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.*;
/*
NOTES:
-more than 10 digits (10^10), use long
-prefix sum of 1's and -1's to control flow of highlighting a region
-log(a)/log(b) = log base b of a
-create a position array if trying to see if elements show up next to each other sequentially
 i.e. elements 1,2,3 are within 3 spots of each other in an array
      arr[scannedInt - 1] = i; (position of i at p[i] in original)
- use a freq array when counting instances of an element occurring
- increment the counter first before simulating anything turn-based (two player game)

 */

public class Main {
    static long mod = (long)(Math.pow(10, 9) + 7);
    public static String alpha = "zabcdefghijklmnopqrstuvwxy";
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t =1;
        while (t-- > 0) {
            int ans= 0;
            String a = scan.next();
            String b = scan.next();
            for(int i = 0; i<a.length(); i++){
                if(a.charAt(i) != b.charAt(i)) ans++;
            }
            System.out.println(ans);

        }
        scan.close();
    }

    static ArrayList factors(int n ){
        ArrayList<Integer> facts = new ArrayList<>();
        for(int i = 1; i<=(int)Math.sqrt(n); i++){
            if(n%i == 0){
                if( i == Math.sqrt(n)) facts.add(i);
                else{
                    facts.add(i);
                    facts.add(n/i);
                }
            }
        }
        Collections.sort(facts);
        return facts;
    }

    static boolean isPrime (int n){
        if(factors(n).size() ==2) return true;
        else return false;
    }





    static int GCD(int a, int b) {
        // Everything divides 0
        if (a == 0) return b;
        if (b == 0) return a;

        // base case
        if (a == b) return a;

        // a is greater
        if (a > b) return GCD(a-b, b);
        return GCD(a, b-a);
    }
    static int LCM(int a, int b){
        return (a*b)/(GCD(a,b));
    }
    static String rev(String s){
        char[] arr = s.toCharArray();
        for(int i = 0; i<(s.length()+1)/2; i++){
            char temp = arr[i];
            arr[i] = arr[s.length()-i-1];
            arr[s.length()-i-1] = temp;
        }
        String fin = new String(arr);
        return fin;
    }
    static long pow(long a, long N) {
        if (N == 0) return 1;
        else if (N == 1) return a;
        else {
            long R = pow(a,N/2);
            if (N % 2 == 0) {
                return R*R;
            }
            else {
                return R*R*a;
            }
        }
    }

    static long powMod(long a, long N) {
        if (N == 0) return 1;
        else if (N == 1) return a % mod;
        else {
            long R = powMod(a,N/2) % mod;
            R *= R;
            R %= mod;
            if (N % 2 == 1) {
                R *= a;
                R %= mod;
            }
            return R % mod;
        }
    }

}