# Nth Square free number

• Difficulty Level : Hard
• Last Updated : 10 Mar, 2022

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.

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 implementation:

## C++

 // Program to find the nth square free number #include 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; }

## Java

 // Java Program to find the nth square // free number import java.io.*;   class GFG {           // Function to find nth square free     // number     public static int squareFree(int n)     {                   // To maintain count of square         // free number         int cnt = 0;               // Loop for square free numbers         for (int i = 1; ; i++) {                           boolean 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;             }         }     }           // driven code     public static void main(String[] args) {                   int n = 10;                   System.out.println("" + squareFree(n));     } }   // This code is contributed by sunnysingh

## Python3

 # Python3 Program to find the nth # square free number   # Function to find nth square # free number def squareFree(n):           # To maintain count of     # square free number     cnt = 0;       # Loop for square free numbers     i = 1;     while (True):         isSqFree = True;         j = 2;         while (j * j <= i):                           # Checking whether square of a number             # is divisible by any number which is             # a perfect square             if (i % (j * j) == 0):                 isSqFree = False;                 break;             j += 1;           # If number is square free         if (isSqFree == True):             cnt += 1;                   # If the cnt becomes n, return the number             if (cnt == n):                 return i;         i += 1;       return 0;   # Driver Code n = 10; print(squareFree(n));   # This code is contributed by mits

## C#

 // C# Program to find the // nth square free number using System;   class GFG {           // Function to find nth     // square free number     public static 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;             }         }     }           // Driver code     public static void Main()     {         int n = 10;         Console.Write("" + squareFree(n));     } }   // This code is contributed by nitin mittal



## Javascript



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 Inclusionâ€“exclusion principle. Using Inclusionâ€“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 implementation of above algorithm:

## C++

 // Program to find the nth square free number #include 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 &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 whose square to be checked. long long countSquares(long long i, long long cur,                        long long k, vector &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 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; }

## Java

 // Java Program to find the nth square free number import java.util.*;   class GFG {   // Maximum prime number to be considered for square // divisibility static int MAX_PRIME = 100000;   // Maximum value of result. We do // binary search from 1 to MAX_RES static int MAX_RES = 2000000000;   static void SieveOfEratosthenes(Vector 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.     boolean prime[] = new boolean[MAX_PRIME + 1];     Arrays.fill(prime, true);       for (int 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 (int i = p * 2; i <= MAX_PRIME; i += p)                 prime[i] = false;         }     }       // Store all prime numbers in a[]     for (int p = 2; p <= MAX_PRIME; p++)         if (prime[p])             a.add((long)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 whose square to be checked. static long countSquares(long i, long cur,                     long k, Vector a) {     // variable to store square of prime     long square = a.get((int) i)*a.get((int)(i));       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 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 static long squareFree(long n) {     // Computing primes and storing it in an array a[]     Vector a = new Vector<>();     SieveOfEratosthenes(a);       // Applying binary search     long low = 1;     long high = MAX_RES;       while (low < high)     {         long mid = low + (high - low)/2;           // 'c' contains Number of square free numbers         // less than or equal to 'mid'         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 code public static void main(String[] args) {     int n = 10;     System.out.println(squareFree(n)); } }   // This code has been contributed by 29AjayKumar

## Python3

 # Python 3 Program to find the nth square free number import sys sys.setrecursionlimit(15000)   # Maximum prime number to be considered for square # divisibility MAX_PRIME = 100000;   # Maximum value of result. We do binary search from 1 # to MAX_RES MAX_RES = 2000000000   def SieveOfEratosthenes(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.     prime = [True for i in range(MAX_PRIME + 1)];     p = 2       while(p * p <= MAX_PRIME):                   # If prime[p] is not changed, then it is a prime         if (prime[p] == True):                       # Update all multiples of p             for i in range(p * 2, MAX_PRIME + 1, p):                            prime[i] = False;         p += 1               # Store all prime numbers in a[]     for p in range(2, MAX_PRIME + 1):           if (prime[p]):             a.append(p);        return a   # 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 whose square to be checked. def countSquares(i, cur, k, a):       # variable to store square of prime     square = a[i]*a[i];     newCur = square*cur;       # If value of greatest integer becomes zero     if (newCur > k):         return 0, a;       # Applying inclusion-exclusion principle       # Counting integers with squares as factor     cnt = k//(newCur);       # Inclusion (Recur for next prime number)     tmp, a = countSquares(i + 1, cur, k, a);       cnt += tmp           # Exclusion (Recur for next prime number)     tmp, a = countSquares(i + 1, newCur, k, a);     cnt -= tmp           # Final count     return cnt,a;   # Function to return nth square free number def squareFree(n):       # Computing primes and storing it in an array a[]     a = SieveOfEratosthenes([]);       # Applying binary search     low = 1;     high = MAX_RES;     while (low < high):          mid = low + (high - low)//2;           # 'c' contains Number of square free numbers         # less than or equal to 'mid'         c,a = countSquares(0, 1, mid, a);         c = mid - c           # 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 if __name__=='__main__':       n = 10;     print(squareFree(n))       # This code is contributed by rohitsingh07052.

## C#

 // C# Program to find the nth square free number using System; using System.Collections.Generic;   class GFG {   // Maximum prime number to be considered // for square divisibility static int MAX_PRIME = 100000;   // Maximum value of result. We do // binary search from 1 to MAX_RES static int MAX_RES = 2000000000;   static void SieveOfEratosthenes(List 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 = new bool[MAX_PRIME + 1];     for(int i = 0; i < MAX_PRIME + 1; i++)         prime[i] = true;       for (int 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 (int i = p * 2; i <= MAX_PRIME; i += p)                 prime[i] = false;         }     }       // Store all prime numbers in a[]     for (int p = 2; p <= MAX_PRIME; p++)         if (prime[p])             a.Add((long)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 whose square to be checked. static long countSquares(long i, long cur,                     long k, List a) {     // variable to store square of prime     long square = a[(int) i]*a[(int)(i)];       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 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 static long squareFree(long n) {     // Computing primes and storing it in an array a[]     List a = new List();     SieveOfEratosthenes(a);       // Applying binary search     long low = 1;     long high = MAX_RES;       while (low < high)     {         long mid = low + (high - low)/2;           // 'c' contains Number of square free numbers         // less than or equal to 'mid'         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 code public static void Main() {     int n = 10;     Console.WriteLine(squareFree(n)); } }   /* This code contributed by PrinciRaj1992 */

## Javascript



Output:

14

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