Count perfect square fractions from given array

Last Updated : 28 Oct, 2021

Given two arrays arr1[] and arr2[] of length N which contains Numerator and Denominator of N fractions respectively, the task is to count the number of fractions from the array which is a perfect square of a fraction.

Examples:

Input: arr1[] = {3, 25, 49, 9}, arr2[] = {27, 64, 7, 3}
Output: 2
Explanation:
(arr1[0] / arr2[0]) = (3 / 27) = (1 / 9) = (1 / 3)²
(arr1[1] / arr2[1]) = (25 / 64) = (5 / 8) ²
(arr1[0] / arr2[0]) and (arr1[1] / arr2[1]) are perfect square fractions. Therefore, the required output is 2.

Input: arr1[] = {1, 11, 3, 9}, arr2[] = {9, 11, 5, 1}
Output: 3

Approach: Follow the steps below to solve the problem:

Initialize a variable cntPerfNum to store the count of perfect square fractions.
Traverse both the arrays and for each array elements, check if the value of (arr1[i] / GCD(arr1[i], arr2[i])) and (arr2[i] / GCD(arr1[i], arr2[i])) is a perfect square or not. If found to be true then increment the value of cntPerfNum by 1.
Finally, print the value of cntPerfNum.

Below is the implementation of the above approach:

C++

// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the GCD
// of two numbers
int GCD(int a,int b)
{
    // If b is 0
    if (b ==0 ) {
      return a;
    }
    return GCD(b,a%b);
}

// Function to check if N
// is perfect square
bool isPerfectSq(int N)
{
   // Stores square root
   // of N 
   int x = sqrt(N);
   
   // Check if N is a 
   // perfect square
   if (x * x == N) {
       return 1;
   }
   
   return 0;
}

// Function to count perfect square fractions
int cntPerSquNum(int arr1[], int arr2[], 
                                  int N)
{
    // Stores count of perfect square
    // fractions in both arrays
    int cntPerfNum = 0;
    
    // Traverse both the arrays
    for (int i = 0; i < N; i++) {
        
        // Stores gcd of (arr1[i], arr2[i])
        int gcd = GCD(arr1[i], arr2[i]);
        
        // If numerator and denomerator of a
        // reduced fraction is a perfect square
        if (isPerfectSq(arr1[i]/ gcd) &&
           isPerfectSq(arr2[i]/ gcd)) {
               
            // Update cntPerfNum
            cntPerfNum++;       
        }
        
    }
    
    return cntPerfNum;
}


//Driver Code
int main() {

    int arr1[]={ 3, 25, 49, 9 };
    int arr2[]={ 27, 64, 7, 3 };
  
    int N = sizeof(arr1) / sizeof(arr1[0]);
  
    cout<<cntPerSquNum(arr1, arr2, N);
}

Java

// Java implementation of the 
// above approach
import java.lang.Math; 

class GFG{
    
// Function to find the GCD 
// of two numbers 
public static int GCD(int a, int b) 
{ 
    
    // If b is 0 
    if (b == 0) 
    { 
      return a; 
    } 
    return GCD(b, a % b); 
} 
  
// Function to check if N 
// is perfect square 
public static boolean isPerfectSq(int N) 
{ 
    
    // Stores square root 
    // of N  
    int x = (int)Math.sqrt(N); 
    
    // Check if N is a  
    // perfect square 
    if (x * x == N) 
    { 
        return true; 
    } 
    
    return false; 
} 
  
// Function to count perfect square fractions 
public static int cntPerSquNum(int arr1[],
                               int arr2[],  
                               int N) 
{ 
    
    // Stores count of perfect square 
    // fractions in both arrays 
    int cntPerfNum = 0; 
      
    // Traverse both the arrays 
    for(int i = 0; i < N; i++) 
    { 
        
        // Stores gcd of (arr1[i], arr2[i]) 
        int gcd = GCD(arr1[i], arr2[i]); 
          
        // If numerator and denomerator of a 
        // reduced fraction is a perfect square 
        if (isPerfectSq(arr1[i] / gcd) && 
            isPerfectSq(arr2[i] / gcd))
        { 
            
            // Update cntPerfNum 
            cntPerfNum++;        
        } 
    } 
      
    return cntPerfNum; 
}

// Driver code
public static void main(String[] args) 
{
    int arr1[] = { 3, 25, 49, 9 }; 
    int arr2[] = { 27, 64, 7, 3 }; 
    
    int N = arr1.length; 
    
    System.out.println(cntPerSquNum(arr1, arr2, N));
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation of the
# above approach

