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: 

Examples: 
 

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

Input: N = 5184 
Output:
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++

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// 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); 
}

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Java

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// 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

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Python3

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# 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_

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C#

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// 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

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Output:

10

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

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Improved By : VirusBuddah_, vikas_g