Count operations of the given type required to reduce N to 0

Given an integer n. The task is to count the number of operations required to reduce n to 0. In every operation, n can be updated as n = n – d where d is the smallest prime divisor of n.

Examples:

Input: n = 5
Output: 1
5 is the smallest prime divisor, thus it gets subtracted and n gets reduced to 0.



Input: n = 25
Output: 11
5 is the smallest prime divisor, thus it gets subtracted and n gets reduced to 20. Then 2 is the smallest divisor and so on.

Input: n = 4
Output: 2

Approach:

  1. When n is even then the smallest prime divisor of n will be 2 and subtracting 2 from n will again give an even integer i.e. gain 2 will be chosen as the smallest prime divisor and these steps will repeat until n gets reduced to 0.
  2. When n is odd then the smallest prime divisor of n will also be odd and subtracting an odd integer from another odd integer will give an even integer as the result and then the result can be found out by repeating step 1 for the current even integer.
  3. Thus, the task is to find the smallest divisor d, subtract it, n = n – d and print 1 + ((n – d) / 2).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
  
// Function to return the required 
// number of operations
int countOperations (int n)
{
    int i = 2;
      
    // Finding the smallest divisor
    while ((i * i) < n && (n % i))
        i += 1;
      
    if ((i * i) > n)
        i = n;
      
    // Return the count of operations
    return (1 + (n - i)/2);
}
  
// Driver code
int main()
{
    int n = 5;
    cout << countOperations(n);
}
  
//This code is contributed by Shivi_Aggarwal

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Java

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// Java implementation of the approach
class GFG
{
  
// Function to return the required 
// number of operations
static int countOperations (int n)
{
    int i = 2;
      
    // Finding the smallest divisor
    while ((i * i) < n && (n % i) > 0)
        i += 1;
      
    if ((i * i) > n)
        i = n;
      
    // Return the count of operations
    return (1 + (n - i) / 2);
}
  
// Driver code
public static void main(String[] args)
{
    int n = 5;
    System.out.println(countOperations(n));
}
}
  
// This code is contributed
// by Akanksha Rai

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Python3

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# Python3 implementation of the approach
  
# Function to return the required 
# number of operations
def countOperations(n):
    i = 2
      
    # Finding the smallest divisor
    while ((i * i) < n and (n % i)):
        i += 1
      
    if ((i * i) > n):
        i = n
      
    # Return the count of operations
    return (1 + (n - i)//2)
  
# Driver code
n = 5
print(countOperations(n))

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C#

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// C# implementation of the approach
using System;
class GFG
{
  
// Function to return the required 
// number of operations
static int countOperations (int n)
{
    int i = 2;
      
    // Finding the smallest divisor
    while ((i * i) < n && (n % i) > 0)
        i += 1;
      
    if ((i * i) > n)
        i = n;
      
    // Return the count of operations
    return (1 + (n - i) / 2);
}
  
// Driver code
static void Main()
{
    int n = 5;
    Console.WriteLine(countOperations(n));
}
}
  
// This code is contributed by mits

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PHP

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<?php
// PHP implementation of the approach 
  
// Function to return the required 
// number of operations 
function countOperations($n)
    $i = 2;
      
    # Finding the smallest divisor 
    while (($i * $i) < $n and ($n % $i))
        $i += 1;
          
    if (($i * $i) > $n
        $i = $n;
      
    # Return the count of operations 
    return 1 + floor(($n - $i) / 2);
}
  
// Driver code 
$n = 5 ;
echo countOperations($n); 
  
// This code is contributed by Ryuga
?>

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Output:

1

Time Complexity: O(\sqrt{n})



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