THIS code is not getting submitted but it is correct for most of the test cases so please tell me where I am wrong

#1

static ArrayList max_of_subarrays(int arr[], int n, int k)
{
// Your code here
Queue pq = new LinkedList<>();
ArrayList ar = new ArrayList();
int i = 0;
int j = 0;
int max = -100000;
while(j < arr.length){
if(arr[j] > max){
while(!pq.isEmpty()){
pq.remove();
}
pq.add(arr[j]);
max = arr[j];

        }
        else
        pq.add(arr[j]);
        
        if(j == k+i-1){
            ar.add(pq.peek());
            if(arr[i] == pq.peek()){
            pq.remove();
            }
            i++;
            
        }
        j++;
        
                
    }
    return ar;

}

#2

Link of my code --> https://ide.codingblocks.com/s/609883

#3

Question Link–> https://practice.geeksforgeeks.org/problems/maximum-of-all-subarrays-of-size-k3101/1#

#4

Hey @ayush_r18,

This logic used for adding the number as window maximum is incorrect as it isn’t giving the maximum element at the peek position.

Your code fails for this testcase.
10 3
1 3 1 2 1 4 5 2 3 6

The output obtained is : 3 3 1 4 5 5 5 6
which is incorrect as it should have been 2 instead of 1 but 1 lies as the peek element, thus you are getting the wrong answer.

The expected output is: 3 3 2 4 5 5 5 6

You can refer to the theory video to understand the correct approach.
Hope it helps.

1 Like
#5

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