# 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.
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

 `// C++ implementation of the approach ` `#include ` `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, 5 }, { 3, 15 } }; ` `    ``int` `q = ``sizeof``(queries) / ``sizeof``(queries); ` ` `  `    ``// Perform queries ` `    ``for` `(``int` `i = 0; i < q; i++) ` `        ``cout << solveQuery(queries[i], ` `                           ``queries[i]) ` `             ``<< endl; ` ` `  `    ``return` `0; ` `} `

 `// 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 `

 `# 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 `

 `// 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.. `

Output:
```7
11
```

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Improved By : mohit kumar 29, AnkitRai01, vt_m

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