Given an integer N, below operations can be performed any number of times on N:
- Multiply N by any positive integer X i.e. N = N * X.
- Replace N with square root of N (N must be an integer) i.e. N = sqrt(N).
The task is to find the minimum integer to which N can be reduced with the above operations.
Examples:
Input: N = 20
Output: 10
We can multiply 20 by 5, then take sqrt(20*5) = 10, this is the minimum number that 20 can be reduced to with the given operations.Input: N = 36
Output: 6
Take sqrt(36). Number 6 can’t be reduced further.
Approach:
- First factorize the number N.
- Say, 12 has factors 2, 2 and 5. Only the factors that are repeating can be reduced with sqrt(n) i.e. sqrt(2*2) = 2.
- The numbers appearing only once in the factors cannot be further reduced.
- So, the final answer will be the product of all the distinct prime factors of number N
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>#define ll long long intusing namespace std;// function to return the product of// distinct prime factors of a numberll minimum(ll n){ ll product = 1; // find distinct prime for (int i = 2; i * i <= n; i++) { if (n % i == 0) { while (n % i == 0) n = n / i; product = product * i; } } if (n >= 2) product = product * n; return product;}// Driver codeint main(){ ll n = 20; cout << minimum(n) << endl; return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class solution{ // function to return the product of // distinct prime factors of a numberstatic int minimum(int n){ int product = 1; // find distinct prime for (int i = 2; i * i <= n; i++) { if (n % i == 0) { while (n % i == 0) n = n / i; product = product * i; } } if (n >= 2) product = product * n; return product;}// Driver codepublic static void main(String arr[]){ int n = 20; System.out.println(minimum(n));}}//This code is contributed by//Surendra_Gangwar |
Python3
# Python3 implementation of above approach# function to return the product # of distinct prime factors of a # numberdef minSteps(str):def minimum(n): product = 1 # find distinct prime i = 2 while i * i <= n: if n % i == 0: while n % i == 0: n = n / i product = product * i i = i + 1 if n >= 2: product = product * n return product# Driver code# Get the binary stringn = 20print(minimum(n)) # This code is contributed# by Shashank_Sharma |
C#
// C# implementation of the above approachclass GFG{// function to return the product of// distinct prime factors of a numberstatic int minimum(int n){ int product = 1; // find distinct prime for (int i = 2; i * i <= n; i++) { if (n % i == 0) { while (n % i == 0) n = n / i; product = product * i; } } if (n >= 2) product = product * n; return product;}// Driver codestatic void Main(){ int n = 20; System.Console.WriteLine(minimum(n));}}// This code is contributed by mits |
PHP
<?php// PHP implementation of the// above approach// function to return the product of// distinct prime factors of a numberfunction minimum($n){ $product = 1; // find distinct prime for ($i = 2; $i * $i <= $n; $i++) { if ($n % $i == 0) { while ($n % $i == 0) $n = $n / $i; $product = $product * $i; } } if ($n >= 2) $product = $product * $n; return $product;}// Driver code$n = 20;echo minimum($n),"\n";// This code is contributed by ANKITRAI1?> |
Javascript
<script>// JavaScript implementation of the above approach// function to return the product of// distinct prime factors of a numberfunction minimum( n){ let product = 1; // find distinct prime for (let i = 2; i * i <= n; i++) { if (n % i == 0) { while (n % i == 0) n = n / i; product = product * i; } } if (n >= 2) product = product * n; return product;}// Driver code let n = 20; document.write(minimum(n));// This code is contributed by sravan kumar</script> |
Output:
10
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
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