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

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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

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:

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

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