Given an array of N numbers. Find out the number of pairs i and j such that i < j and Ai and Aj 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 falsebool 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 pairsint 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 functionsint 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 digitimport 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 falsedef 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 pairsdef 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 Codeif __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 digitusing 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 falsefunction 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 pairsfunction 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 falsefunction 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 pairsfunction 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 << ith 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 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++
// 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 arrayvoid 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 pairsint 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 functionsint 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 digitimport 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 arraydef 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 pairsdef 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 Codearr = [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 digitusing 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 frequencieslet freq = new Map();// This function calculates the mask frequencies // for every present in the arrayfunction 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 pairsfunction 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 Codelet 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 )
