Merge pull request codeharborhub#1995 from Saipradyumnagoud/main

Added 1438-longest-continuous-subarray-with-absolute-diff-less-than-o…

Author: ajay-dhangar

dsa-solutions/lc-solutions/1400-1499/1438-longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit.md

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+---
+id: longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit
+title: Longest Continuous Subarray with Absolute Diff Less Than or Equal to Limit
+level: hard
+sidebar_label: Longest Continuous Subarray with Absolute Diff Less Than or Equal to Limit
+tags:
+  - Array
+  - Sliding Window
+  - Deque
+  - Java
+description: "This document provides solutions for the Longest Continuous Subarray with Absolute Diff Less Than or Equal to Limit problem."
+---
+
+## Problem Statement
+
+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`
+
+## Solutions
+
+### Approach
+
+To determine the length of the longest subarray where the absolute difference between any two elements is less than or equal to `limit`, follow these steps:
+
+1. **Sliding Window with Deque:**
+   - Use two deques to maintain the maximum and minimum values in the current window.
+   - Slide the window across the array and adjust the window size to ensure the absolute difference condition is met.
+
+### Java 
+
+```java
+
+
+class Solution {
+    public int longestSubarray(int[] nums, int limit) {
+        LinkedList<Integer> increase = new LinkedList<>();
+        LinkedList<Integer> decrease = new LinkedList<>();
+
+        int max = 0;
+        int left = 0;
+
+        for (int i = 0; i < nums.length; i++) {
+            int n = nums[i];
+
+            while (!increase.isEmpty() && n < increase.getLast()) {
+                increase.removeLast();
+            }
+            increase.add(n);
+
+            while (!decrease.isEmpty() && n > decrease.getLast()) {
+                decrease.removeLast();
+            }
+            decrease.add(n);
+
+            while (decrease.getFirst() - increase.getFirst() > limit) {
+                if (nums[left] == decrease.getFirst()) {
+                    decrease.removeFirst();
+                }
+                if (nums[left] == increase.getFirst()) {
+                    increase.removeFirst();
+                }
+                left++;
+            }
+
+            int size = i - left + 1;
+            max = Math.max(max, size);
+        }
+
+        return max;
+    }
+}
+```
+### Python 
+```Python 
+class Solution:
+    def longestSubarray(self, nums: List[int], limit: int) -> int:
+        increase = deque()
+        decrease = deque()
+        max_length = 0
+        left = 0
+
+        for i in range(len(nums)):
+            n = nums[i]
+
+            while increase and n < increase[-1]:
+                increase.pop()
+            increase.append(n)
+
+            while decrease and n > decrease[-1]:
+                decrease.pop()
+            decrease.append(n)
+
+            while decrease[0] - increase[0] > limit:
+                if nums[left] == decrease[0]:
+                    decrease.popleft()
+                if nums[left] == increase[0]:
+                    increase.popleft()
+                left += 1
+
+            max_length = max(max_length, i - left + 1)
+
+        return max_length
+```