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:

1. Keep replacing N to sqrt(N) until N is a perfect square.
2. 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.
3. 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

Javascript

<script>

// Javascript program for above approach

// Function to reduce N to its minimum
// possible value by the given operations
function minValue(n)
{

    // Keep replacing n until is
    // an integer
    while (parseInt(Math.sqrt(n)) == Math.sqrt(n)
        && n > 1)
    {
        n = parseInt(Math.sqrt(n));
    }

    // Keep replacing n until n
    // is divisible by i * i
    for (var i = parseInt(Math.sqrt(n));
        i > 1; i--) {

        while (n % (i * i) == 0)
            n /= i;
    }

    // Print the answer
    document.write(n);
}

// Driver Code

// Given N
var N = 20;

// Function Call
minValue(N);

// This code is contributed by rutvik_56.
</script>

Output:

10

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