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/*
    What we should do next is to shrink left pointer because of limit. 
    If current.max - current.min > limit.
    we need to dequeue the "tail" element who is smaller than the nums[right]. 
    Since, those "old small tail" elements will never be the range maximum from now on. 
    After "clean up" the "old small tail" elements, add nums[right] into the deque, 
    and then, the head of deque is the current maximum.
    
    Time O(N)
    Space O(N)
*/

class Solution {
    public int longestSubarray(int[] A, int limit) {
        Deque<Integer> maxd = new ArrayDeque<>();
        Deque<Integer> mind = new ArrayDeque<>();
        int i = 0, res = 0;
        for (int j = 0; j < A.length; ++j) {
            while (!maxd.isEmpty() && A[j] > maxd.peekLast())   maxd.pollLast();
            maxd.add(A[j]);
            while (!mind.isEmpty() && A[j] < mind.peekLast())   mind.pollLast();
            mind.add(A[j]);
            if (maxd.peek() - mind.peek() > limit) {
                if (maxd.peek() == A[i]) maxd.poll();
                if (mind.peek() == A[i]) mind.poll();
                i++;
            }
            res = Math.max(res, j - i + 1);
        }
        return res;
    }
}