# Function to find the GCD
# of two numbers
def GCD(a, b):
    
    # If b is 0
    if (b == 0):
        return a
        
    return GCD(b, a % b)

# Function to check if N
# is perfect square
def isPerfectSq(N):

    # Stores square root
    # of N
    x = (int)(pow(N, 1 / 2))
    
    # Check if N is a
    # perfect square
    if (x * x == N):
        return True
        
    return False

# Function to count perfect square 
# fractions
def cntPerSquNum(arr1, arr2, N):
    
    # Stores count of perfect square
    # fractions in both arrays
    cntPerfNum = 0
    
    # Traverse both the arrays
    for i in range(N):

        # Stores gcd of (arr1[i], arr2[i])
        gcd = GCD(arr1[i], arr2[i])

        # If numerator and denomerator of a
        # reduced fraction is a perfect square
        if (isPerfectSq(arr1[i] / gcd) and
            isPerfectSq(arr2[i] / gcd)):
                
            # Update cntPerfNum
            cntPerfNum += 1

    return cntPerfNum

# Driver code
if __name__ == '__main__':
    
    arr1 = [ 3, 25, 49, 9 ]
    arr2 = [ 27, 64, 7, 3 ]

    N = len(arr1)

    print(cntPerSquNum(arr1, arr2, N))

# This code is contributed by Princi Singh

// C# implementation of the 
// above approach
using System;
class GFG{
    
// Function to find the GCD 
// of two numbers 
public static int GCD(int a, 
                      int b) 
{     
  // If b is 0 
  if (b == 0) 
  { 
    return a; 
  } 
  return GCD(b, a % b); 
} 

// Function to check if N 
// is perfect square 
public static bool isPerfectSq(int N) 
{ 
  // Stores square root 
  // of N  
  int x = (int)Math.Sqrt(N); 

  // Check if N is a  
  // perfect square 
  if (x * x == N) 
  { 
    return true; 
  } 

  return false; 
} 
  
// Function to count perfect 
// square fractions 
public static int cntPerSquNum(int []arr1,
                               int []arr2,  
                               int N) 
{     
  // Stores count of perfect square 
  // fractions in both arrays 
  int cntPerfNum = 0; 

  // Traverse both the arrays 
  for(int i = 0; i < N; i++) 
  { 
    // Stores gcd of (arr1[i], arr2[i]) 
    int gcd = GCD(arr1[i], arr2[i]); 

    // If numerator and denomerator
    // of a reduced fraction is a 
    // perfect square 
    if (isPerfectSq(arr1[i] / gcd) && 
        isPerfectSq(arr2[i] / gcd))
    { 
      // Update cntPerfNum 
      cntPerfNum++;        
    } 
  } 

  return cntPerfNum; 
}

// Driver code
public static void Main(String[] args) 
{
  int []arr1 = {3, 25, 49, 9}; 
  int []arr2 = {27, 64, 7, 3}; 
  int N = arr1.Length; 
  Console.WriteLine(cntPerSquNum(arr1, 
                                 arr2, N));
}
}

// This code is contributed by shikhasingrajput

JavaScript

<script>

// Function to find the GCD
// of two numbers
function GCD(a,b)
{
    // If b is 0
    if (b ==0 ) {
      return a;
    }
    return GCD(b,a%b);
}

// Function to check if N
// is perfect square
function isPerfectSq( N)
{
   // Stores square root
   // of N 
  var x = Math.sqrt(N);
   
   // Check if N is a 
   // perfect square
   if (x * x == N) {
       return 1;
   }
   
   return 0;
}

// Function to count perfect square fractions
function cntPerSquNum(arr1,arr2,N)
{

    // Stores count of perfect square
    // fractions in both arrays
    var cntPerfNum = 0;
    
    // Traverse both the arrays
    for (var i = 0; i < N; i++) {
        
        // Stores gcd of (arr1[i], arr2[i])
        var gcd = GCD(arr1[i], arr2[i]);
        
        // If numerator and denomerator of a
        // reduced fraction is a perfect square
        if (isPerfectSq(arr1[i]/ gcd) &&
           isPerfectSq(arr2[i]/ gcd)) {
               
            // Update cntPerfNum
            cntPerfNum++;       
        }
        
    }    
    return cntPerfNum;
}

var arr1 = [ 3, 25, 49, 9 ];
    var arr2 = [ 27, 64, 7, 3 ];  
    var N = 4;
    document.write(cntPerSquNum(arr1, arr2, N));

// This code is contributed by akshitsaxenaa09.
</script>

Output:

Time Complexity: O(N * log(M)), where M is the maximum element from both the arrays.
Auxiliary Space: O(1)

Count perfect square fractions from given array

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