Count array elements having modular inverse under given prime number P equal to itself

Given an array arr[] of size N and a prime number P, the task is to count the elements of the array such that modulo multiplicative inverse of the element under modulo P is equal to the element itself.

Examples:

Input: arr[] = {1, 6, 4, 5}, P = 7
Output: 2
Explanation:
Modular multiplicative inverse of arr[0](=1) under modulo P(= 7) is arr[0](= 1) itself.
Modular multiplicative inverse of arr1](= 6) under modulo P(= 7) is arr[1](= 6) itself.
Therefore, the required output is 2.

Input: arr[] = {1, 3, 8, 12, 12}, P = 13
Output: 3

Naive Approach: The simplest approach is to solve this problem is to traverse the array and print the count of array elements such that modulo multiplicative inverse of the element under modulo P is equal to the element itself.

Time Complexity: O(N * log P)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the following observations:

If X and Y are two numbers such that (X × Y) % P = 1, then Y is modulo inverse of X.
Therefore, If Y is X itself, then (X × X) % P must be 1.

Follow the steps below to solve the problem:

• Initialize a variable, say cntElem to store the count of elements that satisfy the given condition.
• Traverse the given array and check if (arr[i] * arr[i]) % P equal to 1 or not. If found to be true then increment the count of cntElem by 1.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to get the count
// of elements that satisfy
// the given condition.
int equvInverse(int arr[],
                int N, int P)
{
    // Stores count of elements
    // that satisfy the condition
    int cntElem = 0;

    // Traverse the given array.
    for (int i = 0; i < N; i++) {

        // If square of current
        // element is equal to 1
        if ((arr[i] * arr[i]) % P
            == 1) {
            cntElem++;
        }
    }
    return cntElem;
}

// Driver Code
int main()
{
    int arr[] = { 1, 6, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int P = 7;
    cout << equvInverse(arr, N, P);
}

Java

// Java program to implement
// the above approach
import java.io.*;

class GFG{

// Function to get the count
// of elements that satisfy
// the given condition.
static int equvInverse(int[] arr,
                       int N, int P)
{

    // Stores count of elements
    // that satisfy the condition
    int cntElem = 0;

    // Traverse the given array.
    for(int i = 0; i < N; i++)
    {

        // If square of current
        // element is equal to 1
        if ((arr[i] * arr[i]) % P == 1)
        {
            cntElem++;
        }
    }
    return cntElem;
}

// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 6, 4, 5 };
    int N = arr.length;
    int P = 7;

    System.out.println(equvInverse(arr, N, P));
}
}

// This code is contributed by akhilsaini

Python3

# Python3 program to implement
# the above approach

# Function to get the count
# of elements that satisfy
# the given condition.
def equvInverse(arr, N, P):

    # Stores count of elements
    # that satisfy the condition
    cntElem = 0

    # Traverse the given array.
    for i in range(0, N):

        # If square of current
        # element is equal to 1
        if ((arr[i] * arr[i]) % P == 1):
            cntElem = cntElem + 1

    return cntElem

# Driver Code
if __name__ == "__main__":

    arr = [ 1, 6, 4, 5 ]
    N = len(arr)
    P = 7

    print(equvInverse(arr, N, P))

# This code is contributed by akhilsaini

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to get the count
// of elements that satisfy
// the given condition.
static int equvInverse(int[] arr, int N, int P)
{

    // Stores count of elements
    // that satisfy the condition
    int cntElem = 0;

    // Traverse the given array.
    for(int i = 0; i < N; i++)
    {

        // If square of current
        // element is equal to 1
        if ((arr[i] * arr[i]) % P == 1)
        {
            cntElem++;
        }
    }
    return cntElem;
}

// Driver Code
public static void Main()
{
    int[] arr = { 1, 6, 4, 5 };
    int N = arr.Length;
    int P = 7;

    Console.WriteLine(equvInverse(arr, N, P));
}
}

// This code is contributed by akhilsaini

Javascript

<script>
// Javascript program to implement
// the above approach

// Function to get the count
// of elements that satisfy
// the given condition.
function equvInverse(arr, N, P)
{
    // Stores count of elements
    // that satisfy the condition
    let cntElem = 0;

    // Traverse the given array.
    for (let i = 0; i < N; i++) {

        // If square of current
        // element is equal to 1
        if ((arr[i] * arr[i]) % P
            == 1) {
            cntElem++;
        }
    }
    return cntElem;
}

// Driver Code
let arr = [ 1, 6, 4, 5 ];
let N = arr.length;
let P = 7;
document.write(equvInverse(arr, N, P));

// This code is contributed by subham348.
</script>

Output:

2

Time Complexity: O(N)
Auxiliary Space: O(1)