        if (a[i] == 1):
            b[i] = 0
        else:
            b[i] = 1
  
  
    for i in range(min(size_a, L),L):
        b[i] = 1
  
    b[L - 1] = 1
  
    # Initializing variable which gives
    # maximum xor
    temp = 0
    p = 1
  
    for i in range(L):
        temp += b[i] * p
        p *= 2
  
    # Getting the maximum xor value
    res = temp ^ X
  
    # Return the result
    return res
  
# Driver code
queries= [ [ 2, 5 ], [ 3, 15 ] ]
  
q = len(queries)
  
# Perform queries
for i in range(q):
    print(solveQuery(queries[i][0],queries[i][1]))
  
# This code is contributed by mohit kumar 29
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// C# implementation of the approach
using System;
  
class GFG
{
      
    static int MAXN = 60; 
      
    // Function to solve queries of the maximum xor value 
    // between the nodes in a given level L of a 
    // perfect binary tree and a given value X 
    static int solveQuery(int L, int X) 
    
        // Initialize result 
        int res; 
      
        // Initialize array to store bits 
        int []a = new int [MAXN];
        int []b = new int[L]; 
      
        // Initialize a copy of X 
        // and size of array 
        int refer = X, size_a = 0; 
      
        // Storing the bits of X 
        // in the array a[] 
        while (refer > 0)
        
            a[size_a] = refer % 2; 
            refer /= 2; 
            size_a++; 
        
      
        // Filling the array b[] 
        for (int i = 0; i < Math.Min(size_a, L); i++) 
        
            if (a[i] == 1) 
                b[i] = 0; 
            else
                b[i] = 1; 
        
      
        for (int i = Math.Min(size_a, L); i < L; i++) 
            b[i] = 1; 
      
        b[L - 1] = 1; 
      
        // Initializing variable which gives 
        // maximum xor 
        int temp = 0, p = 1; 
      
        for (int i = 0; i < L; i++) 
        
            temp += b[i] * p; 
            p *= 2; 
        
      
        // Getting the maximum xor value 
        res = temp ^ X; 
      
        // Return the result 
        return res; 
    
      
    // Driver code 
    static public void Main ()
    
        int [,]queries= { { 2, 5 }, { 3, 15 } }; 
        int q = queries.Length; 
      
        // Perform queries 
        for (int i = 0; i < q; i++) 
            Console.WriteLine( solveQuery(queries[i,0], 
                            queries[i,1]) );
          
    
}
  
// This code is contributed by anuj_67..
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Output:
7
11

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Improved By : mohit kumar 29, AnkitRai01, vt_m

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