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Number of common digits present in two given numbers
  • Difficulty Level : Medium
  • Last Updated : 22 Apr, 2021

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 M
int 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 Code
int main()
{
    // Input
    int N = 748294;
    int M = 34298156;
 
    cout << CommonDigits(N, M);
 
    return 0;
}
Java



// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to count number of digits
// that are common in both N and M
static 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 Code
public 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 M
def 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 Code
if __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 approach
using System;
 
class GFG{
     
// Function to count number of digits
// that are common in both N and M
static 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 code
static 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 M
function 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)
 
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vikkycirus
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