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; }