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 coden = 20# Function callMinValue(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)
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