# XOR of array elements whose modular inverse with a given number exists

Given an array arr[] of length N and a positive integer M, the task is to find the Bitwise XOR of all the array elements whose modular inverse with M exists.

Examples:

Input: arr[] = {1, 2, 3}, M = 4
Output: 2
Explanation:
Initialize the value xor with 0:
For element indexed at 0 i.e., 1, its mod inverse with 4 is 1 because (1 * 1) % 4 = 1 i.e., it exists. Therefore, xor = (xor ^ 1) = 1.
For element indexed at 1 i.e., 2, its mod inverse does not exist.
For element indexed at 2 i.e., 3, its mod inverse with 4 is 3 because (3 * 3) % 4 = 1 i.e., it exists. Therefore, xor = (xor ^ 3) = 2.
Hence, xor is 2.

Input: arr[] = {3, 6, 4, 5, 8}, M = 9
Output: 9
Explanation:
Initialize the value xor with 0:
For element indexed at 0 i.e., 3, its mod inverse does not exist.
For element indexed at 1 i.e., 6, its mod inverse does not exist.
For element indexed at 2 i.e., 4, its mod inverse with 9 is 7 because (4 * 7) % 9 = 1 i.e., it exists. Therefore, xor = (xor ^ 4) = 4.
For element indexed at 3 i.e., 5, its mod inverse with 9 is 2 because (5 * 2) % 9 = 1 i.e., it exists. Therefore, xor = (xor ^ 5) = 1.
For element indexed at 4 i.e., 8, its mod inverse with 9 is 8 because (8 * 8) % 9 = 1 i.e., it exists. Therefore, xor = (xor ^ 8) = 9.
Hence, xor is 9.

Naive Approach: The simplest approach is to print the XOR of all the elements of the array for which there exists any j where (1 <= j < M) such that (arr[i] * j) % M = 1 where 0 ? i < N.

Time Complexity: O(N * M)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to use the property that the modular inverse of any number X under mod M exists if and only if the GCD of M and X is 1 i.e., gcd(M, X) is 1. Follow the steps below to solve the problem:

1. Initialize a variable xor with 0, to store the xor of all the elements whose modular inverse under M exists.
2. Traverse the array over the range [0, N – 1].
3. If gcd(M, arr[i]) is 1 then update xor as xor = (xor^arr[i]).
4. After traversing, print the value xor as the required result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to return the gcd of a & b` `int` `gcd(``int` `a, ``int` `b)` `{` `    ``// Base Case` `    ``if` `(a == 0)` `        ``return` `b;` `    `  `     ``// Recursively calculate GCD` `    ``return` `gcd(b % a, a);` `}`   `// Function to print the Bitwise XOR of` `// elements of arr[] if gcd(arr[i], M) is 1` `void` `countInverse(``int` `arr[], ``int` `N, ``int` `M)` `{` `    ``// Initialize xor` `    ``int` `XOR = 0;`   `    ``// Traversing the array` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// GCD of M and arr[i]` `        ``int` `gcdOfMandelement` `          ``= gcd(M, arr[i]);`   `        ``// If GCD is 1, update xor` `        ``if` `(gcdOfMandelement == 1) {`   `            ``XOR ^= arr[i];` `        ``}` `    ``}`   `    ``// Print xor` `    ``cout << XOR << ``' '``;` `}`   `// Drive Code` `int` `main()` `{` `    ``// Given array arr[]` `    ``int` `arr[] = { 1, 2, 3 };`   `    ``// Given number M` `    ``int` `M = 4;`   `    ``// Size of the array` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function Call` `    ``countInverse(arr, N, M);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;`   `class` `GFG{`   `// Function to return the gcd of a & b` `static` `int` `gcd(``int` `a, ``int` `b)` `{` `    `  `    ``// Base Case` `    ``if` `(a == ``0``)` `        ``return` `b;`   `    ``// Recursively calculate GCD` `    ``return` `gcd(b % a, a);` `}`   `// Function to print the Bitwise XOR of` `// elements of arr[] if gcd(arr[i], M) is 1` `static` `void` `countInverse(``int``[] arr, ``int` `N, ``int` `M)` `{` `    `  `    ``// Initialize xor` `    ``int` `XOR = ``0``;`   `    ``// Traversing the array` `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{` `        `  `        ``// GCD of M and arr[i]` `        ``int` `gcdOfMandelement = gcd(M, arr[i]);`   `        ``// If GCD is 1, update xor` `        ``if` `(gcdOfMandelement == ``1``) ` `        ``{` `            ``XOR ^= arr[i];` `        ``}` `    ``}`   `    ``// Print xor` `    ``System.out.println(XOR);` `}`   `// Drive Code` `public` `static` `void` `main(String[] args)` `{`   `    ``// Given array arr[]` `    ``int``[] arr = { ``1``, ``2``, ``3` `};`   `    ``// Given number M` `    ``int` `M = ``4``;`   `    ``// Size of the array` `    ``int` `N = arr.length;`   `    ``// Function Call` `    ``countInverse(arr, N, M);` `}` `}`   `// This code is contributed by akhilsaini`

