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 answer to the query.
Below is the implementation of the above approach:
C++
// 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 Xint 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 codeint 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;} |
Java
// Java implementation of the approachclass 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 |
Python3
# Python3 implementation of the approachMAXN = 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 Xdef 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 codequeries= [ [ 2, 5 ], [ 3, 15 ] ]q = len(queries)# Perform queriesfor i in range(q): print(solveQuery(queries[i][0],queries[i][1]))# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approachusing 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.. |
Javascript
<script>// Javascript implementation of the approachvar 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 Xfunction solveQuery(L, X){ // Initialize result var res; // Initialize array to store bits var a = Array(MAXN), b = Array(L); // Initialize a copy of X // and size of array var ref = X, size_a = 0; // Storing the bits of X // in the array a[] while (ref > 0) { a[size_a] = ref % 2; ref = parseInt(ref / 2); size_a++; } // Filling the array b[] for(var i = 0; i < Math.min(size_a, L); i++) { if (a[i] == 1) b[i] = 0; else b[i] = 1; } for(var i = Math.min(size_a, L); i < L; i++) b[i] = 1; b[L - 1] = 1; // Initializing variable which gives // maximum xor var temp = 0, p = 1; for(var 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 codevar queries = [ [ 2, 5 ], [ 3, 15 ] ];var q = queries.length;// Perform queriesfor(var i = 0; i < q; i++) document.write(solveQuery(queries[i][0], queries[i][1]) + "<br>"); // This code is contributed by rutvik_56</script> |
Output:
7 11
Time Complexity: O(q * MAXN)
Auxiliary Space: O(MAXN)
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