[Leetcode] Problem 1438 - Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit

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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

No.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.

No.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.

No.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

Code

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public int longestSubarray(int[] nums, int limit) {
    int result = 0;
    int start = 0;
    int end = 0;
    Deque<Integer> max = new LinkedList<>();
    Deque<Integer> min = new LinkedList<>();

    while (end < nums.length) {
        while (!max.isEmpty() && nums[end] > max.peekLast())
            max.pollLast();
        while (!min.isEmpty() && nums[end] < min.peekLast())
            min.pollLast();

        max.offerLast(nums[end]);
        min.offerLast(nums[end]);

        while (max.peekFirst() - min.peekFirst() > limit) {
            if (nums[start] == max.peekFirst())
                max.pollFirst();

            if (nums[start] == min.peekFirst())
                min.pollFirst();
            
            start++;
        }

        result = Math.max(result, end - start + 1);
        end++;
    }

    return result;
}