Count Pandigital Fractions pairs in given Array

Given an array arr[], the task is to count the pairs in the array such that arr[i]/arr[j] is a Pandigital Fraction.

A fraction N/D is called a Pandigital Fraction if the fraction contains all the digits from 0 to 9.  

Examples:

Input: arr = [ 12345, 67890, 123, 4567890 ] 
Output: 3 
Explanation:
The fractions are 12345/67890, 12345/4567890, and 123/4567890 

Input: arr = [ 12345, 6789 ] 
Output: 0 

Approach: The idea is to iterate over every possible pair of the array using two nested loops and for every pair concatenate arr[i] and arr[j] into a single number and check if the concatenation of arr[i] and arr[j] is a Pandigital number in base 10 then increment the count.

Below is the implementation of the above approach:

C++

#include <cmath>
#include <cstring>
#include <iostream>

using namespace std;

// Function to concatenate two numbers into one
long long numConcat(long long num1, long long num2)
{

    // Find number of digits in num2
    int digits = to_string(num2).length();

    // Add zeroes to the end of num1
    num1 = num1 * pow(10, digits);

    // Add num2 to num1
    num1 += num2;

    return num1;
}

// Return true if n is pandigital else return false.
int checkPanDigital(long long n)
{
    string str = to_string(n);
    int b = 10;

    // Checking length is less than base
    if (str.length() < b) {
        return 0;
    }

    int hash[b];
    memset(hash, 0, sizeof(hash));

    // Traversing each digit of the number
    for (int i = 0; i < str.length(); i++) {
        // If digit is integer
        if (str[i] >= '0' && str[i] <= '9') {
            hash[str[i] - '0'] = 1;
        }

        // If digit is alphabet
        else if (str[i] - 'A' <= b - 11) {
            hash[str[i] - 'A' + 10] = 1;
        }
    }

    // Checking hash array, if any index is unmarked
    for (int i = 0; i < b; i++) {
        if (hash[i] == 0) {
            return 0;
        }
    }

    return 1;
}

// Returns true if N is a Pandigital Fraction Number
bool isPandigitalFraction(long long N, long long D)
{
    long long join = numConcat(N, D);
    return checkPanDigital(join) == 1;
}

// Returns number pandigital fractions in the array
int countPandigitalFraction(long long v[], int n)
{
    // iterate over all pair of strings
    int count = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (isPandigitalFraction(v[i], v[j])) {
                count++;
            }
        }
    }
    return count;
}

// Driver Code
int main()
{
    long long arr[] = { 12345, 67890, 123, 4567890 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << countPandigitalFraction(arr, n) << endl;

    return 0;
}

Java

import java.util.Arrays;

public class Main {

  // Function to concatenate two numbers into one
  static long numConcat(long num1, long num2) {

    // Find number of digits in num2
    int digits = Long.toString(num2).length();

    // Add zeroes to the end of num1
    num1 = num1 * (long) Math.pow(10, digits);

    // Add num2 to num1
    num1 += num2;

    return num1;
  }

  // Return true if n is pandigital else return false.
  static int checkPanDigital(long n) {
    String str = Long.toString(n);
    int b = 10;

    // Checking length is less than base
    if (str.length() < b) {
      return 0;
    }

    int[] hash = new int[b];
    Arrays.fill(hash, 0);

    // Traversing each digit of the number
    for (int i = 0; i < str.length(); i++) {
      // If digit is integer
      if (str.charAt(i) >= '0' && str.charAt(i) <= '9') {
        hash[str.charAt(i) - '0'] = 1;
      }

      // If digit is alphabet
      else if (str.charAt(i) - 'A' <= b - 11) {
        hash[str.charAt(i) - 'A' + 10] = 1;
      }
    }

    // Checking hash array, if any index is unmarked
    for (int i = 0; i < b; i++) {
      if (hash[i] == 0) {
        return 0;
      }
    }

    return 1;
  }

  // Returns true if N is a Pandigital Fraction Number
  static boolean isPandigitalFraction(long N, long D) {
    long join = numConcat(N, D);
    return checkPanDigital(join) == 1;
  }

  // Returns number pandigital fractions in the array
  static int countPandigitalFraction(long[] v, int n) {
    // iterate over all pair of strings
    int count = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (isPandigitalFraction(v[i], v[j])) {
          count++;
        }
      }
    }
    return count;
  }

  // Driver Code
  public static void main(String[] args) {
    long[] arr = {12345, 67890, 123, 4567890};
    int n = arr.length;

    System.out.println(countPandigitalFraction(arr, n));
  }
}

Python3

# Python3 implementation of the
# above approach

import math

# Function to concatenate
# two numbers into one
def numConcat(num1, num2):

     # Find number of digits in num2
     digits = len(str(num2))

     # Add zeroes to the end of num1
     num1 = num1 * (10**digits)

     # Add num2 to num1
     num1 += num2

     return num1

# Return true if n is pandigit
# else return false.
def checkPanDigital(n):
    n = str(n)
    b = 10

    # Checking length is
    # less than base
    if (len(n) < b):
        return 0;

    hash = [0] * b;

