Given a positive integer n. Find the minimum number which divide n to make it a perfect square.
Examples:
Input : n = 50 Output : 2 By Dividing n by 2, we get which is a perfect square. Input : n = 6 Output : 6 By Dividing n by 6, we get which is a perfect square. Input : n = 36 Output : 1
A number is perfect square if all prime factors appear even number of times. The idea is to find the prime factor of n and find each prime factor power. Now, find and multiply all the prime factor whose power is odd. The resultant of the multiplication is the answer.
C++
// C++ program to find minimum number which divide n // to make it a perfect square. #include<bits/stdc++.h> using namespace std; // Return the minimum number to be divided to make // n a perfect square. int findMinNumber(int n) { int count = 0, ans = 1; // Since 2 is only even prime, compute its // power separately. while (n%2 == 0) { count++; n /= 2; } // If count is odd, it must be removed by dividing // n by prime number. if (count%2) ans *= 2; for (int i = 3; i <= sqrt(n); i += 2) { count = 0; while (n%i == 0) { count++; n /= i; } // If count is odd, it must be removed by // dividing n by prime number. if (count%2) ans *= i; } if (n > 2) ans *= n; return ans; } // Driven Program int main() { int n = 72; cout << findMinNumber(n) << endl; return 0; } |
Java
// Java program to find minimum number // which divide n to make it a perfect square. class GFG { // Return the minimum number to be // divided to make n a perfect square. static int findMinNumber(int n) { int count = 0, ans = 1; // Since 2 is only even prime, // compute its power separately. while (n % 2 == 0) { count++; n /= 2; } // If count is odd, it must be removed by dividing // n by prime number. if (count % 2 == 1) ans *= 2; for (int i = 3; i <= Math.sqrt(n); i += 2) { count = 0; while (n % i == 0) { count++; n /= i; } // If count is odd, it must be removed by // dividing n by prime number. if (count % 2 == 1) ans *= i; } if (n > 2) ans *= n; return ans; } // Driver code public static void main (String[] args) { int n = 72; System.out.println(findMinNumber(n)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python program to find # minimum number which # divide n to make it a # perfect square. import math # Return the minimum # number to be divided # to make n a perfect # square. def findMinNumber(n): count = 0 ans = 1 # Since 2 is only # even prime, compute # its power separately. while n % 2 == 0: count += 1 n //= 2 # If count is odd, # it must be removed # by dividing n by # prime number. if count % 2 is not 0: ans *= 2 for i in range(3, (int)(math.sqrt(n)) + 1, 2): count = 0 while n % i == 0: count += 1 n //= i # If count is odd, it # must be removed by # dividing n by prime # number. if count % 2 is not 0: ans *= i if n > 2: ans *= n return ans # Driver Code n = 72print(findMinNumber(n)) # This code is contributed # by Sanjit_Prasad. |
C#
// C# program to find minimum // number which divide n to // make it a perfect square. using System; class GFG { // Return the minimum number // to be divided to make // n a perfect square. static int findMinNumber(int n) { int count = 0, ans = 1; // Since 2 is only even prime, // compute its power separately. while (n % 2 == 0) { count++; n /= 2; } // If count is odd, it must // be removed by dividing // n by prime number. if (count % 2 == 1) ans *= 2; for (int i = 3; i <= Math.Sqrt(n); i += 2) { count = 0; while (n % i == 0) { count++; n /= i; } // If count is odd, it must // be removed by dividing n // by prime number. if (count % 2 == 1) ans *= i; } if (n > 2) ans *= n; return ans; } // Driver code public static void Main () { int n = 72; Console.WriteLine(findMinNumber(n)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to find minimum // number which divide n // to make it a perfect square. // Return the minimum number // to be divided to make // n a perfect square. function findMinNumber($n) { $count = 0; $ans = 1; // Since 2 is only // even prime, // compute its // power separately. while ($n % 2 == 0) { $count++; $n /= 2; } // If count is odd, // it must be removed // by dividing n by // prime number. if ($count % 2) $ans *= 2; for ($i = 3; $i <= sqrt($n); $i += 2) { $count = 0; while ($n % $i == 0) { $count++; $n /= $i; } // If count is odd, // it must be removed // by dividing n by // prime number. if ($count % 2) $ans *= $i; } if ($n > 2) $ans *= $n; return $ans; } // Driver Code $n = 72; echo findMinNumber($n), "\n"; // This code is contributed by ajit. ?> |
Javascript
<script> // Javascript program to find minimum // number which divide n // to make it a perfect square. // Return the minimum number // to be divided to make // n a perfect square. function findMinNumber(n) { let count = 0; let ans = 1; // Since 2 is only // even prime, // compute its // power separately. while (n % 2 == 0) { count++; n /= 2; } // If count is odd, // it must be removed // by dividing n by // prime number. if (count % 2) ans *= 2; for (let i = 3; i <= Math.sqrt(n); i += 2) { count = 0; while (n % i == 0) { count++; n /= i; } // If count is odd, // it must be removed // by dividing n by // prime number. if (count % 2) ans *= i; } if (n > 2) ans *= n; return ans; } // Driver Code let n = 72; document.write(findMinNumber(n) + "\n"); // This code is contributed by _saurabh_jaiswal. </script> |
Output:
2
Time Complexity: O(?n log n)
Auxiliary Space: O(1)
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