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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 equation yi + 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;

impl Solution {
    pub fn find_max_value_of_equation(points: Vec<Vec<i32>>, k: i32) -> i32 {
        let mut deque = VecDeque::new();
        let mut ret = i32::MIN;

        for j in 0..points.len() {
            let (xj, yj) = (points[j][0], points[j][1]);

            while xj - deque.front().unwrap_or(&(xj, 0)).0 > k {
                deque.pop_front();
            }

            if let Some(&(xi, yi)) = deque.front() {
                ret = ret.max(yi + yj + xj - xi);
            }

            while let Some(&(xi, yi)) = deque.back() {
                if yi - xi <= yj - xj {
                    deque.pop_back();
                } else {
                    break;
                }
            }

            deque.push_back((xj, yj));
        }

        ret
    }
}