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
.
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.
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.
2 <= points.length <= 105
points[i].length == 2
-108 <= xi, yi <= 108
0 <= k <= 2 * 108
xi < xj
for all1 <= i < j <= points.length
xi
form a strictly increasing sequence.
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
}
}