1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit

Given an array of integers nums and an integer limit, return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal to limit .

Example 1:

Input: nums = [8,2,4,7], limit = 4
Output: 2
Explanation: All subarrays are:
[8] with maximum absolute diff |8-8| = 0 <= 4.
[8,2] with maximum absolute diff |8-2| = 6 > 4.
[8,2,4] with maximum absolute diff |8-2| = 6 > 4.
[8,2,4,7] with maximum absolute diff |8-2| = 6 > 4.
[2] with maximum absolute diff |2-2| = 0 <= 4.
[2,4] with maximum absolute diff |2-4| = 2 <= 4.
[2,4,7] with maximum absolute diff |2-7| = 5 > 4.
[4] with maximum absolute diff |4-4| = 0 <= 4.
[4,7] with maximum absolute diff |4-7| = 3 <= 4.
[7] with maximum absolute diff |7-7| = 0 <= 4.
Therefore, the size of the longest subarray is 2.

Example 2:

Input: nums = [10,1,2,4,7,2], limit = 5
Output: 4
Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.

Example 3:

Input: nums = [4,2,2,2,4,4,2,2], limit = 0
Output: 3

Constraints:

• 1 <= nums.length <= 10^5
• 1 <= nums[i] <= 10^9
• 0 <= limit <= 10^9

# @lc code=start

using LeetCode

function longest_subarray(nums::Vector{Int}, limit::Int)
    queMin, queMax = Deque{Int}(), Deque{Int}()
    n = length(nums)
    left = right = 1
    res = 0

    while right <= n
        while !isempty(queMin) && !isempty(queMax) && last(queMax) < nums[right]
            pop!(queMax)
        end
        while !isempty(queMin) && !isempty(queMax) && last(queMin) > nums[right]
            pop!(queMin)
        end
        push!(queMax, nums[right])
        push!(queMin, nums[right])

        while !isempty(queMin) && !isempty(queMax) && first(queMax) - first(queMin) > limit
            if nums[left] == first(queMin)
                popfirst!(queMin)
            end
            if nums[left] == first(queMax)
                popfirst!(queMax)
            end
            left += 1
        end

        res = max(res, right - left + 1)
        right += 1

    end

    return res
end
# @lc code=end

longest_subarray (generic function with 1 method)

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