Deque-STL C++


Problem Statement :


Double ended queue or Deque(part of C++ STL) are sequence containers with dynamic sizes that can be expanded or contracted on both ends (either its front or its back). The member functions of deque that are mainly used are:

    Deque Template:

    std::deque<value_type>

    Declaration:

    deque<int> mydeque; //Creates a double ended queue of deque of int type

    Size

    int length = mydeque.size(); //Gives the size of the deque

    Push

    mydeque.push_back(1); //Pushes element at the end
    mydeque.push_front(2); //Pushes element at the beginning

    Pop

    mydeque.pop_back(); //Pops element from the end
    mydeque.pop_front(); //Pops element from the beginning

    Empty

    mydeque.empty() //Returns a boolean value which tells whether the deque is empty or not

To know more about deque, click here

Given a set of arrays of size N and an integer K, you have to find the maximum integer for each and every contiguous subarray of size K for each of the given arrays.

Input Format

First line of input will contain the number of test cases T. For each test case, you will be given the size of array N and the size of subarray to be used K. This will be followed by the elements of the array Ai.

Constraints

  1 <=  T  <= 1000
  1 <=  N  <= 10000
  1 <=  K  <= N
  1 <=  Ai  <= 10000  , where Ai is ith  the element in the array  A.

Output Format

For each of the contiguous subarrays of size K of each array, you have to print the maximum integer.


Solution :



title-img 


                            Solution in C :

#include<bits/stdc++.h>
using namespace std;
int main(){

  int t;
  cin >> t;
  while(t--){
       int n , k , x ;
       cin >> n >> k ;

       vector<int> vec;
       for(int i =  0 ; i < n ; ++i ){
          cin >> x;
          vec.push_back(x);
       }
        int mx = 0;
        vector<int> nval;
       for(int i =  0 ; i < n ; ++i ){

                  if(i%k == 0)
                   mx = vec[i];
                else
                mx = max( mx , vec[i] );
                nval.push_back(mx);
       }
       mx = 0;
       vector<int> val;
       for(int i = n-1 ; i >= 0 ; --i ){
             if(i%k == (k-1))
               mx = vec[i];
               else
               mx = max( mx , vec[i]);
               val.push_back(mx);

       }
       for(int i = 0 ; i < vec.size()-k+1  ; ++i )
             cout << max( nval[i+k-1] , val[val.size()-i-1]) <<" ";
             cout << endl;
             }
             return 0;

}








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