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Copy pathJobBounties(DP).cpp
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Copy pathJobBounties(DP).cpp
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112 lines (95 loc) · 3.25 KB
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#include "algorithm"
#include "iostream"
#include "numeric"
#include "iomanip"
#include "cstring"
#include "math.h"
#include "bitset"
#include "string"
#include "vector"
#include "ctime"
#include "queue"
#include "stack"
#include "map"
#include "set"
#include "ext/pb_ds/assoc_container.hpp" // Common file
#include "ext/pb_ds/tree_policy.hpp" // Including tree_order_statistics_node_update
#include "ext/pb_ds/detail/standard_policies.hpp"
using namespace std;
using namespace __gnu_pbds;
#define f first
#define lgn 25
#define endl '\n'
#define sc second
#define pb push_back
#define N (int)2e5+5
#define PI acos(-1.0)
#define int long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define eb emplace_back
#define mii map<int,int>
#define vpii vector<pii>
#define pii pair<int,int>
#define pq priority_queue
#define BLOCK (int)sqrt(N)
#define test(x) while(x--)
#define all(x) begin(x),end(x)
#define allr(x) rbegin(x),rend(x)
#define fo(i,a,n) for(int i=a;i<n;i++)
#define rfo(i,n,a) for(int i=n;i>=a;i--)
#define FAST ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define time() cerr << "Time : " << (double)clock() / (double)CLOCKS_PER_SEC << "s\n"
#define bug(...) __f (#__VA_ARGS__, __VA_ARGS__)
typedef tree< int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update >
OS ;
template <typename Arg1>
void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f (const char* names, Arg1&& arg1, Args&&... args)
{
const char* comma = strchr (names + 1, ',');
cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
}
const int inf = 0x3f3f3f3f;
const int INF = 0x3f3f3f3f3f3f3f3f;
int n,m,k,q;
string s;
int dp[N];
/*
dp[i] -> denotes maximum lenght valid string upto ith index
Case 1 - if we encounter this "()......." tthen dp[i] = 2 + dp[i-2];
Case 2 - if we encounter this "......))" then we have to check for a '(' previously encounter but not in maximum lenght till (i-1)
so we check at ( i - dp[i-1] - 1) -> here dp[i-1] is maximum lenght valid string till (i-1)th index , so we have to check
at just previous of its starting that's why ( dp[i-1] - 1) from ith index
, so if we encounter '(' then we can extend dp[i-1] answer by 2 and also add then maximum lenght valid string just previous of encountered '('
i.e dp[ i - dp[i-1] - 2].
*/
void go()
{
cin >> s;
n = s.size();
s = " " + s;
int ans = 0;
fo(i,2,n+1)
{
if( s[i-1] == '(' and s[i] == ')' )
{
dp[i] = dp[i-2] + 2;
}
if( s[i] == ')' and s[i-1] == ')' )
{
if( s[ i - dp[i-1] - 1] == '(' ) dp[i] = dp[i-1] + 2 + dp[i - dp[i-1] - 2];
}
ans = max( ans , dp[i] );
}
cout << ans << endl;
}
int32_t main()
{
FAST;
int t=1;
// cin>>t;
test(t) go();
}