## Editorial for Bedao Mini Contest 01 - FCHAR

Remember to use this editorial only when stuck, and not to copy-paste code from it. Please be respectful to the problem author and editorialist.
Submitting an official solution before solving the problem yourself is a bannable offence.

#### Code mẫu

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define sz(a) (int)a.size()

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair <int, int> ii;

int p[26][100007];

int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0), cout.tie(0);

string s;
int m;
cin >> s >> m;
int n = sz(s);
s = " " + s;
for (int i = 1; i <= n; ++i)
for (int j = 0; j < 26; ++j)
p[j][i] = p[j][i - 1] + (s[i] - 'a' == j);

int ans = n;
for (int i = 1; i <= n; ++i)
{
int l = i, r = n, cur = INT_MAX;
while (l <= r)
{
int mid = (l + r) >> 1;
if (p[0][mid] - p[0][i - 1] >= m && p[1][mid] - p[1][i - 1] >= m
&& p[2][mid] - p[2][i - 1] >= m && p[3][mid] - p[3][i - 1] >= m
&& p[4][mid] - p[4][i - 1] >= m) cur = mid, r = mid - 1;
else l = mid + 1;
}
ans = min(ans, cur - i + 1);
}

if (p[0][n] < m || p[1][n] < m || p[2][n] < m || p[3][n] < m || p[4][n] < m) ans = -1;
cout << ans;
return 0;
}