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++ program for the above approach
#include <bits/stdc++.h>
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 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 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# 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

<script>

// Javascript program for the above approach

// Function to return the gcd of a & b
function 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
function countInverse(arr, N, M)
{
    
    // Initialize xor
    var XOR = 0;

    // Traversing the array
    for(var i = 0; i < N; i++) 
    {
        
        // GCD of M and arr[i]
        var gcdOfMandelement = gcd(M, arr[i]);

        // If GCD is 1, update xor
        if (gcdOfMandelement == 1) 
        {
            XOR ^= arr[i];
        }
    }

    // Print xor
    document.write(XOR);
}

// Driver Code

// Given array arr[]
var arr = [ 1, 2, 3 ];

// Given number M
var M = 4;

// Size of the array
var N = arr.length;

// Function Call
countInverse(arr, N, M);

// This code is contributed by Kirti

</script>                    

2

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