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XOR of array elements whose modular inverse with a given number exists
• Difficulty Level : Medium
• Last Updated : 19 Apr, 2021

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);` `    ``// 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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