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> 
    #include<ext/pb_ds/assoc_container.hpp> 
    #include<ext/pb_ds/tree_policy.hpp>
    using namespace __gnu_pbds;
    using namespace std;
    template <typename T>
    using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
    #define int long long  
    typedef long double ld;
    #define pb push_back
    #define f(i,a,b) for(int i=a;i<b;i++)
    #define fd(i,a,b) for(int i=a-1;i>=b;i--)
    #define pf push_front
    #define fi first
    #define se second
    #define INF 1e10
    const double pi=3.14159265358979323;
    const int mod=1e9+7;
    const int MAX=1e7+5;
    int gcd(int a, int b) 
    { 
    if (b == 0) 
        return a; 
    return gcd(b, a % b);  
      
    } 
    void swap(int a,int b)
    {
      int temp=a;
      a=b;
      b=temp;
    } 
    int  binpow(int  a,int  b) 
    {
    int res = 1;
    while (b > 0) {
        if (b%2 == 1)
            res = (res * a)%mod;
        a = (a* a)% mod;
        b >>= 1;
    }
    return res;
}
    int modinv(int y)
    {
        return binpow(y,mod-2);
    }

    int fact[500000+5];
    void precal()
    {
    fact[0]=1;
    for(int i=1;i<500000+5;i++)
    {
        fact[i]=(i*fact[i-1])%mod;
    }
}   
    bool isPrime(int n) 
    { 
    // Corner cases 
    if (n <= 1) 
        return false; 
    if (n <= 3) 
        return true; 
   
    if (n % 2 == 0 || n % 3 == 0) 
        return false; 
  
    for (int i = 5; i * i <= n; i = i + 6) 
        if (n % i == 0 || n % (i + 2) == 0) 
            return false; 
  
    return true; 
    } 
    
    signed main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);  
    #ifndef ONLINE_JUDGE
     if(fopen("INPUT.txt","r"))
      {
      freopen ("INPUT.txt" , "r" , stdin);
      freopen ("OUTPUT.txt" , "w" , stdout); 
      }
    #endif  
    // -------------------------------------Code starts here--------------------------------------------------------------------- 
    int t=1; 
    //cin>>t;
    while(t--)
    {   
        string a,b;
        cin>>a>>b;
        int n=a.length();
        int cnt=0;
        f(i,0,n)
        {
            if(a[i]!=b[i])
                cnt++;
        }
        cout<<cnt<<endl;
    }
}