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 Ai and Aj have atleast 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 atleast one common digit, by extracting every digit of 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 naive implementation.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right



Output

2

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

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 -  1  1  1  1  1  1  1  1  1  1 

Now we just have to extract every digit of a number and make the corresponding bit set (1 << ith digit) and store the whole number as mask. Careful analysis suggests that the maximum value of mask is 1023 in decimal (which contains all the digits from 0 – 9). Since the same set of digits can exists 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 freqi and freqj respectively,
If(i AND j) return true, means ith and jth 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 += freqi * freqj [ if i != j ]
ans += (freqi * (freqi – 1)) / 2 [ if j == i ]

Below is the implementation of this efficient approach.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right



Output

2

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



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.