SPOJ ARRAYSUB – subarrays

Problem Link : http://www.spoj.com/problems/ARRAYSUB

Solution Idea:

This is a RMQ (Range Max/Min Query) problem. This types of problem can be solved in many ways like segment tree, sliding window etc. But in this problem time limit is very tight and for this reason we have to use Sliding window technique which complexity is O(n) for solving this problem.



/*
         +-+ +-+ +-+
         |R| |.| |S|
         +-+ +-+ +-+
 */

#include <bits/stdc++.h>

#define pii             pair <int,int>
#define sc              scanf
#define pf              printf
#define Pi              2*acos(0.0)
#define ms(a,b)         memset(a, b, sizeof(a))
#define pb(a)           push_back(a)
#define MP              make_pair
#define db              double
#define ll              long long
#define EPS             10E-10
#define ff              first
#define ss              second
#define sqr(x)          (x)*(x)
#define D(x)            cout<<#x " = "<<(x)<<endl
#define VI              vector <int>
#define DBG             pf("Hi\n")
#define MOD             100007
#define MAX             10000
#define CIN             ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)           (int)a.size()
#define sf(a)           scanf("%d",&a)
#define sfl(a)          scanf("%lld",&a)
#define sff(a,b)        scanf("%d %d",&a,&b)
#define sffl(a,b)       scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)     scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)    scanf("%lld %lld %lld",&a,&b,&c)
#define loop(i,n)       for(int i=0;i<n;i++)
#define REP(i,a,b)      for(int i=a;i<b;i++)
#define TEST_CASE(t)    for(int z=1;z<=t;z++)
#define PRINT_CASE      printf("Case %d: ",z)
#define all(a)          a.begin(),a.end()
#define intlim          2147483648
#define inf             1000000
#define ull             unsigned long long

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

int ara[1000005];
deque<int> big,small;

int main()
{
    ///freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    CIN;
    int n,d;
    cin>>n;
    loop(i,n) cin>>ara[i];
    cin>>d;

    if(d==1)
    {
        cout<<ara[0];
        REP(i,1,n) cout<<" "<<ara[i];
        cout<<endl;
    }

    else
    {

        big.pb(0);
        int ans=0;
        bool test=0;
        REP(i,1,n)
        {
            while(!big.empty() && ara[big.back()]<=ara[i])
                big.pop_back();
            big.pb(i);
            while(i-big.front()>=d)
                big.pop_front();

            if(i>=d-1)
            {
                if(test)
                    cout<<" "<<ara[big.front()];
                else
                    cout<<ara[big.front()];
                test=1;

            }
        }

        cout<<endl;
    }
    return 0;
}


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