# Nth Square free number

Given a number n, find the n-th square-free number. A number is square-free if it is not divisible by a perfect square other than 1.

Examples:

```Input : n = 2
Output : 2

Input : 5
Output : 6
There is one number (in range from 1 to 6)
that is divisible by a square. The number
is 4.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 (Brute Force):
The idea is to iteratively check every number whether it is divisible by any perfect square number and increase the count whenever an square free number is encountered and returning the nth square free number.

Following is the C++ implementation:

```// Program to find the nth square free number
#include<bits/stdc++.h>
using namespace std;

// Function to find nth square free number
int squareFree(int n)
{
// To maintain count of square free number
int cnt = 0;

// Loop for square free numbers
for (int i=1;; i++)
{
bool isSqFree = true;
for (int j=2; j*j<=i; j++)
{
// Checking whether square of a number
// is divisible by any number which is
// a perfect square
if (i % (j*j) == 0)
{
isSqFree = false;
break;
}
}

// If number is square free
if (isSqFree == true)
{
cnt++;

// If the cnt becomes n, return
// the number
if (cnt == n)
return i;
}
}
return 0;
}

// Driver Program
int main()
{
int n = 10;
cout << squareFree(n) << endl;
return 0;
}
```

Output:

```14
```

Method 2 (Better approach):
Idea is to count square free numbers less than or equal to upper limit ‘k’ and then apply binary search to find n-th square free number. First, we calculate count of square numbers (numbers with squares as factors) upto ‘k’ and then subtracting that count from total numbers to get count of square free numbers upto ‘k’.

Explanation:

1. If any integer has a perfect square as factor, then it is guaranteed that it has a square of prime as a factor too. So we need to count the integers less than or equal to ‘k’ which have square of primes as a factor.
For example, find number of integers who have either 4 or 9 as a factor upto ‘k’. It can done using Inclusion–exclusion principle. Using Inclusion–exclusion principle, total number of integers are [k/4] + [k/9] -[k/36] where [] is greatest integer function.
2. Recursively apply inclusion and exclusion till the value of greatest integer becomes zero. This step will return count of numbers with squares as factors.
3. Apply binary search to find the nth square free number.

Following is the C++ implementation of above algorithm:

```// Program to find the nth square free number
#include<bits/stdc++.h>
using namespace std;

// Maximum prime number to be considered for square
// divisibility
const int MAX_PRIME = 100000;

// Maximum value of result. We do binary search from 1
// to MAX_RES
const int MAX_RES = 2000000000l;

void SieveOfEratosthenes(vector<long long> &a)
{
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
bool prime[MAX_PRIME + 1];
memset(prime, true, sizeof(prime));

for (long long p=2; p*p<=MAX_PRIME; p++)
{
// If prime[p] is not changed, then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (long long i=p*2; i<=MAX_PRIME; i += p)
prime[i] = false;
}
}

// Store all prime numbers in a[]
for (long long p=2; p<=MAX_PRIME; p++)
if (prime[p])
a.push_back(p);
}

// Function to count integers upto k which are having
// perfect squares as factors. i is index of next
// prime number whose square needs to be checked.
// curr is current number whos square to be checked.
long long countSquares(long long i, long long cur,
long long k, vector<long long> &a)
{
// variable to store square of prime
long long square = a[i]*a[i];

long long newCur = square*cur;

// If value of greatest integer becomes zero
if (newCur > k)
return 0;

// Applying inclusion-exclusion principle

// Counting integers with squares as factor
long long cnt = k/(newCur);

// Inclusion (Recur for next prime number)
cnt += countSquares(i+1, cur, k, a);

// Exclusion (Recur for next prime number)
cnt -= countSquares(i+1, newCur, k, a);

// Final count
return cnt;
}

// Function to return nth square free number
long long squareFree(long long n)
{
// Computing primes and storing it in an array a[]
vector <long long> a;
SieveOfEratosthenes(a);

// Applying binary search
long long low = 1;
long long high = MAX_RES;

while (low < high)
{
long long mid = low + (high - low)/2;

// 'c' contains Number of square free numbers
// less than or equal to 'mid'
long long c = mid - countSquares(0, 1, mid, a);

// If c < n, then search right side of mid
// else search left side of mid
if (c < n)
low = mid+1;
else
high = mid;
}

// nth square free number
return low;
}

// Driver Program
int main()
{
int n = 10;
cout << squareFree(n) << endl;
return 0;
}
```

Output:

```14
```

This article is contributed by Rahul Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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