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 conditionInput: 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 codeint main(){ int N = 123, M = 321; cout << countDigits(N, M); return 0;} |
Java
// Java implementation of the above approachimport java.io.*;public 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 codeN = 123M = 321;print(countDigits(N, M));# This code is contributed by Saurabh Jaiswal |
C#
// C# implementation of the above approachusing 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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