Count array elements having modular inverse under given prime number P equal to itself
Last Updated :
21 Jun, 2022
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 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 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# 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
<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>
Time Complexity: O(N)
Auxiliary Space: O(1)
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