Skip to content
Related Articles

Related Articles

Number of pairs with Pandigital Concatenation
  • Difficulty Level : Expert
  • Last Updated : 31 Oct, 2020

A pair of strings when concatenated is said to be a ‘Pandigital Concatenation’ if their concatenation consists of all digits from (0 – 9) in any order at least once.The task is, given N strings, compute the number of pairs resulting in a ‘Pandigital Concatenation’. 

Examples: 

Input  : num[] = {"123567", "098234", "14765", "19804"}
Output : 3
The pairs, 1st and 2nd giving 
(123567098234),1st and 4rd giving(12356719804) and 
2nd and 3rd giving (09823414765),
on concatenation result in Pandigital Concatenations. 

Input : num[] =  {"56789", "098345", "1234"}
Output : 0
None of the pairs on concatenation result in Pandigital 
Concatenations.

Method 1 (Brute Force): A possible brute-force solution is to form all possible concatenations by forming all pairs in O(n2 and using a frequency array for digits (0 – 9), we check if each digit exists at least once in each concatenation formed for every pair. 

C++




// C++ program to find all
// Pandigital concatenations
// of two strings.
#include <bits/stdc++.h>
using namespace std;
 
// Checks if a given
// string is Pandigital
bool isPanDigital(string s)
{
    bool digits[10] = {false};
    for (int i = 0; i < s.length(); i++)
        digits[s[i] - '0'] = true;
 
    // digit i is not present
    // thus not pandigital
    for (int i = 0; i <= 9; i++)
        if (digits[i] == false)
            return false;
 
    return true;
}
 
// Returns number of pairs
// of strings resulting in
// Pandigital Concatenations
int countPandigitalPairs(vector<string> &v)
{
    // iterate over all
    // pair of strings
    int pairs = 0;
    for (int i = 0; i < v.size(); i++)
        for (int j = i + 1; j < v.size(); j++)
            if (isPanDigital(v[i] + v[j]))
                pairs++;
    return pairs;
}
 
// Driver code
int main()
{
    vector<string> v = {"123567", "098234",
                        "14765", "19804"};
    cout << countPandigitalPairs(v) << endl;
    return 0;
}

Java




// Java program to find all
// Pandigital concatenations
// of two strings.
import java.io.*;
import java.util.*;
 
class GFG
{
    static ArrayList<String> v =
                  new ArrayList<String>();
                   
    // Checks if a given
    // string is Pandigital
    static int isPanDigital(String s)
    {
        int digits[] = new int[10];
         
        for (int i = 0; i < s.length(); i++)
            digits[s.charAt(i) -
                        (int)'0'] = 1;
     
        // digit i is not present
        // thus not pandigital
        for (int i = 0; i <= 9; i++)
            if (digits[i] == 0)
                return 0;
     
        return 1;
    }
     
    // Returns number of pairs
    // of strings resulting in
    // Pandigital Concatenations
    static int countPandigitalPairs()
    {
        // iterate over all
        // pair of strings
        int pairs = 0;
        for (int i = 0; i < v.size(); i++)
            for (int j = i + 1;
                     j < v.size(); j++)
                if (isPanDigital(v.get(i) +
                                 v.get(j)) == 1)
                    pairs++;
        return pairs;
    }
     
    // Driver code
    public static void main(String args[])
    {
        v.add("123567");
        v.add("098234");
        v.add("14765");
        v.add("19804");
        System.out.print(countPandigitalPairs());
    }
}
 
// This code is contributed
// by Manish Shaw(manishshaw1)

Python3




# Python3 program to find all
# Pandigital concatenations
# of two strings.
 
# Checks if a given
# is Pandigital
def isPanDigital(s) :
 
    digits = [False] * 10;
 
    for i in range(0, len(s)) :
        digits[int(s[i]) -
               int('0')] = True
 
