Given two positive numbers N and M, the task is to count the number of digits that are present in both N and M.
Examples:
Input: N = 748294, M = 34298156
Output: 4
Explanation: The digits that are present in both the numbers are {4, 8, 2, 9}. Therefore, the required count is 4.Input: N = 111222, M = 333444
Output: 0
Explanation: No common digits present in the two given numbers.
Approach: The given problem can be solved using Hashing. Follow the steps below to solve the problem:
- Initialize a variable, say count as 0, to store the number of digits that are common in both the numbers.
- Initialize two arrays, say freq1[10] and freq2[10] as {0}, to store the count of digits present in the integers N and M respectively.
- Iterate over the digits of the integer N and increment the count of each digit in freq1[] by 1.
- Iterate over the digits of the integer M and increment the count of each digit in freq2[] by 1.
- Iterate over the range [0, 9] and increment the count by 1 if freq1[i] and freq2[i] both exceeds 0.
- Finally, after completing the above steps, print the count obtained as the required answer.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count number of digits// that are common in both N and Mint CommonDigits(int N, int M){ // Stores the count of common digits int count = 0; // Stores the count of digits of N int freq1[10] = { 0 }; // Stores the count of digits of M int freq2[10] = { 0 }; // Iterate over the digits of N while (N > 0) { // Increment the count of // last digit of N freq1[N % 10]++; // Update N N = N / 10; } // Iterate over the digits of M while (M > 0) { // Increment the count of // last digit of M freq2[M % 10]++; // Update M M = M / 10; } // Iterate over the range [0, 9] for (int i = 0; i < 10; i++) { // If freq1[i] and freq2[i] both exceeds 0 if (freq1[i] > 0 & freq2[i] > 0) { // Increment count by 1 count++; } } // Return the count return count;}// Driver Codeint main(){ // Input int N = 748294; int M = 34298156; cout << CommonDigits(N, M); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to count number of digits// that are common in both N and Mstatic int CommonDigits(int N, int M){ // Stores the count of common digits int count = 0; // Stores the count of digits of N int freq1[] = new int[10]; // Stores the count of digits of M int freq2[] = new int[10]; // Iterate over the digits of N while (N > 0) { // Increment the count of // last digit of N freq1[N % 10]++; // Update N N = N / 10; } // Iterate over the digits of M while (M > 0) { // Increment the count of // last digit of M freq2[M % 10]++; // Update M M = M / 10; } // Iterate over the range [0, 9] for(int i = 0; i < 10; i++) { // If freq1[i] and freq2[i] both exceeds 0 if (freq1[i] > 0 & freq2[i] > 0) { // Increment count by 1 count++; } } // Return the count return count;}// Driver Codepublic static void main(String[] args){ // Input int N = 748294; int M = 34298156; System.out.print(CommonDigits(N, M));}}// This code is contributed by gauravrajput1 |
Python3
# Python3 program for the above approach# Function to count number of digits# that are common in both N and Mdef CommonDigits(N, M): # Stores the count of common digits count = 0 # Stores the count of digits of N freq1 = [0] * 10 # Stores the count of digits of M freq2 = [0] * 10 # Iterate over the digits of N while (N > 0): # Increment the count of # last digit of N freq1[N % 10] += 1 # Update N N = N // 10 # Iterate over the digits of M while (M > 0): # Increment the count of # last digit of M freq2[M % 10] += 1 # Update M M = M // 10 # Iterate over the range [0, 9] for i in range(10): # If freq1[i] and freq2[i] both exceeds 0 if (freq1[i] > 0 and freq2[i] > 0): # Increment count by 1 count += 1 # Return the count return count# Driver Codeif __name__ == '__main__': # Input N = 748294 M = 34298156 print (CommonDigits(N, M))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{ // Function to count number of digits// that are common in both N and Mstatic int CommonDigits(int N, int M){ // Stores the count of common digits int count = 0; // Stores the count of digits of N int[] freq1 = new int[10]; // Stores the count of digits of M int[] freq2 = new int[10]; // Iterate over the digits of N while (N > 0) { // Increment the count of // last digit of N freq1[N % 10]++; // Update N N = N / 10; } // Iterate over the digits of M while (M > 0) { // Increment the count of // last digit of M freq2[M % 10]++; // Update M M = M / 10; } // Iterate over the range [0, 9] for(int i = 0; i < 10; i++) { // If freq1[i] and freq2[i] // both exceeds 0 if (freq1[i] > 0 & freq2[i] > 0) { // Increment count by 1 count++; } } // Return the count return count;}// Driver codestatic void Main(){ // Input int N = 748294; int M = 34298156; Console.WriteLine(CommonDigits(N, M));}}// This code is contributed by sanjoy_62 |
Javascript
