Hackerrank Deque-STL Solution
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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 typeSize
int length = mydeque.size(); //Gives the size of the dequePush
mydeque.push_back(1); //Pushes element at the end
mydeque.push_front(2); //Pushes element at the beginningPop
mydeque.pop_back(); //Pops element from the end
mydeque.pop_front(); //Pops element from the beginningEmpty
mydeque.empty() //Returns a boolean value which tells whether the deque is empty or notTo know more about deque, click here
Given a set of arrays of size and an integer , you have to find the maximum integer for each and every contiguous subarray of size 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
, where is the element in the array .
Output Format
For each of the contiguous subarrays of size of each array, you have to print the maximum integer.
Sample Input
2
5 2
3 4 6 3 4
7 4
3 4 5 8 1 4 10Sample Output
4 6 6 4
8 8 8 10Explanation
For the first case, the contiguous subarrays of size 2 are {3,4},{4,6},{6,3} and {3,4}. The 4 maximum elements of subarray of size 2 are: 4 6 6 4.
For the second case,the contiguous subarrays of size 4 are {3,4,5,8},{4,5,8,1},{5,8,1,4} and {8,1,4,10}. The 4 maximum element of subarray of size 4 are: 8 8 8 10.
Solution in cpp#
Approach 1.
#include <iostream>
#include <deque>
using namespace std;
void printKMax(int arr[], int n, int k){
deque<pair<int, int>> dq;
for(int a=0; a<n; a++){
while(!dq.empty() && a - dq.front().second >= k){
dq.pop_front();
}
while(!dq.empty() && dq.back().first <= arr[a]){
dq.pop_back();
}
dq.push_back({arr[a], a});
if(a >= k - 1){
cout << dq.front().first << " ";
}
}
cout << "\n";
}
int main(){
int t;
cin >> t;
while(t>0) {
int n,k;
cin >> n >> k;
int i;
int arr[n];
for(i=0;i<n;i++)
cin >> arr[i];
printKMax(arr, n, k);
t--;
}
return 0;
}Approach 2.
#include <iostream>
#include <deque>
using namespace std;
void printKMax(int arr[], int n, int k){
if (n == 0) {
return;
}
if (n == 1 && k == 1) {
cout << arr[1] << "\n";
}
deque<int> maxs;
for(int i = 0; i < n; i++) {
while(!maxs.empty() && arr[i] > arr[maxs.back()]) {
maxs.pop_back();
}
maxs.push_back(i);
while(!maxs.empty() && maxs.front() <= (i - k)) {
maxs.pop_front();
}
if (i >= k-1) {
cout << arr[maxs.front()] << " ";
}
}
cout << "\n";
}
int main(){
int t;
cin >> t;
while(t>0) {
int n,k;
cin >> n >> k;
int i;
int arr[n];
for(i=0;i<n;i++)
cin >> arr[i];
printKMax(arr, n, k);
t--;
}
return 0;
}Approach 3.
#include <iostream>
#include <deque>
#include <algorithm>
using namespace std;
void printKMax(int arr[], int n, int k){
//Write your code here.
deque<int> deq(k); int i;
for(i=0;i<k;i++)
{
while(!deq.empty()&&arr[i]>=arr[deq.back()])
{
deq.pop_back();
}
deq.push_back(i);
}
for(i=k;i<n;i++)
{
cout<<arr[deq.front()]<<" ";
while(!deq.empty() && deq.front()<=(i-k))
{
deq.pop_front();
}
while(!deq.empty()&&arr[i]>=arr[deq.back()])
{
deq.pop_back();
}
deq.push_back(i);
}
cout<<arr[deq.front()]<<endl;
}
int main(){
int t;
cin >> t;
while(t>0) {
int n,k;
cin >> n >> k;
int i;
int arr[n];
for(i=0;i<n;i++)
cin >> arr[i];
printKMax(arr, n, k);
// printf("\n");
t--;
}
return 0;
}
