README.md

1499. Max Value of Equation

You are given an array points containing the coordinates of points on a 2D plane, sorted by the x-values, where points[i] = [xi, yi] such that xi < xj for all 1 <= i < j <= points.length. You are also given an integer k.

Return the maximum value of the equationyi + yj + |xi - xj| where |xi - xj| <= k and 1 <= i < j <= points.length.

It is guaranteed that there exists at least one pair of points that satisfy the constraint |xi - xj| <= k.

Example 1:

Input: points = [[1,3],[2,0],[5,10],[6,-10]], k = 1 Output: 4 Explanation: The first two points satisfy the condition |xi - xj| <= 1 and if we calculate the equation we get 3 + 0 + |1 - 2| = 4. Third and fourth points also satisfy the condition and give a value of 10 + -10 + |5 - 6| = 1. No other pairs satisfy the condition, so we return the max of 4 and 1. 

Example 2:

Input: points = [[0,0],[3,0],[9,2]], k = 3 Output: 3 Explanation: Only the first two points have an absolute difference of 3 or less in the x-values, and give the value of 0 + 0 + |0 - 3| = 3. 

Constraints:

• 2 <= points.length <= 105
• points[i].length == 2
• -108 <= xi, yi <= 108
• 0 <= k <= 2 * 108
• xi < xj for all 1 <= i < j <= points.length
• xi form a strictly increasing sequence.

Solutions (Rust)

1. Solution

use std::collections::VecDeque;implSolution{pubfnfind_max_value_of_equation(points:Vec<Vec<i32>>,k:i32) -> i32{letmut deque = VecDeque::new();letmut ret = i32::MIN;for j in0..points.len(){let(xj, yj) = (points[j][0], points[j][1]);while xj - deque.front().unwrap_or(&(xj,0)).0 > k { deque.pop_front();}ifletSome(&(xi, yi)) = deque.front(){ ret = ret.max(yi + yj + xj - xi);}whileletSome(&(xi, yi)) = deque.back(){if yi - xi <= yj - xj { deque.pop_back();}else{break;}} deque.push_back((xj, yj));} ret }}