Problem:
Score : 200 points

Problem Statement

Given are strings S and T. Consider changing S to T by repeating the operation below. Find the minimum number of operations required to do so.

Operation: Choose one character of S and replace it with a different character.

Constraints

S and T have lengths between 1 and 2\times 10^5 (inclusive).

S and T consists of lowercase English letters.

S and T have equal lengths.

Input

Input is given from Standard Input in the following format:

S
T

Output

Print the answer.

Sample Input 1

cupofcoffee
cupofhottea

Sample Output 1

4

We can achieve the objective in four operations, such as the following:

First, replace the sixth character c with h.

Second, replace the eighth character f with t.

Third, replace the ninth character f with t.

Fourth, replace the eleventh character e with a.

Sample Input 2

abcde
bcdea

Sample Output 2

5

Sample Input 3

apple
apple

Sample Output 3

0

No operations may be needed to achieve the objective.

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

		public static void main(String[] args) {

			Scanner sc = new Scanner(System.in);

			String S = sc.next();
			String T = sc.next();

			char[] Schar = new char[S.length()];
			char[] Tchar = new char[T.length()];

			for(int i =0; i < Schar.length; i++) {
				Schar[i] = S.charAt(i);
				Tchar[i] = T.charAt(i);
			}
			int count = 0;

			for(int i = 0; i < Schar.length; i++) {
				if(Schar[i]!=Tchar[i] ){
					Tchar[i] = Schar[i];
					count++;
				}
			}

			System.out.print(count);

		}

//	    public static int[] shuffle(int[] nums, int n) {
//
//
//	    	int[] xAray = new int[nums.length];
//	    	int[] yAray = new int[nums.length];
//	    	int[] xyAray = new int[nums.length];
//
//	    	for(int i = 0;i<n;i++) {
//
//	    		xAray[i] = nums[i];
//
//	    	}
//
//	    	for(int i = n;i<nums.length;i++) {
//	    		yAray[i] = nums[i];
//	    	}
//
//	    	for(int i = 0; i<n;i++) {
//	    		if(i == 0) {
//	    			xyAray[i]=xAray[i];
//	    		}if(i %2 == 0 ) {
//	    			xyAray[i] = xAray[i];
//	    		}else if(i % 2 != 0){
//	    			xyAray[i] = yAray[i+n];
//	    		}
//	    	}
//
//	    	return xyAray;
//
//
//	    }
}