PROBLEM LINK:
Author: sid_rasa
Tester: akay_99
Editorialist: sid_rasa
DIFFICULTY:
EASY
PREREQUISITES:
None
PROBLEM:
Given a previous score X, add 1 if the answer is correct and subtract only when the answer is wrong and the current score is greater than 0.
EXPLANATION:
If the current score is greater than -1 then you can add and subtract and else you can only add.
NOTE -
Be careful with integer overflows.
TIME COMPLEXITY
O(N) i.e, linear time as we need to iterate on the whole string.
SOLUTIONS:
Setter's Solution
#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define pb push_back
#define mp make_pair
#define fl(i, a, b) for (int i = a; i < b; i++)
#define vi vector<int>
#define e1(a) int a; cin>>a;
#define e2(a,b) int a,b; cin>>a>>b;
#define e3(a,b,c) int a,b,c; cin>>a>>b>>c;
#define av(arr,n) vector<int> arr(n); fl(i,0,n) cin>>arr[i];
#define aa(arr,n) int arr[n]; fl(i,0,n) cin>>arr[i];
#define es(s) string(s); cin >> (s);
#define rz resize
#define vvi vector<vector<int>>
#define sz(s) s.size()
#define mod 1000000007
#define ff first
#define ss second
#define tn -10e15
#define bg 10e15
#define all(x) (x).begin(), (x).end()
#define pii pair<int, int>
#define mii map<int,int>
#define st(a, n) memset(a, n, sizeof(a))
#define vl(n) cout << n << "\n"
#define vs(n) cout << n << " "
#define el cout<<"\n"
#define vn(n) cout << n
#define dsc greater<int>()
#define ps(x,y) fixed<<setprecision(y)<<x
#define R return
#define B break
#define C continue
#define YC cout<<"YES"<<"\n"
#define YS cout<<"Yes"<<"\n"
#define NC cout<<"NO"<<"\n"
#define NS cout<<"No"<<"\n"
#define lcm(a,b) (a/__gcd(a,b))*b
#define pa(a) for(auto e:a)cout<<e<<" "
const int N = 1e5 + 5;
#ifndef ONLINE_JUDGE
#define pra(a,n)cerr<<#a<<":";for(int i=0;i<n;i++)cerr<<a[i]<<" ";cerr<<endl;
#define prm(mat,row,col)cerr<<#mat<<":\n";for(int i=0;i<row;i++){for(int j=0;j<col;j++)cerr<<mat[i][j]<<" ";cerr<<endl;}
#define pr(...)dbs(#__VA_ARGS__,__VA_ARGS__)
template<class S, class T>ostream &operator<<(ostream &os, const pair<S, T> &p) {return os << "(" << p.first << "," << p.second << ")";}
template<class T>ostream &operator<<(ostream &os, const vector<T> &p) {os << "["; for (auto&it : p)os << it << " "; return os << "]";}
template<class T>ostream &operator<<(ostream &os, const set<T>&p) {os << "["; for (auto&it : p)os << it << " "; return os << "]";}
template<class T>ostream &operator<<(ostream &os, const multiset<T>&p) {os << "["; for (auto&it : p)os << it << " "; return os << "]";}
template<class S, class T>ostream &operator<<(ostream &os, const map<S, T>&p) {os << "["; for (auto&it : p)os << it << " "; return os << "]";}
template<class T>void dbs(string str, T t) {cerr << str << ":" << t << "\n";}
template<class T, class...S>void dbs(string str, T t, S... s) {int idx = str.find(','); cerr << str.substr(0, idx) << ":" << t << ","; dbs(str.substr(idx + 1), s...);}
#else
#define pr(...){}
#define pra(a,n){}
#define prm(mat,row,col){}
#endif
void solve()
{
e2(n, x);
es(s);
fl(i, 0, n) {
if (s[i] == '1') {
x++;
}
else {
if (x > 0)
x--;
}
}
vl(x);
}
int32_t main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
solve();
return 0;
}