LeetCode-in-All
239. Sliding Window Maximum
Hard
You are given an array of integers nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.
Return the max sliding window.
Example 1:
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [3,3,5,5,6,7]
Explanation:
Window position Max
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Example 2:
Input: nums = [1], k = 1
Output: [1]
Constraints:
1 <= nums.length <= 105-104 <= nums[i] <= 1041 <= k <= nums.length
Solution
-spec max_sliding_window(Nums :: [integer()], K :: integer()) -> [integer()].
max_sliding_window(Nums, K) ->
{Results, _} = lists:mapfoldl(
fun({X, I}, Q) ->
Q2 = enqueue(Q, X, I),
get_max(Q2, I - K)
end,
queue:new(),
lists:zip(Nums, lists:seq(0, length(Nums)-1))
),
lists:nthtail(K-1, Results).
-spec enqueue(queue:queue(), integer(), integer()) -> queue:queue().
enqueue(Q, X, I) ->
case queue:peek_r(Q) of
{value, {Y, _}} when Y =< X ->
enqueue(queue:drop_r(Q), X, I);
_ ->
queue:in({X, I}, Q)
end.
-spec get_max(queue:queue(), integer()) -> {integer(), queue:queue()}.
get_max(Q, DropUntil) ->
case queue:peek(Q) of
{value, {_, J}} when J =< DropUntil ->
get_max(queue:drop(Q), DropUntil);
{value, {Y, _}} ->
{Y, Q}
end.