Code with Jess - Leetcode #239 Sliding Window Maximum

Thanks for watching!
Please let me know what other problems do you want me to do :)

The link to the problem:

[abhilashsulibela]

Such a clear train of thought! Thanks for explaining. It was very helpful

[arif]

Here is my solution - class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {


if(nums == null || nums.length == 0 || k > nums.length){
return new int[0];
}

// create a max heap priority queue
PriorityQueue<Integer> pq = new PriorityQueue<Integer>(k, Collections.reverseOrder());

int[] result = new int[nums.length - k + 1];

int windowStart = 0;
for (int windowEnd = 0; windowEnd < nums.length; windowEnd++) {
// keep adding the elements from the array till it hit the window size of 'k'
pq.add(nums[windowEnd]);


// slide the window, we don't need to slide if we've not hit the required window size of 'k'
if (windowEnd >= k - 1) {
// Max heap will return maximum number as max value will be at head
result[windowStart] = pq.peek();
// remove the element at start index of the current window from the max heap before sliding to next window
pq.remove(nums[windowStart]);
// slide to the next window
windowStart++;
}
}


return result;
}
}

[hiimnormal]

Thank you! It's the only video that's helped me actually understand sliding window with deque

[Sohanbadaya]

Nice video, @5:45 you said you are keeping ascending order deque. I think you are keeping values in descending order in deque. so you can find max element at 0 index. Please correct me if I am wrong.

[weizhou]

Wow, Jess, this is amazing! Please make more videos like this!

[ahmadalk]

So all this time I was pronouncing "deQewww" in a wrong way. 😅😅

[PrathamMantri]

It was very clear, checked other videos with same logic, but yours is pretty neat.

[lugiadark]

Thank you so much, I was having a lot of trouble trying to understand this problem! You should solve more problems

[snaily]

If the window is larger than the array, wouldn't we just return the array with the largest element? I think instead the edge case should be if K < 1. Then there's no window, thus returning empty array.

[NikhilYadav-ylhf]

if we were to find the minimum out of the result array then what would be the optimal solution. I thought of using a minHeap instead of an array to store the integers as the window slides and then take to top from minHeap at the end. is there an optimal way to solve this?

[sumitghewade]

Thanks for the explanation.. It helped to at least wrap my head around the solution

[xueli]

Hi Jess, thanks for your videos. Are you planning to make more?

[rajathanand]

Why are we adding the indexes to the deque and not the actual array elements?

[siobhanahbois]

Thanks, great video. Also TIL deque is pronounced like deck.

[sarthakbhatt]

How did you get idea of deque? I was thinking more like max heap (O(1) lookup time for max element).

[pruthvirajk]

Graphs!Do something on how to begin with graphs

[shashankshekhar]

if( !dq.isEmpty() && dq.peekFirst() == i - k) {
dq.pollFirst();
}

what does it really do? Can you please explain. Thanks :)

[tankman]

Attractive and beautiful sound, must be native American ^_^

[headyshotta]

I don't understand line 20 here. if(!dq.isEmpty() && dq.peekFirst() == i - k).... peekFirst() returns the actual deque value, not its index, so why are we comparing it to the index minus the window size (i - k)? I see that it works but I am having trouble following the logic here.