Count pairs in an array which have at least one digit common

Given an array of N numbers. Find out the number of pairs i and j such that i < j and A_i and A_j have at least one digit common (For e.g. (11, 19) have 1 digit common but (36, 48) have no digit common)

Examples:

Input: A[] = { 10, 12, 24 } 
Output: 2 
Explanation: Two valid pairs are (10, 12) and (12, 24) which have atleast one digit common 

Method 1 (Brute Force):

A naive approach to solve this problem is just by running two nested loops and consider all possible pairs. We can check if the two numbers have at least one common digit, by extracting every digit of the first number and try to find it in the extracted digits of second number. The task would become much easier we simply convert them into strings.

Below is the implementation of the above approach:

C++

// CPP Program to count pairs in an array
// with some common digit
#include <bits/stdc++.h>

using namespace std;

// Returns true if the pair is valid,
// otherwise false
bool checkValidPair(int num1, int num2)
{
    // converting integers to strings
    string s1 = to_string(num1);
    string s2 = to_string(num2);

    // Iterate over the strings and check
    // if a character in first string is also
    // present in second string, return true
    for (int i = 0; i < s1.size(); i++)
        for (int j = 0; j < s2.size(); j++)
            if (s1[i] == s2[j])
                return true;

    // No common digit found
    return false;
}

// Returns the number of valid pairs
int countPairs(int arr[], int n)
{
    int numberOfPairs = 0;

    // Iterate over all possible pairs
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (checkValidPair(arr[i], arr[j]))
                numberOfPairs++;

    return numberOfPairs;
}

// Driver Code to test above functions
int main()
{
    int arr[] = { 10, 12, 24 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countPairs(arr, n) << endl;
    return 0;
}

Java

// Java Program to count
// pairs in an array
// with some common digit
import java.io.*;

class GFG
{

    // Returns true if the pair
    // is valid, otherwise false
    static boolean checkValidPair(int num1,
                                  int num2)
    {
        // converting integers
        // to strings
        String s1 = Integer.toString(num1);
        String s2 = Integer.toString(num2);

        // Iterate over the strings
        // and check if a character
        // in first string is also
        // present in second string,
        // return true
        for (int i = 0; i < s1.length(); i++)
            for (int j = 0; j < s2.length(); j++)
                if (s1.charAt(i) == s2.charAt(j))
                    return true;

        // No common
        // digit found
        return false;
    }

    // Returns the number
    // of valid pairs
    static int countPairs(int []arr, int n)
    {
        int numberOfPairs = 0;

        // Iterate over all
        // possible pairs
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                if (checkValidPair(arr[i], arr[j]))
                    numberOfPairs++;

        return numberOfPairs;
    }

    // Driver Code
    public static void main(String args[])
    {
        int []arr = new int[]{ 10, 12, 24 };
        int n = arr.length;
        System.out.print(countPairs(arr, n));
    }
}

// This code is contributed
// by manish shaw.

Python3

# Python3 Program to count pairs in
# an array with some common digit

# Returns true if the pair is
# valid, otherwise false
def checkValidPair(num1, num2) :

    # converting integers to strings
    s1 = str(num1)
    s2 = str(num2)

    # Iterate over the strings and check if
    # a character in first string is also
    # present in second string, return true
    for i in range(len(s1)) :
        for j in range(len(s2)) :
            if (s1[i] == s2[j]) :
                return True;

    # No common digit found
    return False;

# Returns the number of valid pairs
def countPairs(arr, n) :

    numberOfPairs = 0

    # Iterate over all possible pairs
    for i in range(n) :
        for j in range(i + 1, n) :
            if (checkValidPair(arr[i], arr[j])) :
                numberOfPairs += 1

    return numberOfPairs

# Driver Code
if __name__ == "__main__" :
    arr = [ 10, 12, 24 ]
    n = len(arr)
    print(countPairs(arr, n))

# This code is contributed by Ryuga

C#

// C# Program to count pairs in an array
// with some common digit
using System;

class GFG {

    // Returns true if the pair is valid,
    // otherwise false
    static bool checkValidPair(int num1, int num2)
    {
        // converting integers to strings
        string s1 = num1.ToString();
        string s2 = num2.ToString();