    # digit i is not present
    # thus not pandigital
    for i in range(0, 10) :
        if (digits[i] == False) :
            return False
 
    return True
 
# Returns number of pairs
# of strings resulting in
# Pandigital Concatenations
def countPandigitalPairs(v) :
 
    # iterate over all
    # pair of strings
    pairs = 0
    for i in range(0, len(v)) :
 
        for j in range (i + 1,
                        len(v)) :
         
            if (isPanDigital(v[i] +
                             v[j])) :
                pairs = pairs + 1
    return pairs
 
# Driver code
v = ["123567", "098234",
        "14765", "19804"]
 
print (countPandigitalPairs(v))
 
# This code is contributed by
# Manish Shaw(manishshaw1)

C#




// C# program to find all Pandigital
// concatenations of two strings.
using System;
using System.Collections.Generic;
 
class GFG
{
    // Checks if a given
    // string is Pandigital
    static int isPanDigital(string s)
    {
        int []digits = new int[10];
        Array.Clear(digits, 0, 10);
        for (int i = 0; i < s.Length; i++)
            digits[s[i] - (int)'0'] = 1;
     
        // digit i is not present
        // thus not pandigital
        for (int i = 0; i <= 9; i++)
            if (digits[i] == 0)
                return 0;
     
        return 1;
    }
     
    // Returns number of pairs
    // of strings resulting in
    // Pandigital Concatenations
    static int countPandigitalPairs(ref List<string> v)
    {
        // iterate over all
        // pair of strings
        int pairs = 0;
        for (int i = 0; i < v.Count; i++)
            for (int j = i + 1; j < v.Count; j++)
                if (isPanDigital(v[i] + v[j]) == 1)
                    pairs++;
        return pairs;
    }
     
    // Driver code
    static void Main()
    {
        List<string> v = new List<string>{"123567", "098234",
                                          "14765", "19804"};
        Console.WriteLine(countPandigitalPairs(ref v));
    }
}
 
// This code is contributed
// by Manish Shaw(manishshaw1)

PHP




<?php
// PHP program to find all
// Pandigital concatenations
// of two strings.
 
// Checks if a given
// $is Pandigital
function isPanDigital($s)
{
    $digits = array();
    $digits = array_fill(0, 10, false);
 
    for ($i = 0; $i < strlen($s); $i++)
        $digits[ord($s[$i]) -
                ord('0')] = true;
 
    // digit i is not present
    // thus not pandigital
    for ($i = 0; $i <= 9; $i++)
        if ($digits[$i] == false)
            return false;
 
    return true;
}
 
// Returns number of pairs
// of strings resulting in
// Pandigital Concatenations
function countPandigitalPairs(&$v)
{
    // iterate over all
    // pair of strings
    $pairs = 0;
    for ($i = 0;
         $i < count($v); $i++)
    {
        for ($j = $i + 1;
             $j < count($v); $j++)
        {
            if (isPanDigital($v[$i].$v[$j]))
            {
                $pairs++;
            }
        }
    }
    return $pairs;
}
 
// Driver code
$v = array("123567", "098234",
           "14765", "19804");
 
echo (countPandigitalPairs($v));
 
// This code is contributed by
// Manish Shaw(manishshaw1)
?>

Output: 

3



Method 2 (Efficient): 
Now we look for something better than the brute-force discussed above. Careful analysis suggests that, for every digit 0 – 9 to be present we have a mask as 1111111111 (i.e. all numbers 0-9 exist in the array of numbers



Digits -  0  1  2  3  4  5  6  7  8  9
          |  |  |  |  |  |  |  |  |  |
Mask   -  1  1  1  1  1  1  1  1  1  1 

Here 1 denotes that the corresponding digits
exists at-least once thus for all such Pandigital 
Concatenations, this relationship should hold.
So we can represent 11...11 as a valid mask for
pandigital concatenations.



So now the approach is to represent every string as a mask of 10 bits where the ith bit is set if the ith digit exists in the string. 

E.g., "11405" can be represented as
Digits -           0  1  2  3  4  5  6  7  8  9
                   |  |  |  |  |  |  |  |  |  |
Mask for 11405 -   1  1  0  0  1  1  0  0  0  0



The approach though may look complete is still not efficient as we still have to iterate over all pairs and check if the OR of these two strings results in the mask of a valid Pandigital Concatenation. 

