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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define si(X) scanf("%d", &(X))
#define sll(X) scanf("%lld",&(X))
#define INFL 0x3f3f3f3f3f3f3f3fLL
const ll mod = 1e9+7;

//-------------------------<MATH>-------------------------------------------------------//
ll gcd(ll a,ll b){if(b==0)return a;return gcd(b,a%b);}
ll expo(ll base,ll pow){ll ans = 1;
   while(pow!=0){if(pow&1==1){ans = ans*base;ans = ans%mod;}
   base *= base;base%=mod;pow/=2;}return ans;}
ll inv(ll x){return expo(x,mod-2);}
ll mult_mod(ll a , ll b){a = a%mod; b = b%mod;
ll c = (a * b)%mod;
//cout << a << " " << b << endl;
return c;}
ll add_mod(ll a , ll b){a = (a + mod)%mod;b = (b + mod)%mod;ll c = (a + b)%mod;return c;}
ll fac[(int)(5e5 + 5)] , inv_fac[(int)(5e5 + 5)];
void precomp_factorial(int M){fac[0] = 1;for(int i = 1 ; i <= M ; i++) fac[i] = mult_mod(i , fac[i - 1]);
    inv_fac[M] = inv(fac[M]);for(int i = M - 1 ; i >= 0 ; i--) inv_fac[i] = mult_mod(i + 1 , inv_fac[i + 1]);}
ll nCk(ll n , ll k){ll ans = fac[n];ans = mult_mod(ans , inv_fac[n - k]);ans = mult_mod(ans , inv_fac[k]);return ans;}
int lo[(int)(1e3 + 6)];
void sieve(int M){for(int i = 2 ; i < M ; i++){ if(lo[i]) continue; for(int j = i ; j < M ; j += i)lo[j] = i;}}
int max_expo_prime_n_fac(ll N , int prime){int ress = 0;while(N){ress += (N/prime);N /= prime;}return ress;}
//-------------------------</MATH>-------------------------------------------------------//

//-------------------------<STRING>-------------------------------------------------------//
string get_prefix(string x , int MAX){return x.substr(0 , min((int)x.size() , MAX));}
string get_suffix(string x,int MAX){int dn=1;for(int i=x.size()-1;i>=0;i--){if(dn==MAX)return x.substr(i);dn++;}return x;}
string convert_to_binary(int x,int len) {string ans=""; while(x){if(x&1){ans='1'+ans;}
    else{ans ='0'+ans;}x/=2;}int k = ans.size();while(k < len){ans = '0' + ans;k++;}return ans;}
//-------------------------</STRING>-------------------------------------------------------//

ull llrand() {ull r = 0;for(int i = 0; i < 5; ++i){r = (r << 15)|(rand() & 0x7FFF);}return r & 0xFFFFFFFFFFFFFFFFULL;}
int dx[8] = {1 , 0 , -1 , 0 , 1 , -1 , -1 , 1};    // last 4 diagonal
int dy[8] = {0 , 1 , 0 , -1 , 1 , 1 , -1 , -1};
const int M = 60;

int arr[M][M];
bool vis[M][M];

int n, m;
int till = 0;

void reset(){
    till = 0;
    for(int i = 0 ; i <= n + 2 ; i++){
        for(int j = 0 ; j <= m + 2 ; j++){
            vis[i][j] = 0;
            arr[i][j] = 0;
        }
    }
}

bool inside(int x, int y){
    return (x >= 1 && x <= n && y >= 1 && y <= m);
}

void dfs(int x, int y){
    if(vis[x][y]) return;
    vis[x][y] = 1;
    if(arr[x][y] == 2) till++;
    for(int d = 0 ; d < 4 ; d++){
        int nx = x + dx[d];
        int ny = y + dy[d];
        if(inside(nx, ny) && (arr[nx][ny] == 0 || arr[nx][ny] == 2)){
            dfs(nx, ny);
        }
    }

}
int main(){

    string s, t;
    cin >> s >> t;
    int cnt = 0;
    for(int i = 0; i < s.size(); i++){
        if(s[i] != t[i]) cnt++;
    }
    cout << cnt;
}