        // Iterate over the strings and check
        // if a character in first string is also
        // present in second string, return true
        for (int i = 0; i < s1.Length; i++)
            for (int j = 0; j < s2.Length; j++)
                if (s1[i] == s2[j])
                    return true;

        // No common digit found
        return false;
    }

    // Returns the number of valid pairs
    static int countPairs(int []arr, int n)
    {
        int numberOfPairs = 0;

        // Iterate over all possible pairs
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                if (checkValidPair(arr[i], arr[j]))
                    numberOfPairs++;

        return numberOfPairs;
    }

    // Driver Code to test above functions
    static void Main()
    {
        int []arr = new int[]{ 10, 12, 24 };
        int n = arr.Length;
        Console.WriteLine(countPairs(arr, n));
    }
}

// This code is contributed by manish shaw.

PHP

<?php
// PHP Program to count pairs in an array
// with some common digit

// Returns true if the pair is valid,
// otherwise false
function checkValidPair($num1, $num2)
{
    // converting integers to strings
    $s1 = (string)$num1;
    $s2 = (string)$num2;

    // Iterate over the strings and check
    // if a character in first string is also
    // present in second string, return true
    for ($i = 0; $i < strlen($s1); $i++)
        for ($j = 0; $j < strlen($s2); $j++)
            if ($s1[$i] == $s2[$j])
                return true;

    // No common digit found
    return false;
}

// Returns the number of valid pairs
function countPairs(&$arr, $n)
{
    $numberOfPairs = 0;

    // Iterate over all possible pairs
    for ($i = 0; $i < $n; $i++)
        for ($j = $i + 1; $j < $n; $j++)
            if (checkValidPair($arr[$i],
                               $arr[$j]))
                $numberOfPairs++;

    return $numberOfPairs;
}

// Driver Code
$arr = array(10, 12, 24 );
$n = sizeof($arr);
echo (countPairs($arr, $n));

// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>

// Javascript Program to count pairs in an array
// with some common digit

// Returns true if the pair is valid,
// otherwise false
function checkValidPair(num1, num2)
{
    // converting integers to strings
    var s1 = num1.toString();
    var s2 = num2.toString();

    var i,j;
    // Iterate over the strings and check
    // if a character in first string is also
    // present in second string, return true
    for(i = 0; i < s1.length; i++)
        for(j = 0; j < s2.length; j++)
            if (s1[i] == s2[j])
                return true;

    // No common digit found
    return false;
}

// Returns the number of valid pairs
function countPairs(arr, n)
{
    var numberOfPairs = 0;

    // Iterate over all possible pairs
    for(i = 0; i < n; i++)
      for(j = i + 1; j < n; j++)
         if(checkValidPair(arr[i], arr[j]))
            numberOfPairs++;

    return numberOfPairs;
}

// Driver Code to test above functions
    var arr = [10, 12, 24];
    var n = arr.length;;
    document.write(countPairs(arr, n));

</script>

Output

2

Time Complexity: O(N²) where N is the size of array.
Space Complexity : O( 1 )

Method 2 (Bit Masking):

An efficient approach of solving this problem is creating a bit mask for every digit present in a particular number. Thus, for every digit to be present in a number if we have a mask of 1111111111. 

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 ith digit is set. 
For e.g. 1235 can be represented as
Digits -         0  1  2  3  4  5  6  7  8  9
                 |  |  |  |  |  |  |  |  |  |
Mask for 1235 -  0  1  1  1  0  1  0  0  0  0

Now we just have to extract every digit of a number and make the corresponding bit set (1 << i^th digit) and store the whole number as a mask. Careful analysis suggests that the maximum value of the mask is 1023 in decimal (which contains all the digits from 0 – 9). Since the same set of digits can exist in more than one number, we need to maintain a frequency array to store the count of mask value. 

