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> |
2
Time Complexity: O(N)
Auxiliary Space: O(1)
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