<script>// javascript program for the above approach// Function to count number of digits// that are common in both N and Mfunction CommonDigits(N,M){ // Stores the count of common digits var count = 0; // Stores the count of digits of N var freq1 = Array(10).fill(0); // Stores the count of digits of M var freq2 = Array(10).fill(0); // Iterate over the digits of N while (N > 0) { // Increment the count of // last digit of N freq1[N % 10]++; // Update N N = Math.floor(N / 10); } // Iterate over the digits of M while (M > 0) { // Increment the count of // last digit of M freq2[M % 10]++; // Update M M = Math.floor(M / 10); } var i; // Iterate over the range [0, 9] for (i = 0; i < 10; i++) { // If freq1[i] and freq2[i] both exceeds 0 if (freq1[i] > 0 & freq2[i] > 0) { // Increment count by 1 count++; } } // Return the count return count;}// Driver Code // Input var N = 748294; var M = 34298156; document.write(CommonDigits(N, M));</script> |
Output
4
Time Complexity: O(digits(N)+digits(M))
Auxiliary Space: O(10)
Using Sets in Python:
Approach:
One of the simplest approaches is to convert both numbers to sets and find the intersection of the two sets. The length of the intersection set will be the number of common digits.
- Define the function count_common_digits(n, m) that takes two integer inputs n and m.
- Convert both n and m to sets of digits using the set() function and str() function. Store the sets in variables n_set and m_set, respectively.
- Find the intersection of the two sets using the intersection() method of sets. Store the result in a variable common_digits.
- Return the length of the common_digits set using the len() function as the output of the function
C++
#include <iostream>#include <string>#include <unordered_set>using namespace std;int count_common_digits(int n, int m) { string n_str = to_string(n); string m_str = to_string(m); unordered_set<char> n_set(n_str.begin(), n_str.end()); unordered_set<char> m_set(m_str.begin(), m_str.end()); int common_digits = 0; for (char digit : n_set) { if (m_set.count(digit) > 0) { common_digits++; } } return common_digits;}int main() { int n = 748294; int m = 34298156; cout << count_common_digits(n, m) << endl; // Output: 4 return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;import java.util.HashSet;import java.util.Set;public class Main { public static int countCommonDigits(int n, int m) { String nStr = Integer.toString(n); String mStr = Integer.toString(m); Set<Character> nSet = new HashSet<>(); Set<Character> mSet = new HashSet<>(); for (char digit : nStr.toCharArray()) { nSet.add(digit); } for (char digit : mStr.toCharArray()) { mSet.add(digit); } int commonDigits = 0; for (char digit : nSet) { if (mSet.contains(digit)) { commonDigits++; } } return commonDigits; } public static void main(String[] args) { int n = 748294; int m = 34298156; System.out.println(countCommonDigits(n, m)); // Output- 4 }}// This code is contributed by guptapratik |
Python3
def count_common_digits(n, m): n_set = set(str(n)) m_set = set(str(m)) return len(n_set.intersection(m_set))# Example usagen = 748294m = 34298156print(count_common_digits(n, m)) # Output: 4 |
C#
using System;using System.Collections.Generic;using System.Linq;class Program { // Function to count the number of common digits in two // integers static int CountCommonDigits(int n, int m) { // Convert integers to strings to work with their // digits string nStr = n.ToString(); string mStr = m.ToString(); // Create sets to store unique digits from both // integers HashSet<char> nSet = new HashSet<char>(nStr); HashSet<char> mSet = new HashSet<char>(mStr); int commonDigits = 0; // Iterate through the digits of one integer and // check if they exist in the other foreach(char digit in nSet) { if (mSet.Contains(digit)) { commonDigits++; } } return commonDigits; } static void Main() { int n = 748294; int m = 34298156; int result = CountCommonDigits(n, m); Console.WriteLine(result); // Output: 4 }} |
Output
4
Time Complexity: O(m + n), where m and n are the numbers of digits in the two numbers respectively.
Space Complexity: O(m + n)
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