Min operations to reduce N by multiplying by any number or taking square root

Given a number N, the task is to find the minimum value of N by applying below operations any number of times:

Multiply N by any positive integer
Replace N with sqrt(N), only if N is a perfect square.

Examples:

Input: N = 20
Output: 10
Explanation:
Multiply -> 20 * 5 = 100
sqrt(100) = 10, which is the minimum value obtainable.

Input: N = 5184
Output: 6
Explanation:
sqrt(5184) = 72.
Multiply -> 72*18 = 1296
sqrt(1296) = 6, which is the minimum value obtainable.

Approach: This problem can be solved using Greedy Approach. Below are the steps:

Keep replacing N to sqrt(N) until N is a perfect square.
After the above step, iterate from sqrt(N) to 2, and for every, i keep replacing N with N / i if N is divisible by i².
The value of N after the above step will be the minimum possible value.

Below is the implementation of the above approach:

C++

// C++ program for above approach  
#include <bits/stdc++.h>  
using namespace std;  
  
// Function to reduce N to its minimum  
// possible value by the given operations  
void minValue(int n)  
{  
    // Keep replacing n until is  
    // an integer  
    while (int(sqrt(n)) == sqrt(n)  
        && n > 1) {  
        n = sqrt(n);  
    }  
  
    // Keep replacing n until n  
    // is divisible by i * i  
    for (int i = sqrt(n);  
        i > 1; i--) {  
  
        while (n % (i * i) == 0)  
            n /= i;  
    }  
  
    // Print the answer  
    cout << n;  
}  
  
// Driver Code  
int main()  
{  
    // Given N  
    int N = 20;  
  
    // Function Call  
    minValue(N);  
} 

Java

// Java implementation of the above approach  
import java.lang.Math;  
  
class GFG{  
  
// Function to reduce N to its minimum  
// possible value by the given operations  
static void minValue(int n)  
{  
      
    // Keep replacing n until is  
    // an integer  
    while ((int)Math.sqrt(n) ==  
                Math.sqrt(n) && n > 1)  
    {  
        n = (int)(Math.sqrt(n));  
    }  
  
    // Keep replacing n until n  
    // is divisible by i * i  
    for(int i = (int)(Math.sqrt(n));  
            i > 1; i--) 
    {  
        while (n % (i * i) == 0)  
            n /= i; 
    }  
      
    // Print the answer  
    System.out.println(n);  
} 
  
// Driver code  
public static void main(String args[]) 
{ 
      
    // Given N  
    int N = 20;  
      
    // Function call  
    minValue(N);  
}  
} 
  
// This code is contributed by vikas_g 

Python3

# Python3 program for the above approach  
import math  
  
# Function to reduce N to its minimum  
# possible value by the given operations  
def MinValue(n): 
      
    # Keep replacing n until is  
    # an integer  
    while(int(math.sqrt(n)) == 
              math.sqrt(n) and n > 1): 
        n = math.sqrt(n) 
          
    # Keep replacing n until n  
    # is divisible by i * i  
    for i in range(int(math.sqrt(n)), 1, -1): 
        while (n % (i * i) == 0): 
            n /= i 
              
    # Print the answer  
    print(n) 
  
# Driver code 
n = 20
  
# Function call 
MinValue(n) 
  
# This code is contributed by virusbuddah_ 

C#

// C# implementation of the approach  
using System;  
  
class GFG{  
      
// Function to reduce N to its minimum  
// possible value by the given operations  
static void minValue(int n)  
{  
      
    // Keep replacing n until is  
    // an integer  
    while ((int)Math.Sqrt(n) ==  
                Math.Sqrt(n) && n > 1) 
    {  
        n = (int)(Math.Sqrt(n));  
    }  
      
    // Keep replacing n until n  
    // is divisible by i * i  
    for (int i = (int)(Math.Sqrt(n)); 
             i > 1; i--) 
    {  
        while (n % (i * i) == 0)  
            n /= i;  
    }  
      
    // Print the answer  
    Console.Write(n);  
} 
  
// Driver code  
public static void Main()  
{ 
      
    // Given N  
    int N = 20;  
      
    // Function call  
    minValue(N); 
} 
} 
  
// This code is contributed by vikas_g 

Output:

Time Complexity: O(N)
Auxiliary Space: O(1)

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My Personal Notes

C++

Java

Python3

C#

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