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>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define all(x) (x).begin(),(x).end()
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[8]={1,0,-1,0,1,1,-1,-1};
const ll dx[8]={0,-1,0,1,1,-1,1,-1};
using Graph = vector<vector<ll>>;
double nCk(int n, int k) {
double res=1.0;
for(int i=0; i<n; i++){
res*=0.5;}
for(int i=0; i<k; i++){
res*=(double)(n-i);
res/=(double)(k-i);
}
return res;}
struct edge{ll to, cost;};
typedef pair<ll,ll> P;
struct graph{
  ll V;
  vector<vector<edge> > G;
  vector<ll> d;
 
  graph(ll n){
    init(n);
  }
 
  void init(ll n){
    V = n;
    G.resize(V);
    d.resize(V);
    rep(i,V){
      d[i] = INF;
    }
  }
 
  void add_edge(ll s, ll t, ll cost){
    edge e;
    e.to = t, e.cost = cost;
    G[s].push_back(e);
  }
 
  void dijkstra(ll s){
    rep(i,V){
      d[i] = INF;
    }
    d[s] = 0;
    priority_queue<P,vector<P>, greater<P> > que;
    que.push(P(0,s));
    while(!que.empty()){
      P p = que.top(); que.pop();
      ll v = p.second;
      if(d[v]<p.first) continue;
      for(auto e : G[v]){
        if(d[e.to]>d[v]+e.cost){
          d[e.to] = d[v]+e.cost;
          que.push(P(d[e.to],e.to));
        }
      }
    }
  }
};
class UnionFind
{
public:
  ll par[100005];
  ll depth[100005];
  ll nGroup[100005];
 
  UnionFind(ll n) {
    init(n);
  }
 
  void init(ll n) {
    for(ll i=0; i<n; i++) {
      par[i] = i;
      depth[i] = 0;
      nGroup[i] = 1;
    }
  }
 
  ll root(ll x) {
    if(par[x] == x) {
      return x;
    } else {
      return par[x] = root(par[x]);
    }
  }
 
  bool same(ll x, ll y) {
    return root(x) == root(y);
  }
 
  void unite(ll x, ll y) {
    x = root(x);
    y = root(y);
    if(x == y) return;
 
    if(depth[x] < depth[y]) {
      par[x] = y;
      nGroup[y] += nGroup[x];
      nGroup[x] = 0;
    } else {
      par[y] = x;
      nGroup[x] += nGroup[y];
      nGroup[y] = 0;
      if(depth[x] == depth[y])
        depth[x]++;
    }
  }
};
const ll MAX = 510000;
ll fac[MAX], finv[MAX], inv[MAX];
 
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
ll COM(ll n, ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
} 
 
vector<ll> depth;
vector<ll> subtree_size;
void dfs(const Graph &G, ll v, ll p, ll d) {
    depth[v] = d;
    for (auto nv : G[v]) {
        if (nv == p) continue; // nv が親 p だったらダメ
        dfs(G, nv, v, d+1);
    }
 
    // 帰りがけ時に、部分木サイズを求める
    subtree_size[v] = INF; // 自分自身
    vl q;
    for (auto c : G[v]) {
        if (c == p) continue;
        q.pb(subtree_size[c]);// 子のサイズを加える
    }
  ll qq = q.size();
  //cout << v << " " << qq << endl;
  if(qq==0) subtree_size[v]=0;
  else{
  sort(all(q));
    //rep(i,qq)cout << q[i];
    //cout << endl;
  ll caa = 0;
  rep(i,qq){
    caa = max(qq+q[i]-i,caa);}
    subtree_size[v]=caa;
  }
}
template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};
using mint = Fp<MOD>;

mint calc(long long N, long long K) {
    mint res = 1;
    for (long long n = 0; n < K; ++n) {
        res *= (N - n);
        res /= (n + 1);
    }
    return res;
}
ll dp[45][2][2];
int main() {
string s,t; cin>>s>>t;
  ll ls = s.size();
  ll lt = t.size();
  ll ans = 100000;
  for(ll i=0; i<=ls-lt; i++){
    ll ca = 0;
    for(ll j=0; j<lt; j++){
      if(t.at(j)!=s.at(i+j))ca++;}
    ans = min(ca,ans);
  }
  cout << ans << endl;}