ACM CodeFest 2.0 - Play with strings Editorial

Author: return1201

EASY

PROBLEM:

Play with strings | Codechef

EXPLANATION:

There are at most 6 different ways to assign the values A, B, C to numeric 0,1,2. Each of these assignments will result in the formation of a different Sequence.

After we have formed a sequence we need to find the Maximum sum contiguous sequence of the formed sequence, which can be found using Kadane’s Algorithm.

Time complexity: O(n)

SOLUTIONS:

Setter's Solution
``````#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define pb push_back
#define pf push_front
#define all(t) t.begin(), t.end()
#define inrange(i, a, b) (((i)>= min((a), (b))) && ((i) <= max((a), (b))))
typedef vector<ll> vi;
typedef pair<ll, ll> pii;
#define fi first
#define se second
#define upb upper_bound
#define lwb lower_bound
#define show(x) cout << #x << " is " << x << "\n";
const ll inf = 2e18;
const ll mod = 1e9 + 7;
const ld pi = 3.141592653589793238462643383279502884;

void print(ll a[], ll n){for(ll i=0;i<n;i++){cout<<a[i]<<" ";}cout<<"\n";}
ll power(ll x, ll y){
if(y<=0) return 1;
ll ans = 1;
x %= mod;
while(y){
if(y&1)
ans = (x * ans) % mod;
x = (x * x) % mod;
y >>= 1;
}
return ans;
}
ll modInverse(ll n) {return power(n, mod-2);}
inline ll mul(ll a, ll b){ return (a * b) % mod; }
inline ll sub(ll a, ll b){ ll c = (a - b); if(c < 0) c += mod; return c; }
inline ll add(ll a, ll b){ ll c = (a + b); if(c >= mod) c -= mod; return c; }
inline ll divi(ll a, ll b){ return mul(a, modInverse(b)); }

//------------------------------------------------------------------------------------------

//const ll N = 2e5 + 1;
//bool visited[N];

ll ma = -inf;
ll sum = 0;

for(int i=0;i<n;i++){
sum = sum + a[i];
ma = max(ma, sum);
sum = max(0ll, sum);
}

return ma;
}

int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif

ios_base::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout<<fixed<<setprecision(10);

ll n;
cin>>n;
ll a[3];
int numeric[3];
for(int i=0;i<3;i++){
cin>>a[i];
numeric[i] = i;
}
string s;
cin>>s;
ll ans=0;
do{
ll b[n];
for(int i=0;i<n;i++){
int ch = s[i]-'a';
b[i] = a[numeric[ch%3]];
}