Given a prime number N, the task is to find the closest smaller number than N such that modulo multiplicative inverse of a number under modulo N is equal to the number itself.
Examples:
Input: N = 7
Output: 6
Explanation:
Modulo multiplicative inverse of all possible natural numbers from 1 to less than N are:
Modulo multiplicative inverse of 1 under modulo N(=7) is 1.
Modulo multiplicative inverse of 2 under modulo N(=7) is 4.
Modulo multiplicative inverse of 3 under modulo N(=7) is 5.
Modulo multiplicative inverse of 4 under modulo N(=7) is 2.
Modulo multiplicative inverse of 5 under modulo N(=7) is 3.
Modulo multiplicative inverse of 6 under modulo N(=7) is 6.
Therefore, the nearest smaller number to N(= 7) having modulo inverse equal to the number itself is 6.Input: N = 11
Output: 10
Naive Approach: The simplest approach to solve this problem is to traverse all natural numbers from 1 to N and find the largest number such that modulo multiplicative inverse of the number under modulo N is equal to the number itself.
Time Complexity: O(N * log N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach the idea is based on the following observations:
The nearest smaller number to N having modulo multiplicative inverse equal to the number itself is (N – 1).
Mathematical proof:
If X and Y are two numbers such that (X * Y) % N = 1 mod(N), then Y is modulo inverse of X.
Put X = N – 1 then
=>((N – 1) * Y) % N = 1 mod(N)
=>(N × Y) % N – Y % N = 1 mod(N)
=> Y = N – 1
Therefore, for X = N – 1 the value of Y is equal to X.
Therefore, to solve the problem, simply print N – 1 as the required answer.
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the nearest // smaller number satisfying // the condition int clstNum( int N)
{ return (N - 1);
} // Driver Code int main()
{ int N = 11;
cout << clstNum(N);
} |
// Java program to implement // the above approach import java.io.*;
class GFG{
// Function to find the nearest // smaller number satisfying // the condition static int clstNum( int N){ return (N - 1 ); }
// Driver Code public static void main(String[] args)
{ int N = 11 ;
System.out.println(clstNum(N));
} } // This code is contributed by akhilsaini |
# Python3 program to implement # the above approach # Function to find the nearest # smaller number satisfying # the condition def clstNum(N):
return (N - 1 )
# Driver Code if __name__ = = '__main__' :
N = 11
print (clstNum(N))
# This code is contributed by akhilsaini |
// C# program to implement // the above approach using System;
class GFG{
// Function to find the nearest // smaller number satisfying // the condition static int clstNum( int N){ return (N - 1); }
// Driver Code public static void Main()
{ int N = 11;
Console.Write(clstNum(N));
} } // This code is contributed by akhilsaini |
<script> // Javascript program to implement // the above approach // Function to find the nearest // smaller number satisfying // the condition function clstNum(N)
{ return (N - 1);
} // Driver Code let N = 11; document.write(clstNum(N)); // This code is contributed by subham348. </script> |
10
Time Complexity: O(1)
Auxiliary Space: O(1)