Let the frequencies of two masks i and j be freq_i and freq_j respectively, 
If(i AND j) return true, means i_th and j_th mask contains atleast one common set bit which in turn implies that the numbers from which these masks have been built also contain a common digit 
then, 
increment the answer 
ans += freq_i * freq_j [ if i != j ] 
ans += (freq_i * (freq_i – 1)) / 2 [ if j == i ]  

Below is the implementation of this efficient approach:

C++

// CPP Program to count pairs in an array with
// some common digit
#include <bits/stdc++.h>
using namespace std;

// This function calculates the mask frequencies
// for every present in the array
void calculateMaskFrequencies(int arr[], int n,
                 unordered_map<int, int>& freq)
{
    // Iterate over the array
    for (int i = 0; i < n; i++) {

        int num = arr[i];

        // Creating an empty mask
        int mask = 0;

        // Extracting every digit of the number
        // and updating corresponding bit in the
        // mask
        while (num > 0) {
            mask = mask | (1 << (num % 10));
            num /= 10;
        }

        // Update the frequency array
        freq[mask]++;
    }
}

// Function return the number of valid pairs
int countPairs(int arr[], int n)
{
    // Store the mask frequencies
    unordered_map<int, int> freq;

    calculateMaskFrequencies(arr, n, freq);

    long long int numberOfPairs = 0;

    // Considering every possible pair of masks
    // and calculate pairs according to their
    // frequencies
    for (int i = 0; i < 1024; i++) {
        numberOfPairs += (freq[i] * (freq[i] - 1)) / 2;
        for (int j = i + 1; j < 1024; j++) {

            // if it contains a common digit
            if (i & j)
                numberOfPairs += (freq[i] * freq[j]);
        }
    }
    return numberOfPairs;
}

// Driver Code to test above functions
int main()
{
    int arr[] = { 10, 12, 24 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countPairs(arr, n) << endl;
    return 0;
}

Java

// Java Program to count pairs in an array with
// some common digit
import java.io.*;
import java.util.*;
class GFG
{

  // Store the mask frequencies
  public static Map<Integer, Integer> freq = new HashMap<Integer, Integer>();

  // This function calculates the mask frequencies
  // for every present in the array
  public static void calculateMaskFrequencies(int[] arr,int n)
  {

    // Iterate over the array
    for(int i = 0; i < n; i++)
    {
      int num = arr[i];

      // Creating an empty mask
      int mask = 0;

      // Extracting every digit of the number
      // and updating corresponding bit in the
      // mask
      while(num > 0)
      {
        mask = mask | (1 << (num % 10));
        num /= 10;
      }

      // Update the frequency array
      if(freq.containsKey(mask))
      {
        freq.put(new Integer(mask), freq.get(mask) + 1);
      }
      else
      {
        freq.put(new Integer(mask), 1);
      }
    }
  }

  // Function return the number of valid pairs
  public static int countPairs(int[] arr, int n)
  {
    calculateMaskFrequencies(arr, n);
    int numberOfPairs = 0;

    // Considering every possible pair of masks
    // and calculate pairs according to their
    // frequencies
    for(int i = 0; i < 1024; i++)
    {
      int x = 0;
      if(freq.containsKey(i))
      {
        x = freq.get(i);

      }
      numberOfPairs += ((x) * (x - 1)) / 2;
      for(int j = i + 1; j < 1024; j++)
      {
        int y = 0;

        // if it contains a common digit
        if((i & j) != 0)
        {
          if(freq.containsKey(j))
          {
            y = freq.get(j);
          }
          numberOfPairs += x * y;
        }
      }
    }
    return numberOfPairs;
  }

  // Driver Code
  public static void main (String[] args)
  {
    int[] arr = {10, 12, 24};
    int n = arr.length;
    System.out.println(countPairs(arr, n));
  }
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python3 Program to count pairs in an array
# with some common digit

# This function calculates the mask frequencies
# for every present in the array
def calculateMaskFrequencies(arr, n, freq):

    # Iterate over the array
    for i in range(n):

        num = arr[i]

        # Creating an empty mask
        mask = 0

        # Extracting every digit of the number
        # and updating corresponding bit in the
        # mask
        while (num > 0):
            mask = mask | (1 << (num % 10))
            num //= 10

        # Update the frequency array
        freq[mask] = freq.get(mask, 0) + 1

# Function return the number of valid pairs
def countPairs(arr, n):

    # Store the mask frequencies
    freq = dict()

    calculateMaskFrequencies(arr, n, freq)

    numberOfPairs = 0

    # Considering every possible pair of masks
    # and calculate pairs according to their
    # frequencies
    for i in range(1024):

        x = 0

        if i in freq.keys():
            x = freq[i]

        numberOfPairs += (x * (x - 1)) // 2

        for j in range(i + 1, 1024):

            y = 0

            if j in freq.keys():
                y = freq[j]

