Problem:
Score : 200 points

Problem Statement

Given are two strings S and T.

Let us change some of the characters in S so that T will be a substring of S.

At least how many characters do we need to change?

Here, a substring is a consecutive subsequence. For example, xxx is a substring of yxxxy, but not a substring of xxyxx.

Constraints

The lengths of S and T are each at least 1 and at most 1000.

The length of T is at most that of S.

S and T consist of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S
T

Output

Print the minimum number of characters in S that need to be changed.

Sample Input 1

cabacc
abc

Sample Output 1

1

For example, changing the fourth character a in S to c will match the second through fourth characters in S to T.

Since S itself does not have T as its substring, this number of changes - one - is the minimum needed.

Sample Input 2

codeforces
atcoder

Sample Output 2

6

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

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fr first
#define sc second
#define clr(a) memset(a, 0, sizeof(a))
#define sz(x) x.size()
#define printarray(arr) for (auto it = arr.begin(); it != arr.end(); ++it) cout<<*it<<" "; cout<<endl;
#define rep(n) for (ll i = 0; i < n; i++)
#define repc(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, x, y) for (int i = x; i < y; i++)
#define DEC(i, x, y) for (int i = x; i >= y; i--)
#define all(v) v.begin(), v.end()
#define min3(a, b, c) min(a, min(b, c))
#define max3(a, b, c) max(a, max(b, c))
#define alla(a, n) a, a + n
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) (a * b)/gcd(a, b)
#define int long long int
#define ull unsigned long long

using namespace std;
const int MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;

int fastexpo(int b, int exp){
    if(exp == 0) return 1;
    if(exp == 1) return b;
    int ans = (fastexpo(b,exp/2) % MOD);
    ans *= ans;
    ans %= MOD;
    if(exp % 2 == 1){
        ans *= b;
    }
    ans %= MOD;
    return ans;
}

// int a[n];
// int n, k;

void reduceFraction(int x, int y) { 
    int d; 
    d = __gcd(x, y); 
  
    x = x / d; 
    y = y / d; 
  
    cout << "x = " << x << ", y = " << y << endl; 
} 


int finddiff(string a, string b){
	int cnt = 0;
	for(int i= 0;i<a.size();i++){
		cnt += (a[i]!=b[i]);
	}
	return cnt;
 
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    string s, t;
    cin>>s>>t;
    int ans = INT_MAX;
    int n= t.size();
    for(int i= 0;i<=s.size()-n;i++){
        string s1 = s.substr(i, n);
        ans = min(ans, finddiff(s1, t));
    }
    cout<<ans<<endl;
    return 0;
}