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

• Last Updated : 22 Apr, 2021

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++

 // C++ program for above approach #include 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



Output:

10

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

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