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

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Description⚓︎

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:
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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

Solution⚓︎

Way 1⚓︎

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class Solution {
public:
    int longestSubarray(vector<int>& nums, int limit) {
        multiset<int> window;
        int res = 0;

        for (int left = 0, right = 0; right < nums.size(); right++) {
            window.insert(nums[right]);

            while (*window.rbegin() - *window.begin() > limit) {
                window.erase(window.find(nums[left]));
                left++;
            }

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

        return res;
    }
};
  • Time complexity: \(O(n \log n)\);
  • Space complexity: \(O(n)\).

Way 2⚓︎

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class Solution {
public:
    int longestSubarray(vector<int>& nums, int limit) {
        deque<int> maxQueue, minQueue;
        int res = 0;

        for (int left = 0, right = 0; right < nums.size(); right++) {
            while (!maxQueue.empty() && nums[maxQueue.back()] < nums[right]) {
                maxQueue.pop_back();
            }
            while (!minQueue.empty() && nums[minQueue.back()] > nums[right]) {
                minQueue.pop_back();
            }

            maxQueue.push_back(right);
            minQueue.push_back(right);

            while (!maxQueue.empty() && !minQueue.empty() 
                    && nums[maxQueue.front()] - nums[minQueue.front()] > limit) {
                if (nums[left] == nums[maxQueue.front()]) {
                    maxQueue.pop_front();
                } else if (nums[left] == nums[minQueue.front()]) {
                    minQueue.pop_front();
                }
                left++;
            }

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

        return res;
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(n)\).

Another way of writing:

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class Solution {
public:
    int longestSubarray(vector<int>& nums, int limit) {
        deque<int> maxQueue, minQueue;
        int res = 0;

        for (int left = 0, right = 0; right < nums.size(); right++) {
            while (!maxQueue.empty() && nums[maxQueue.back()] <= nums[right]) {
                maxQueue.pop_back();
            }
            while (!minQueue.empty() && nums[minQueue.back()] >= nums[right]) {
                minQueue.pop_back();
            }

            maxQueue.push_back(right);
            minQueue.push_back(right);

            while (!maxQueue.empty() && !minQueue.empty() 
                    && nums[maxQueue.front()] - nums[minQueue.front()] > limit) {
                left++;
                if (maxQueue.front() < left) {
                    maxQueue.pop_front();
                }
                if (minQueue.front() < left) {
                    minQueue.pop_front();
                }
            }

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

        return res;
    }
};