# 1499. Max Value of Equation

## Description

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

• class Solution {
public int findMaxValueOfEquation(int[][] points, int k) {
int ans = -(1 << 30);
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[0] - a[0]);
for (var p : points) {
int x = p[0], y = p[1];
while (!pq.isEmpty() && x - pq.peek()[1] > k) {
pq.poll();
}
if (!pq.isEmpty()) {
ans = Math.max(ans, x + y + pq.peek()[0]);
}
pq.offer(new int[] {y - x, x});
}
return ans;
}
}

• class Solution {
public:
int findMaxValueOfEquation(vector<vector<int>>& points, int k) {
int ans = -(1 << 30);
priority_queue<pair<int, int>> pq;
for (auto& p : points) {
int x = p[0], y = p[1];
while (pq.size() && x - pq.top().second > k) {
pq.pop();
}
if (pq.size()) {
ans = max(ans, x + y + pq.top().first);
}
pq.emplace(y - x, x);
}
return ans;
}
};

• class Solution:
def findMaxValueOfEquation(self, points: List[List[int]], k: int) -> int:
ans = -inf
pq = []
for x, y in points:
while pq and x - pq[0][1] > k:
heappop(pq)
if pq:
ans = max(ans, x + y - pq[0][0])
heappush(pq, (x - y, x))
return ans


• func findMaxValueOfEquation(points [][]int, k int) int {
ans := -(1 << 30)
hp := hp{}
for _, p := range points {
x, y := p[0], p[1]
for hp.Len() > 0 && x-hp[0].x > k {
heap.Pop(&hp)
}
if hp.Len() > 0 {
ans = max(ans, x+y+hp[0].v)
}
heap.Push(&hp, pair{y - x, x})
}
return ans
}

type pair struct{ v, x int }

type hp []pair

func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool {
a, b := h[i], h[j]
return a.v > b.v
}
func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any)   { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any     { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

• function findMaxValueOfEquation(points: number[][], k: number): number {
let ans = -(1 << 30);
const pq = new Heap<[number, number]>((a, b) => b[0] - a[0]);
for (const [x, y] of points) {
while (pq.size() && x - pq.top()[1] > k) {
pq.pop();
}
if (pq.size()) {
ans = Math.max(ans, x + y + pq.top()[0]);
}
pq.push([y - x, x]);
}
return ans;
}

type Compare<T> = (lhs: T, rhs: T) => number;

class Heap<T = number> {
data: Array<T | null>;
lt: (i: number, j: number) => boolean;
constructor();
constructor(data: T[]);
constructor(compare: Compare<T>);
constructor(data: T[], compare: Compare<T>);
constructor(data: T[] | Compare<T>, compare?: (lhs: T, rhs: T) => number);
constructor(
data: T[] | Compare<T> = [],
compare: Compare<T> = (lhs: T, rhs: T) => (lhs < rhs ? -1 : lhs > rhs ? 1 : 0),
) {
if (typeof data === 'function') {
compare = data;
data = [];
}
this.data = [null, ...data];
this.lt = (i, j) => compare(this.data[i]!, this.data[j]!) < 0;
for (let i = this.size(); i > 0; i--) this.heapify(i);
}

size(): number {
return this.data.length - 1;
}

push(v: T): void {
this.data.push(v);
let i = this.size();
while (i >> 1 !== 0 && this.lt(i, i >> 1)) this.swap(i, (i >>= 1));
}

pop(): T {
this.swap(1, this.size());
const top = this.data.pop();
this.heapify(1);
}

top(): T {
return this.data[1]!;
}
heapify(i: number): void {
while (true) {
let min = i;
const [l, r, n] = [i * 2, i * 2 + 1, this.data.length];
if (l < n && this.lt(l, min)) min = l;
if (r < n && this.lt(r, min)) min = r;
if (min !== i) {
this.swap(i, min);
i = min;
} else break;
}
}

clear(): void {
this.data = [null];
}

private swap(i: number, j: number): void {
const d = this.data;
[d[i], d[j]] = [d[j], d[i]];
}
}