Given a positive integer N. In one operation subtract N with its highest divisor other than N and 1. The task is to find minimum operations required to reduce N exactly to a prime number.
Examples:
Input: N = 38
Output: 1
Explanation: Highest divisor of 38 is 19, so subtract it (38 – 19) = 19. 19 is a prime number.
So, number of operations required = 1.Input: N = 69
Output: 2
Approach: This problem can be solved by using simple concepts of maths. Follow the steps below to solve the given problem.
- At first, check whether N is already prime or not.
- If N is already a prime, return 0.
- Else, Initialize a variable say count = 0 to store the number of operations required.
- Initialize a variable say i=2 and run a while loop till N!=i
- Run while loop and on each iteration subtract current value of N with its largest divisor.
- Compute the steps and increment the count by 1.
- Return the count.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function check whether a number // is prime or not bool isPrime( int n)
{ // Corner case
if (n <= 1)
return false ;
// Check from 2 to square root of n
for ( int i = 2; i <= sqrt (n); i++)
if (n % i == 0)
return false ;
return true ;
} // Function to minimum operations required // to reduce the number to a prime number int minOperation( int N)
{ // Because 1 cannot be converted to prime
if (N == 1)
return -1;
// If given number is already prime
// return 0
if (isPrime(N) == true ) {
return 0;
}
// If number is not prime
else {
// Variable for total count
int count = 0;
int i = 2;
// If number is not equal to i
while (N != i) {
// If N is completely divisible by i
while (N % i == 0) {
// Temporary variable to store
// current number
int temp = N;
// Update the number by decrementing
// with highest divisor
N -= (temp / i);
// Increment count by 1
count++;
if (isPrime(N))
return count;
}
i++;
}
// Return the count
return count;
}
} // Driver Code int main()
{ int N = 38;
cout << minOperation(N);
return 0;
} |
// Java program for the above approach import java.util.Arrays;
class GFG {
// Function check whether a number
// is prime or not
public static boolean isPrime( int n) {
// Corner case
if (n <= 1 )
return false ;
// Check from 2 to square root of n
for ( int i = 2 ; i <= Math.sqrt(n); i++)
if (n % i == 0 )
return false ;
return true ;
}
// Function to minimum operations required
// to reduce the number to a prime number
public static int minOperation( int N) {
// Because 1 cannot be converted to prime
if (N == 1 )
return - 1 ;
// If given number is already prime
// return 0
if (isPrime(N) == true ) {
return 0 ;
}
// If number is not prime
else {
// Variable for total count
int count = 0 ;
int i = 2 ;
// If number is not equal to i
while (N != i) {
// If N is completely divisible by i
while (N % i == 0 ) {
// Temporary variable to store
// current number
int temp = N;
// Update the number by decrementing
// with highest divisor
N -= (temp / i);
// Increment count by 1
count++;
if (isPrime(N))
return count;
}
i++;
}
// Return the count
return count;
}
}
// Driver Code
public static void main(String args[]) {
int N = 38 ;
System.out.println(minOperation(N));
}
} // This code is contributed by gfgking. |
# python program for the above approach import math
# Function check whether a number # is prime or not def isPrime(n):
# Corner case
if (n < = 1 ):
return False
# Check from 2 to square root of n
for i in range ( 2 , int (math.sqrt(n)) + 1 ):
if (n % i = = 0 ):
return False
return True
# Function to minimum operations required # to reduce the number to a prime number def minOperation(N):
# Because 1 cannot be converted to prime
if (N = = 1 ):
return - 1
# If given number is already prime
# return 0
if (isPrime(N) = = True ):
return 0
# If number is not prime
else :
# Variable for total count
count = 0
i = 2
# If number is not equal to i
while (N ! = i):
# If N is completely divisible by i
while (N % i = = 0 ):
# Temporary variable to store
# current number
temp = N
# Update the number by decrementing
# with highest divisor
N - = (temp / / i)
# Increment count by 1
count + = 1
if (isPrime(N)):
return count
i + = 1
# Return the count
return count
# Driver Code if __name__ = = "__main__" :
N = 38
print (minOperation(N))
# This code is contributed by rakeshsahni |
// C# program for the above approach using System;
public class GFG {
// Function check whether a number
// is prime or not
public static bool isPrime( int n)
{
// Corner case
if (n <= 1)
return false ;
// Check from 2 to square root of n
for ( int i = 2; i <= Math.Sqrt(n); i++)
if (n % i == 0)
return false ;
return true ;
}
// Function to minimum operations required
// to reduce the number to a prime number
public static int minOperation( int N) {
// Because 1 cannot be converted to prime
if (N == 1)
return -1;
// If given number is already prime
// return 0
if (isPrime(N) == true ) {
return 0;
}
// If number is not prime
else {
// Variable for total count
int count = 0;
int i = 2;
// If number is not equal to i
while (N != i) {
// If N is completely divisible by i
while (N % i == 0) {
// Temporary variable to store
// current number
int temp = N;
// Update the number by decrementing
// with highest divisor
N -= (temp / i);
// Increment count by 1
count++;
if (isPrime(N))
return count;
}
i++;
}
// Return the count
return count;
}
}
// Driver Code
public static void Main( string []args) {
int N = 38;
Console.WriteLine(minOperation(N));
}
} // This code is contributed by AnkThon |
<script> // JavaScript program for the above approach
// Function check whether a number
// is prime or not
const isPrime = (n) => {
// Corner case
if (n <= 1)
return false ;
// Check from 2 to square root of n
for (let i = 2; i <= parseInt(Math.sqrt(n)); i++)
if (n % i == 0)
return false ;
return true ;
}
// Function to minimum operations required
// to reduce the number to a prime number
const minOperation = (N) => {
// Because 1 cannot be converted to prime
if (N == 1)
return -1;
// If given number is already prime
// return 0
if (isPrime(N) == true ) {
return 0;
}
// If number is not prime
else
{
// Variable for total count
let count = 0;
let i = 2;
// If number is not equal to i
while (N != i)
{
// If N is completely divisible by i
while (N % i == 0)
{
// Temporary variable to store
// current number
let temp = N;
// Update the number by decrementing
// with highest divisor
N -= parseInt(temp / i);
// Increment count by 1
count++;
if (isPrime(N))
return count;
}
i++;
}
// Return the count
return count;
}
}
// Driver Code
let N = 38;
document.write(minOperation(N));
// This code is contributed by rakeshsahni
</script> |
1
Time Complexity: O(sqrtN)
Auxiliary Space: O(1)