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1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit

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 <= 105
• 1 <= nums[i] <= 109
• 0 <= limit <= 109

Solutions (Rust)

1. Solution

use std::collections::BinaryHeap;implSolution{pubfnlongest_subarray(nums:Vec<i32>,limit:i32) -> i32{letmut max_heap = BinaryHeap::new();letmut min_heap = BinaryHeap::new();letmut i = 0;letmut ret = 1;for j in0..nums.len(){ max_heap.push((nums[j], j)); min_heap.push((-nums[j], j));whileletSome(&(x, k)) = max_heap.peek(){if x - nums[j] > limit { max_heap.pop(); i = i.max(k + 1);}else{break;}}whileletSome(&(x, k)) = min_heap.peek(){if nums[j] + x > limit { min_heap.pop(); i = i.max(k + 1);}else{break;}} ret = ret.max(j - i + 1);} ret asi32}}