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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output: 

2



 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : akhilsaini