    # Traversing each digit
    # of the number.
    for i in range(len(n)):

        # If digit is integer
        if (n[i] >= '0' and \
            n[i] <= '9'):
            hash[ord(n[i]) - ord('0')] = 1;

        # If digit is alphabet
        elif (ord(n[i]) - ord('A') <= \
                            b - 11):
            hash[ord(n[i]) - \
                 ord('A') + 10] = 1;

    # Checking hash array, if any index is
    # unmarked.
    for i in range(b):
        if (hash[i] == 0):
            return 0;

    return 1;

# Returns true if N is a
# Pandigital Fraction Number
def isPandigitalFraction(N, D):
    join = numConcat(N, D)
    return checkPanDigital(join)

# Returns number pandigital fractions
# in the array
def countPandigitalFraction(v, n) :

    # iterate over all
    # pair of strings
    count = 0
    for i in range(0, n) :

        for j in range (i + 1,
                        n) :

            if (isPandigitalFraction(v[i],
                             v[j])) :
                count = count + 1
    return count


# Driver Code
if __name__ == "__main__":

    arr = [ 12345, 67890, 123, 4567890 ]
    n = len(arr)

    print(countPandigitalFraction(arr, n))

C#

// C# code to implement the approach

using System;
using System.Linq;

class GFG {
    static void Main(string[] args)
    {
        // Input array of integers
        long[] arr = { 12345, 67890, 123, 4567890 };
        int n = arr.Length;

        // Count the number of pandigital fractions
        int count = CountPandigitalFraction(arr, n);

        Console.WriteLine(count);
    }

    // Function to concatenate two numbers into one
    static long NumConcat(long num1, long num2)
    {
        // Find number of digits in num2
        int digits = num2.ToString().Length;

        // Add zeroes to the end of num1
        num1 *= (long)Math.Pow(10, digits);

        // Add num2 to num1
        num1 += num2;

        return num1;
    }

    // Return true if n is pandigital else return false.
    static bool CheckPanDigital(long n)
    {
        string str = n.ToString();
        int b = 10;

        // Checking length is less than base
        if (str.Length < b) {
            return false;
        }

        int[] hash = new int[b];

        // Traversing each digit of the number
        foreach(char ch in str)
        {
            // If digit is integer
            if (ch >= '0' && ch <= '9') {
                hash[ch - '0'] = 1;
            }

            // If digit is alphabet
            else if (ch - 'A' <= b - 11) {
                hash[ch - 'A' + 10] = 1;
            }
        }

        // Checking hash array, if any index is unmarked
        foreach(int val in hash)
        {
            if (val == 0) {
                return false;
            }
        }

        return true;
    }

    // Returns true if N is a Pandigital Fraction Number
    static bool IsPandigitalFraction(long N, long D)
    {
        long join = NumConcat(N, D);
        return CheckPanDigital(join);
    }

    // Returns number pandigital fractions in the array
    static int CountPandigitalFraction(long[] v, int n)
    {
        // Iterate over all pairs of integers
        int count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (IsPandigitalFraction(v[i], v[j])) {
                    count++;
                }
            }
        }
        return count;
    }
}

Javascript

// Function to concatenate two numbers into one
function numConcat(num1, num2) {
    // Find number of digits in num2
    let digits = num2.toString().length;

    // Add zeroes to the end of num1
    num1 = num1 * Math.pow(10, digits);

    // Add num2 to num1
    num1 += num2;

    return num1;
}

// Return true if n is pandigital else return false.
function checkPanDigital(n) {
    n = n.toString();
    let b = 10;

    // Checking length is less than base
    if (n.length < b) {
        return 0;
    }

    let hash = new Array(b).fill(0);

    // Traversing each digit of the number
    for (let i = 0; i < n.length; i++) {
        // If digit is integer
        if (n[i] >= '0' && n[i] <= '9') {
            hash[n.charCodeAt(i) - '0'.charCodeAt(0)] = 1;
        }

        // If digit is alphabet
        else if (n.charCodeAt(i) - 'A'.charCodeAt(0) <= b - 11) {
            hash[n.charCodeAt(i) - 'A'.charCodeAt(0) + 10] = 1;
        }
    }

    // Checking hash array, if any index is unmarked
    for (let i = 0; i < b; i++) {
        if (hash[i] == 0) {
            return 0;
        }
    }

    return 1;
}

// Returns true if N is a Pandigital Fraction Number
function isPandigitalFraction(N, D) {
    let join = numConcat(N, D);
    return checkPanDigital(join);
}

// Returns number pandigital fractions in the array
function countPandigitalFraction(v, n) {
    // iterate over all pair of strings
    let count = 0;
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            if (isPandigitalFraction(v[i], v[j])) {
                count++;
            }
        }
    }
    return count;
}

// Driver Code
let arr = [12345, 67890, 123, 4567890];
let n = arr.length;

console.log(countPandigitalFraction(arr, n));

Output:

3

Time Complexity: O(N²) 
Auxiliary Space: O(1), As constant extra space is used.