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

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  • Last Updated : 09 Feb, 2022

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

Examples:

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

Input: N = 123, M = 111
Output: 1

 

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

  • Convert N and M for ease of traversal
  • Create two pointers where one pointer points to the first digit of N and the other to the first digit of M respectively initially.
  • Now traverse both the numbers left to right and check if the digits on both pointers are same or not.
  • If yes, increase the count.
  • Increase the pointer by 1 on each iteration.
  • Return the final count at the end.

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);
 
    int same_dig_cnt = 0;
 
    // 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++;
 
    return same_dig_cnt;
}
 
// 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);
 
    int same_dig_cnt = 0;
 
    // 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++;
 
    return same_dig_cnt;
  }
 
  // 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)
 
    same_dig_cnt = 0;
 
    # 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
 
    return same_dig_cnt;
 
# Driver code
N = 123
M = 321;
 
print(countDigits(N, M));
 
# This code is contributed by Saurabh Jaiswal

C#




// 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();
 
    int same_dig_cnt = 0;
 
    // 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++;
 
    return same_dig_cnt;
  }
 
  // 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();
 
           let same_dig_cnt = 0;
 
           // 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++;
 
           return same_dig_cnt;
       }
 
       // Driver code
       let N = 123, M = 321;
 
       document.write(countDigits(N, M));
 
      // This code is contributed by Potta Lokesh
   </script>

Output

1

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


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