Number of pairs with Pandigital Concatenation

13 Jun, 2022
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++
Java
Python3
C#
PHP
Javascript
// C++ program to find all// Pandigital concatenations// of two strings.#include <bits/stdc++.h>using namespace std; // Checks if a given// string is Pandigitalbool 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 Concatenationsint 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 codeint main(){    vector<string> v = {"123567", "098234",                        "14765", "19804"};    cout << countPandigitalPairs(v) << endl;    return 0;}
// 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 program to find all# Pandigital concatenations# of two strings. # Checks if a given# is Pandigitaldef 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 Concatenationsdef 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 codev = ["123567", "098234",        "14765", "19804"] print (countPandigitalPairs(v)) # This code is contributed by# Manish Shaw(manishshaw1)
// 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 program to find all// Pandigital concatenations// of two strings. // Checks if a given// $is Pandigitalfunction isPanDigital($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[$i] == false)            return false;     return true;} // Returns number of pairs// of strings resulting in// Pandigital Concatenationsfunction countPandigitalPairs(&$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)?>
<script> // Javascript program to find all// Pandigital concatenations// of two strings. // Checks if a given// is Pandigitalfunction isPanDigital(s){    let digits = new Array(10).fill(false);     for(let i = 0; i < s.length; i++)        digits[s[i].charCodeAt(0) -                '0'.charCodeAt(0)] = true;     // digit i is not present    // thus not pandigital    for(let i = 0; i <= 9; i++)        if (digits[i] == false)            return false;     return true;} // Returns number of pairs// of strings resulting in// Pandigital Concatenationsfunction countPandigitalPairs(v){         // Iterate over all    // pair of strings    let pairs = 0;    for(let i = 0; i < v.length; i++)    {        for(let j = i + 1;                j < v.length; j++)        {            if (isPanDigital(v[i] + v[j]))            {                pairs++;            }        }    }    return pairs;} // Driver codelet v = [ "123567", "098234",          "14765", "19804" ]; document.write(countPandigitalPairs(v)); // This code is contributed by gfgking </script>
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++
Java
Python3
C#
Javascript
// 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 Concatenationsint 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 possible concatenations.    return pandigitalConcatenations(freq);} // Driver codeint main(){    vector<string> v = {"123567", "098234", "14765", "19804"};    cout << countPandigitalPairs(v) << endl;    return 0;}
// Java program to count PanDigital pairsimport 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 Concatenationsstatic 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 possible concatenations.    return pandigitalConcatenations(freq);} // Driver codepublic 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
# Python program to count PanDigital pairspandigitalMask = ((1 << 10) - 1)freq = dict() def computeMaskFrequencies(v):    global freq    for i in range(len(v)):         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)        digits = set()         for j in range(len(v[i])):            digits.add(int(v[i][j]))         # Calculate the mask by considering        # all digits existing atleast once        for it in digits:             digit = it            mask += (1 << digit)         # Increment the frequency of this mask        if mask in freq:            freq[mask] += 1         else:            freq[mask] = 1  # Returns number of pairs of strings resulting# in Pandigital Concatenationsdef pandigitalConcatenations():    global freq     ans = 0     # All possible strings lie between 1 and 1023    # so we iterate over every possible mask    for i in range(1, 1024):        for j in range(1, 1024):             # if the concatenation results in mask of            # Pandigital Concatenation, calculate all            # pairs formed with Masks i and j            if ((i | j) == pandigitalMask and                    i in freq and j in freq):                 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  def countPandigitalPairs(v):     # Find frequencies of all masks in    # given vector of strings    computeMaskFrequencies(v)     # Return all possible concatenations.    return pandigitalConcatenations() # Driver codev = ["123567", "098234", "14765", "19804"]print(countPandigitalPairs(v)) # This code is contributed by phasing17
// C# program to count// PanDigital pairsusing 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// Concatenationsstatic 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 possible concatenations.  return pandigitalConcatenations(freq);} // Driver codepublic 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
<script> // Javascript program to count PanDigital pairsconst pandigitalMask = ((1 << 10) - 1); function computeMaskFrequencies(v, freq){    for(let i = 0; i < v.length; i++)    {        let 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)        let digits = new Set();        for(let j = 0; j < v[i].length; j++)            digits.add((v[i][j]).charCodeAt(0) -                             '0'.charCodeAt(0));         // Calculate the mask by considering        // all digits existing atleast once        for(let it of digits)        {            let digit = it;            mask += (1 << digit);        }         // Increment the frequency of this mask        if (freq.has(mask))        {            freq.set(mask, freq.get(mask) + 1)        }        else        {            freq.set(mask, 1)        }    }} // Returns number of pairs of strings resulting// in Pandigital Concatenationsfunction pandigitalConcatenations(freq){    let ans = 0;     // All possible strings lie between 1 and 1023    // so we iterate over every possible mask    for(let i = 1; i <= 1023; i++)    {        for(let 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.has(i) && freq.has(j))            {                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 Math.floor(ans / 2);} function countPandigitalPairs(v){         // Find frequencies of all masks in    // given vector of strings    let freq = new Map();    computeMaskFrequencies(v, freq);     // Return all possible concatenations.    return pandigitalConcatenations(freq);} // Driver codelet v = [ "123567", "098234", "14765", "19804" ];document.write(countPandigitalPairs(v) + "<br>"); // This code is contributed by gfgking </script>
Output:  
3
Complexity : O(N * |s| + 1023 * 1023) where |s| gives length of strings in the array 
manishshaw1
amit143katiyar
29AjayKumar
gfgking
surinderdawra388
phasing17
cpp-unordered_set
cpp-vector
Bit Magic
Strings
Technical Scripter
Strings
Bit Magic
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