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>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 

using namespace std;
using namespace __gnu_pbds;

#define pbds tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 
#define ll long long 
#define inf 0x3f3f3f3f
#define infll 0x3f3f3f3f3f3f3f3f
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define ii pair<int,int>
#define vii vector<ii>
#define vi vector<int>

#define F_OR(i, a, b, s) for (int i=(a); (s)>0?i<(b):i>(b); i+=(s))
#define F_OR1(e) F_OR(i, 0, e, 1)
#define F_OR2(i, e) F_OR(i, 0, e, 1)
#define F_OR3(i, b, e) F_OR(i, b, e, 1)
#define F_OR4(i, b, e, s) F_OR(i, b, e, s)
#define GET5(a, b, c, d, e, ...) e
#define F_ORC(...) GET5(__VA_ARGS__, F_OR4, F_OR3, F_OR2, F_OR1)
#define FOR(...) F_ORC(__VA_ARGS__)(__VA_ARGS__)
#define EACH(x, a) for (auto& x: a)
 
// debug templates 
#define debug(x)       cerr<< #x <<" : "<< x << endl;
#define debuga(A,N)    cerr<< #A <<" : [";for(int i = 0; i<N;i++) cerr<<A[i]<<" "; cerr<<"]\n";
#define debuga2(A,N,M) cerr<< #A << " : \n"; for(int i=0;i<N;i++){cerr<<"[";for(int j=0;j<M;++j) cerr<<A[i][j]<<" ";cerr<<"]\n";}
#define debugp(p)      cerr<< #p <<" : "<<"("<<(p).first<<","<<(p).second<<")\n";
#define debugv(v)      cerr<< #v <<" : "<<"[";for(int i = 0; i< (v).size(); i++) cerr<<v[i]<<" "; cerr<<"]\n";
#define debugv2(v)     cerr<< #v << " : \n"; for(int i=0;i<v.size();i++){cerr<<"[";for(int j=0;j<(v[0].size());++j) cerr<<v[i][j]<<" ";cerr<<"]\n";}
#define debugs(m)      cerr<< #m <<" : [ "; for(auto itr = m.begin(); itr!=m.end();itr++) cerr<<*itr<<" "; cerr<<"]\n";
#define debugm(m)      cerr<< #m <<" : [ "; for(auto itr = m.begin();itr!=m.end(); itr++) cerr<<"("<<itr->first<<","<<itr->second<<") ";cerr<<"]\n";
 

void solve(){
    string s,t; cin >> s >> t;
    int ans = inf;
    FOR(i,s.size()-t.size()+1){
        int temp_ans = 0;
        FOR(j,t.size()){
            if(t[j] != s[i+j]) ++temp_ans;
        }
        ans = min(ans,temp_ans);
    }
    cout << ans;
}
 
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	solve();
	return 0;
}