Count of digits in N and M that are same but present on different indices

Given two numbers N and M, the task is to find the count of digits in N and M that are the same but present on different indices.

Examples:

Input: N = 123, M = 321
Output: 2
Explanation: Digit 1 and 3 satisfies the condition

Input: N = 123, M = 111
Output: 0

Approach: The problem can be solved using hashing and two-pointer approach.

Convert N and M to string for ease of traversal
Now create 2 hash of size 10 to store frequency of digits in N and M respectively
Now traverse a loop from 0-9 and:
- Add min of hashN[i] and hashM[i] to a variable count
Now traverse the numbers using two pointers to find the count of digits that are same and occur on same indices in both N and M. Store the count in variable same_dig_cnt
Now return the final count as count - same_dig_cnt

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

int countDigits(int N, int M)
{

    // Convert N and M to string
    // for ease of traversal
    string a = to_string(N), b = to_string(M);

    // Create 2 hash of size 10
    // to store frequency of digits
    // in N and M respectively
    vector<int> hashN(10, 0);
    for (char c : a)
        hashN[c - '0']++;
    vector<int> hashM(10, 0);
    for (char c : b)
        hashM[c - '0']++;

    int count = 0, same_dig_cnt = 0;

    // Count of common digits
    // irrespective of their positions
    for (int i = 0; i < 10; i++)
        count += min(hashN[i], hashM[i]);

    // Find the count of digits
    // that are same and occur on same indices
    // in both N and M.
    // Store the count in variable same_dig_cnt
    for (int i = 0; i < a.length() && b.length(); i++)
        if (a[i] == b[i])
            same_dig_cnt++;

    // Remove the count of digits that are
    // not at same indices in both numbers
    count -= same_dig_cnt;

    return count;
}

// Driver code
int main()
{

    int N = 123, M = 321;

    cout << countDigits(N, M);
    return 0;
}

Java

// Java implementation of the above approach
class GFG {

  static int countDigits(int N, int M) {

    // Convert N and M to string
    // for ease of traversal
    String a = Integer.toString(N), b = Integer.toString(M);

    // Create 2 hash of size 10
    // to store frequency of digits
    // in N and M respectively
    int[] hashN = new int[10];

    for (int i = 0; i < 10; i++) {
      hashN[i] = 0;
    }

    for (char c : a.toCharArray()) {
      hashN[c - '0']++;
    }

    int[] hashM = new int[10];

    for (int i = 0; i < 10; i++) {
      hashM[i] = 0;
    }

    for (char c : a.toCharArray()) {
      hashM[c - '0']++;
    }

    int count = 0, same_dig_cnt = 0;

    // Count of common digits
    // irrespective of their positions
    for (int i = 0; i < 10; i++)
      count += Math.min(hashN[i], hashM[i]);

    // Find the count of digits
    // that are same and occur on same indices
    // in both N and M.
    // Store the count in variable same_dig_cnt
    for (int i = 0; i < a.length() && i < b.length(); i++)
      if (a.charAt(i) == b.charAt(i))
        same_dig_cnt++;

    // Remove the count of digits that are
    // not at same indices in both numbers
    count -= same_dig_cnt;

    return count;
  }

  // Driver code
  public static void main(String args[])
  {
    int N = 123, M = 321;
    System.out.println(countDigits(N, M));
  }
}

// This code is contributed by gfgking

Python3

# Python code for the above approach 
def countDigits(N, M):

    # Convert N and M to string
    # for ease of traversal
    a = str(N)
    b = str(M)

    # Create 2 hash of size 10
    # to store frequency of digits
    # in N and M respectively
    hashN = [0] * 10
    for c in range(len(a)):
        hashN[ord(a[c]) - ord('0')] += 1

    hashM = [0] * 10 
    for c in range(len(b)):
        hashM[ord(b[c]) - ord('0')] += 1

    count = 0
    same_dig_cnt = 0;

    # Count of common digits
    # irrespective of their positions
    for i in range(10):
        count += min(hashN[i], hashM[i]);

    # Find the count of digits
    # that are same and occur on same indices
    # in both N and M.
    # Store the count in variable same_dig_cnt
    i = 0
    while(i < len(a) and len(b)):
        if (a[i] == b[i]):
            same_dig_cnt += 1
        i += 1

    # Remove the count of digits that are
    # not at same indices in both numbers
    count -= same_dig_cnt;

    return count;

# Driver code
N = 123
M = 321;

print(countDigits(N, M));

# This code is contributed by Saurabh Jaiswal

// C# implementation of the above approach
using System;
using System.Collections;
class GFG
{

  static int countDigits(int N, int M)
  {

    // Convert N and M to string
    // for ease of traversal
    string a = N.ToString(), b = M.ToString();

    // Create 2 hash of size 10
    // to store frequency of digits
    // in N and M respectively
    int []hashN = new int[10];

    for(int i = 0; i < 10; i++) {
      hashN[i] = 0;
    }

    foreach (char c in a) {
      hashN[c - '0']++;
    }

    int []hashM = new int[10];

    for(int i = 0; i < 10; i++) {
      hashM[i] = 0;
    }

    foreach (char c in a) {
      hashM[c - '0']++;
    }

    int count = 0, same_dig_cnt = 0;

    // Count of common digits
    // irrespective of their positions
    for (int i = 0; i < 10; i++)
      count += Math.Min(hashN[i], hashM[i]);

    // Find the count of digits
    // that are same and occur on same indices
    // in both N and M.
    // Store the count in variable same_dig_cnt
    for (int i = 0; i < a.Length && i < b.Length; i++)
      if (a[i] == b[i])
        same_dig_cnt++;

    // Remove the count of digits that are
    // not at same indices in both numbers
    count -= same_dig_cnt;

    return count;
  }

  // Driver code
  public static void Main()
  {

    int N = 123, M = 321;

    Console.Write(countDigits(N, M));
  }
}

// This code is contributed by Samim Hossain Mondal.

JavaScript

 <script>
        // JavaScript code for the above approach 
        function countDigits(N, M)
        {

            // Convert N and M to string
            // for ease of traversal
            let a = (N).toString(), b = (M).toString();

            // Create 2 hash of size 10
            // to store frequency of digits
            // in N and M respectively
            let hashN = new Array(10).fill(0);
            for (let c = 0; c < a.length; c++)
                hashN[a[c].charCodeAt(0) - '0'.charCodeAt(0)]++;
            let hashM = new Array(10).fill(0);
            for (c = 0; c < b.length; c++)
                hashM[b[c].charCodeAt(0) - '0'.charCodeAt(0)]++;

            let count = 0, same_dig_cnt = 0;

            // Count of common digits
            // irrespective of their positions
            for (let i = 0; i < 10; i++)
                count += Math.min(hashN[i], hashM[i]);

            // Find the count of digits
            // that are same and occur on same indices
            // in both N and M.
            // Store the count in variable same_dig_cnt
            for (let i = 0; i < a.length && b.length; i++)
                if (a[i] == b[i])
                    same_dig_cnt++;

            // Remove the count of digits that are
            // not at same indices in both numbers
            count -= same_dig_cnt;

            return count;
        }

        // Driver code
        let N = 123, M = 321;

        document.write(countDigits(N, M));

       // This code is contributed by Potta Lokesh
    </script>

Output

Time Complexity: O(D), where D is the max count of digits in N or M
Auxiliary Space: O(D), where D is the max count of digits in N or M

Count of digits in N and M that are same but present on different indices

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