# 1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit (M)

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
```

&#x20;

**Constraints:**

* `1 <= nums.length <= 105`
* `1 <= nums[i] <= 109`
* `0 <= limit <= 109`

### Solution:

<https://www.geeksforgeeks.org/longest-subarray-in-which-absolute-difference-between-any-two-element-is-not-greater-than-x/>\
TreeMap/PriorityQueue + Sliding windows

```
class Solution {
    public int longestSubarray(int[] nums, int limit) {
        
        PriorityQueue<Integer> increasing = new PriorityQueue<Integer>();
        //PriorityQueue<Integer> decreasing = new PriorityQueue<Integer>(( a, b) -> {return b-a;});
        PriorityQueue<Integer> decreasing = new PriorityQueue<Integer>(Comparator.reverseOrder());
        int maxLen = 0;
        int start = 0;
        int end = 0;
        while(start<=end && end < nums.length)
        {
            increasing.offer(nums[end]);
            decreasing.offer(nums[end]);
            
            int min = increasing.peek();
            int max = decreasing.peek();
            if(max-min <= limit)
            {
                maxLen = Math.max(maxLen, end-start+1);
                end++;
            }
            else
            {
                increasing.remove(nums[start]);
                decreasing.remove(nums[start]);
                start++;
                end++;
            }
        }
        return maxLen;
    }
}
```


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