            # if it contains a common digit
            if (i & j):
                numberOfPairs += (x * y)

    return numberOfPairs

# Driver Code
arr = [10, 12, 24]
n = len(arr)
print(countPairs(arr, n))

# This code is contributed by mohit kumar

C#

// C# Program to count pairs in an array with
// some common digit
using System;
using System.Collections.Generic;

public class GFG
{

    // Store the mask frequencies
    static Dictionary<int, int> freq =  new Dictionary<int, int>();

    // This function calculates the mask frequencies
    // for every present in the array
    public static void calculateMaskFrequencies(int[] arr,int n)
    {

        // Iterate over the array
        for(int i = 0; i < n; i++)
        {
            int num = arr[i];

            // Creating an empty mask
            int mask = 0;

            // Extracting every digit of the number
            // and updating corresponding bit in the
            // mask
            while(num > 0)
            {
                mask = mask | (1 << (num % 10));
                num /= 10;
            }

            // Update the frequency array
            if(freq.ContainsKey(mask))
            {
                freq[mask]++;
            }
            else
            {
                freq.Add(mask, 1);
            }
        }

    }
    public static int countPairs(int[] arr, int n)
    {
        calculateMaskFrequencies(arr, n);
        int numberOfPairs = 0;

        // Considering every possible pair of masks
        // and calculate pairs according to their
        // frequencies
        for(int i = 0; i < 1024; i++)
        {
            int x = 0;
            if(freq.ContainsKey(i))
            {
                x = freq[i];

            }
            numberOfPairs += ((x) * (x - 1)) / 2;
            for(int j = i + 1; j < 1024; j++)
            {
                int y = 0;

                // if it contains a common digit
                if((i & j) != 0)
                {
                    if(freq.ContainsKey(j))
                    {
                        y = freq[j];
                    }
                    numberOfPairs += x * y;
                }
            }
        }
        return numberOfPairs;

    }

    // Driver Code
    static public void Main ()
    {
        int[] arr = {10, 12, 24};
        int n = arr.Length;
        Console.WriteLine(countPairs(arr, n));
    }
}

// This code is contributed by rag2127

Javascript

<script>
// Javascript Program to count pairs in an array with
// some common digit

// Store the mask frequencies
let freq = new Map();

// This function calculates the mask frequencies
  // for every present in the array
function calculateMaskFrequencies(arr,n)
{
    // Iterate over the array
    for(let i = 0; i < n; i++)
    {
      let num = arr[i];

      // Creating an empty mask
      let mask = 0;

      // Extracting every digit of the number
      // and updating corresponding bit in the
      // mask
      while(num > 0)
      {
        mask = mask | (1 << (num % 10));
        num = Math.floor(num/10);
      }

      // Update the frequency array
      if(freq.has(mask))
      {
        freq.set((mask), freq.get(mask) + 1);
      }
      else
      {
        freq.set((mask), 1);
      }
    }
}

// Function return the number of valid pairs
function countPairs(arr,n)
{
    calculateMaskFrequencies(arr, n);
    let numberOfPairs = 0;

    // Considering every possible pair of masks
    // and calculate pairs according to their
    // frequencies
    for(let i = 0; i < 1024; i++)
    {
      let x = 0;
      if(freq.has(i))
      {
        x = freq.get(i);

      }
      numberOfPairs += Math.floor((x) * (x - 1)) / 2;
      for(let j = i + 1; j < 1024; j++)
      {
        let y = 0;

        // if it contains a common digit
        if((i & j) != 0)
        {
          if(freq.has(j))
          {
            y = freq.get(j);
          }
          numberOfPairs += x * y;
        }
      }
    }
    return numberOfPairs;
}

// Driver Code
let arr=[10, 12, 24];
let n = arr.length;
document.write(countPairs(arr, n));


// This code is contributed by unknown2108
</script>

Output

2

Time Complexity: O(N + 1024 * 1024), where N is the size of the array.

Space Complexity : O( 1 )