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

0 0 5 minutes read

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:

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

// 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

<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>

Output:

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

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!