If we analyze the possible masks of all possible strings we can understand that every single string would be only comprised of digits 0 – 9, so every number can at max contain all digits 0 to 9 at least once thus the mask of such a number would be 1111111111 (1023 in decimal). Thus, in the decimal system all masks exit in (0 – 1023]. 

Now we just have to maintain a frequency array to store the number of times a mask exists in the array of strings.

Let two masks be i and j with frequencies freqi and freqj respectively,
If (i OR j) = Maskpandigital concatenation 
Then, 
Number of Pairs = freqi * freqj  

C++




// C++ program to count PanDigital pairs
#include <bits/stdc++.h>
using namespace std;
 
const int pandigitalMask = ((1 << 10) - 1);
 
void computeMaskFrequencies(vector<string> v, map<int,
                                        int>& freq)
{
    for (int i = 0; i < v.size(); i++) {
        int mask = 0;
 
        // Stores digits present in string v[i]
        // atleast once. We use a set as we only
        // need digits which exist only once
        // (irrespective of reputation)
        unordered_set<int> digits;
        for (int j = 0; j < v[i].size(); j++)
            digits.insert(v[i][j] - '0');
 
        // Calculate the mask by considering all digits
        // existing atleast once
        for (auto it = digits.begin(); it != digits.end(); it++) {
            int digit = (*it);
            mask += (1 << digit);
        }
 
        // Increment the frequency of this mask
        freq[mask]++;
    }
}
 
// Returns number of pairs of strings resulting
// in Pandigital Concatenations
int pandigitalConcatenations(map<int, int> freq)
{
    int ans = 0;
 
    // All possible strings lie between 1 and 1023
    // so we iterate over every possible mask
    for (int i = 1; i <= 1023; i++) {
        for (int j = 1; j <= 1023; j++) {
 
            // if the concatenation results in mask of
            // Pandigital Concatenation, calculate all
            // pairs formed with Masks i and j
            if ((i | j) == pandigitalMask) {
                if (i == j)
                    ans += (freq[i] * (freq[i] - 1));            
                else
                    ans += (freq[i] * freq[j]);            
            }
        }
    }
 
    // since every pair is considers twice,
    // we get rid of half of these
    return ans/2;
}
 
int countPandigitalPairs(vector<string> v)
{
    // Find frequencies of all masks in
    // given vector of strings
    map<int, int> freq;
    computeMaskFrequencies(v, freq);
     
    // Return all possiblg concatenations.
    return pandigitalConcatenations(freq);
}
 
// Driver code
int main()
{
    vector<string> v = {"123567", "098234", "14765", "19804"};
    cout << countPandigitalPairs(v) << endl;
    return 0;
}

Java




// Java program to count PanDigital pairs
import java.util.*;
 
class GFG{
 
static int pandigitalMask = ((1 << 10) - 1);
 
static void computeMaskFrequencies(Vector<String> v,
                         HashMap<Integer, Integer> freq)
{
    for(int i = 0; i < v.size(); i++)
    {
        int mask = 0;
 
        // Stores digits present in String v[i]
        // atleast once. We use a set as we only
        // need digits which exist only once
        // (irrespective of reputation)
        HashSet<Integer> digits = new HashSet<>();
        for(int j = 0; j < v.get(i).length(); j++)
            digits.add(v.get(i).charAt(j) - '0');
 
        // Calculate the mask by considering
        // all digits existing atleast once
        for(int it :digits)
        {
            int digit = (it);
            mask += (1 << digit);
        }
 
        // Increment the frequency of
        // this mask
        if (freq.containsKey(mask))
        {
            freq.put(mask, freq.get(mask) + 1);
        }
        else
        {
            freq.put(mask, 1);
        }
    }
}
 
// Returns number of pairs of Strings
// resulting in Pandigital Concatenations
static int pandigitalConcatenations(
    HashMap<Integer, Integer> freq)
{
    int ans = 0;
 