## Python3

 `# Python3 program for the above approach`   `# Function to return the gcd of a & b` `def` `gcd(a, b):` `    `  `    ``# Base Case` `    ``if` `(a ``=``=` `0``):` `        ``return` `b`   `    ``# Recursively calculate GCD` `    ``return` `gcd(b ``%` `a, a)`   `# Function to print the Bitwise XOR of` `# elements of arr[] if gcd(arr[i], M) is 1` `def` `countInverse(arr, N, M):`   `    ``# Initialize xor` `    ``XOR ``=` `0`   `    ``# Traversing the array` `    ``for` `i ``in` `range``(``0``, N):`   `        ``# GCD of M and arr[i]` `        ``gcdOfMandelement ``=` `gcd(M, arr[i])`   `        ``# If GCD is 1, update xor` `        ``if` `(gcdOfMandelement ``=``=` `1``):` `            ``XOR ``=` `XOR ^ arr[i]`   `    ``# Print xor` `    ``print``(XOR)`   `# Drive Code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``# Given array arr[]` `    ``arr ``=` `[ ``1``, ``2``, ``3` `]`   `    ``# Given number M` `    ``M ``=` `4`   `    ``# Size of the array` `    ``N ``=` `len``(arr)`   `    ``# Function Call` `    ``countInverse(arr, N, M)`   `# This code is contributed by akhilsaini`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG{`   `// Function to return the gcd of a & b` `static` `int` `gcd(``int` `a, ``int` `b)` `{` `    `  `    ``// Base Case` `    ``if` `(a == 0)` `        ``return` `b;`   `    ``// Recursively calculate GCD` `    ``return` `gcd(b % a, a);` `}`   `// Function to print the Bitwise XOR of` `// elements of arr[] if gcd(arr[i], M) is 1` `static` `void` `countInverse(``int``[] arr, ``int` `N, ``int` `M)` `{` `    `  `    ``// Initialize xor` `    ``int` `XOR = 0;`   `    ``// Traversing the array` `    ``for``(``int` `i = 0; i < N; i++)` `    ``{` `        `  `        ``// GCD of M and arr[i]` `        ``int` `gcdOfMandelement = gcd(M, arr[i]);`   `        ``// If GCD is 1, update xor` `        ``if` `(gcdOfMandelement == 1)` `        ``{`   `            ``XOR ^= arr[i];` `        ``}` `    ``}`   `    ``// Print xor` `    ``Console.WriteLine(XOR);` `}`   `// Drive Code` `public` `static` `void` `Main()` `{`   `    ``// Given array arr[]` `    ``int``[] arr = { 1, 2, 3 };`   `    ``// Given number M` `    ``int` `M = 4;`   `    ``// Size of the array` `    ``int` `N = arr.Length;`   `    ``// Function Call` `    ``countInverse(arr, N, M);` `}` `}`   `// This code is contributed by akhilsaini`

## Javascript

 ``

Output:

`2`

Time Complexity: O(N*log M)
Auxiliary Space: O(N)

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