Find the largest good number in the divisors of given number N
Last Updated :
23 Jul, 2022
Given a number N. The task is to find the largest good number among the divisors of a given number N. A number X is defined as the good number if there is no such positive integer a > 1, such that a^2 is a divisor of X.
Examples:
Input: N = 10
Output: 10
In 1, 2, 5, 10.
10 is the largest good number
Input: N = 12
Output: 6
In 1, 2, 3, 4, 6, 12.
6 is the largest good number
Approach: Find all prime divisors of N. Assume they are p1, p2, ..., pk (in O(sqrt(n)) time complexity). If the answer is a, then we know that for each 1 <= I <= k, obviously, a is not divisible by pi^2 (and all greater powers of pi). So a <= p1 × p2 ×... × pk. And we know that p1 × p2 × ... × pk is itself a good number. So, the answer is p1 × p2 ×...× pk.
Below is the implementation of the above approach:
C++
// CPP program to find the largest, good
// number in the divisors of given number N.
#include <bits/stdc++.h>
using namespace std;
// function to return distinct prime factors
vector<int> PrimeFactors(int n)
{
// to store distinct prime factors
vector<int> v;
int x = n;
// run a loop upto sqrt(n)
for (int i = 2; i * i <= n; i++) {
if (x % i == 0) {
// place this prime factor in vector
v.push_back(i);
while (x % i == 0)
x /= i;
}
}
// This condition is to handle the case when n
// is a prime number greater than 1
if (x > 1)
v.push_back(x);
return v;
}
// function that returns good number
int GoodNumber(int n)
{
// distinct prime factors
vector<int> v = PrimeFactors(n);
// to store answer
int ans = 1;
// product of all distinct prime
// factors is required answer
for (int i = 0; i < v.size(); i++)
ans *= v[i];
return ans;
}
// Driver code
int main()
{
int n = 12;
// function call
cout << GoodNumber(n);
return 0;
}
// Java program to find the largest, good
// number in the divisors of given number N.
import java.util.*;
class GFG {
// function to return distinct prime factors
static Vector<Integer> PrimeFactors(int n)
{
// to store distinct prime factors
Vector<Integer> v = new Vector<Integer>();
int x = n;
// run a loop upto sqrt(n)
for (int i = 2; i * i <= n; i++) {
if (x % i == 0) {
// place this prime factor in vector
v.add(i);
while (x % i == 0)
x /= i;
}
}
// This condition is to handle the case when n
// is a prime number greater than 1
if (x > 1)
v.add(x);
return v;
}
// function that returns good number
static int GoodNumber(int n)
{
// distinct prime factors
Vector<Integer> v = new Vector<Integer>(PrimeFactors(n));
// to store answer
int ans = 1;
// product of all distinct prime
// factors is required answer
for (int i = 0; i < v.size(); i++)
ans *= v.get(i);
return ans;
}
// Driver code
public static void main(String[] args)
{
int n = 12;
// function call
System.out.println(GoodNumber(n));
}
}
// This code is contributed by ihritik
# Python 3 program to find the
# largest, good number in the
# divisors of given number N.
# function to return distinct
# prime factors
def PrimeFactors(n):
# to store distinct
# prime factors
v = []
x = n
# run a loop upto sqrt(n)
i = 2
while(i * i <= n):
if (x % i == 0) :
# place this prime factor
# in vector
v.append(i)
while (x % i == 0):
x //= i
i += 1
# This condition is to handle
# the case when n is a prime
# number greater than 1
if (x > 1):
v.append(x)
return v
# function that returns good number
def GoodNumber(n):
# distinct prime factors
v = PrimeFactors(n)
# to store answer
ans = 1
# product of all distinct prime
# factors is required answer
for i in range(len(v)):
ans *= v[i]
return ans
# Driver code
if __name__ == "__main__":
n = 12
# function call
print(GoodNumber(n))
# This code is contributed
# by ChitraNayal
// C# program to find the largest, good
// number in the divisors of given number N.
using System;
using System.Collections.Generic;
public class GFG {
// function to return distinct prime factors
static List<int> PrimeFactors(int n)
{
// to store distinct prime factors
List<int> v = new List<int>();
int x = n;
// run a loop upto sqrt(n)
for (int i = 2; i * i <= n; i++) {
if (x % i == 0) {
// place this prime factor in vector
v.Add(i);
while (x % i == 0)
x /= i;
}
}
// This condition is to handle the case when n
// is a prime number greater than 1
if (x > 1)
v.Add(x);
return v;
}
// function that returns good number
static int GoodNumber(int n)
{
// distinct prime factors
List<int> v = new List<int>(PrimeFactors(n));
// to store answer
int ans = 1;
// product of all distinct prime
// factors is required answer
for (int i = 0; i < v.Count; i++)
ans *= v[i];
return ans;
}
// Driver code
public static void Main(String[] args)
{
int n = 12;
// function call
Console.WriteLine(GoodNumber(n));
}
}
// This code has been contributed by 29AjayKumar
<?php
// PHP program to find the largest, good
// number in the divisors of given number N.
// function to return distinct prime factors
function PrimeFactors($n)
{
// to store distinct prime factors
$v = array();
$x = $n;
// run a loop upto sqrt(n)
for ($i = 2; $i * $i <= $n; $i++)
{
if ($x % $i == 0)
{
// place this prime factor in vector
array_push($v, $i);
while ($x % $i == 0)
$x /= $i;
}
}
// This condition is to handle the case when n
// is a prime number greater than 1
if ($x > 1)
array_push($v, $x);
return $v;
}
// function that returns good number
function GoodNumber($n)
{
// distinct prime factors
$v = PrimeFactors($n);
// to store answer
$ans = 1;
// product of all distinct prime
// factors is required answer
for ($i = 0; $i < count($v); $i++)
$ans *= $v[$i];
return $ans;
}
// Driver code
$n = 12;
// function call
echo GoodNumber($n);
// This code is contributed by Rajput-Ji
?>
<script>
// Javascript program to find the largest, good
// number in the divisors of given number N.
// function to return distinct prime factors
function PrimeFactors(n)
{
// to store distinct prime factors
let v = [];
let x = n;
// run a loop upto sqrt(n)
for (let i = 2; i * i <= n; i++) {
if (x % i == 0) {
// place this prime factor in vector
v.push(i);
while (x % i == 0)
x = Math.floor(x/i);
}
}
// This condition is to handle the case when n
// is a prime number greater than 1
if (x > 1)
v.push(x);
return v;
}
// function that returns good number
function GoodNumber(n)
{
// distinct prime factors
let v = PrimeFactors(n);
// to store answer
let ans = 1;
// product of all distinct prime
// factors is required answer
for (let i = 0; i < v.length; i++)
ans *= v[i];
return ans;
}
// Driver code
let n = 12;
// function call
document.write(GoodNumber(n));
// This code is contributed by rag2127
</script>
Time Complexity : O(sqrt(n) + k) ,where n is given number and k is number of distinct prime factors.
Space Complexity : O(k) ,to store distinct prime factors.
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