    // All possible Strings lie between
    // 1 and 1023 so we iterate over every
    // possible mask
    for(int i = 1; i <= 1023; i++)
    {
        for(int j = 1; j <= 1023; j++)
        {
             
            // If the concatenation results in mask of
            // Pandigital Concatenation, calculate all
            // pairs formed with Masks i and j
            if ((i | j) == pandigitalMask &&
                      freq.containsKey(j) &&
                      freq.containsKey(i))
            {
                if (i == j)
                    ans += (freq.get(i) *
                           (freq.get(i) - 1));            
                else
                    ans += (freq.get(i) *
                            freq.get(j));            
            }
        }
    }
     
    // Since every pair is considers twice,
    // we get rid of half of these
    return ans / 2;
}
 
static int countPandigitalPairs(Vector<String> v)
{
     
    // Find frequencies of all masks in
    // given vector of Strings
    HashMap<Integer,Integer> freq = new HashMap<>();
    computeMaskFrequencies(v, freq);
     
    // Return all possiblg concatenations.
    return pandigitalConcatenations(freq);
}
 
// Driver code
public static void main(String[] args)
{
    Vector<String> v  = new Vector<>();
    v.add("123567");
    v.add("098234");
    v.add("14765");
    v.add("19804");
     
    System.out.print(countPandigitalPairs(v) + "\n");
}
}
 
// This code is contributed by Amit Katiyar

C#




// C# program to count
// PanDigital pairs
using System;
using System.Collections.Generic;
class GFG{
 
static int pandigitalMask =
           ((1 << 10) - 1);
 
static void computeMaskFrequencies(List<String> v,
                                   Dictionary<int,
                                   int> freq)
{
  for(int i = 0; i < v.Count; i++)
  {
    int mask = 0;
 
    // Stores digits present in String v[i]
    // atleast once. We use a set as we only
    // need digits which exist only once
    // (irrespective of reputation)
    HashSet<int> digits = new HashSet<int>();
     
    for(int j = 0; j < v[i].Length; j++)
      digits.Add(v[i][j] - '0');
 
    // Calculate the mask by considering
    // all digits existing atleast once
    foreach(int it in digits)
    {
      int digit = (it);
      mask += (1 << digit);
    }
 
    // Increment the frequency of
    // this mask
    if (freq.ContainsKey(mask))
    {
      freq[mask]++;
    }
    else
    {
      freq.Add(mask, 1);
    }
  }
}
 
// Returns number of pairs of
// Strings resulting in Pandigital
// Concatenations
static int pandigitalConcatenations(Dictionary<int,
                                    int> freq)
{
  int ans = 0;
 
  // All possible Strings lie between
  // 1 and 1023 so we iterate over every
  // possible mask
  for(int i = 1; i <= 1023; i++)
  {
    for(int j = 1; j <= 1023; j++)
    {
      // If the concatenation results in
      // mask of Pandigital Concatenation,
      // calculate all pairs formed with
      // Masks i and j
      if ((i | j) == pandigitalMask &&
          freq.ContainsKey(j) &&
          freq.ContainsKey(i))
      {
        if (i == j)
          ans += (freq[i] *
                  (freq[i] - 1));            
        else
          ans += (freq[i] *
                  freq[j]);            
      }
    }
  }
 
  // Since every pair is considers
  // twice, we get rid of half of
  // these
  return ans / 2;
}
 
static int countPandigitalPairs(List<String> v)
{   
  // Find frequencies of all masks in
  // given vector of Strings
  Dictionary<int,
             int> freq = new Dictionary<int,
                                        int>();
  computeMaskFrequencies(v, freq);
 
  // Return all possiblg concatenations.
  return pandigitalConcatenations(freq);
}
 
// Driver code
public static void Main(String[] args)
{
  List<String> v  = new List<String>();
  v.Add("123567");
  v.Add("098234");
  v.Add("14765");
  v.Add("19804");
  Console.Write(countPandigitalPairs(v) + "\n");
}
}
 
// This code is contributed by 29AjayKumar

Output:  

3



Complexity : O(N * |s| + 1023 * 1023) where |s| gives length of strings in the array
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :