Maximum element in every window of size

#1


give wrong answer

#2

Hey @khemchandrs

You need to replace
while(!(dq.empty()) && A[i]>=dq.back())
with
while(!(dq.empty()) && A[i]>=A[dq.back()])
because dq.back() denotes the index of element at the back(we’re storing indexes in the queue), and we need to compare with element present at that index i. e. A[dq.back()]
Here’s your altered code which should work:

#include
using namespace std;
#include

void maxKthArray(int *A,int n,int k){

 deque<int> dq;
 int i;
 for(i=0;i<k;i++){
         while( (!dq.empty()) && A[i]>=A[dq.back()]){
            dq.pop_back();
         }
         dq.push_back(i);
 }

 for(;i<n;i++){
     cout<<A[dq.front()]<<" ";
     while(!(dq.empty()) && dq.front()<=(i-k) ){
        dq.pop_front();
     }

     while(!(dq.empty()) && A[i]>=A[dq.back()]){
        dq.pop_back();
     }
     dq.push_back(i);
 }

 cout<<A[dq.front()];

}
int main(){
int A[]={3,5,2,2,7};
maxKthArray(A,5,3);

}

#3

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