Question

1499. Max Value of Equation

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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

Solution 1

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

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;
  }
};

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);
    return top!;
  }

  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]];
  }
}

Solution 2

class Solution:
  def findMaxValueOfEquation(self, points: List[List[int]], k: int) -> int:
    ans = -inf
    q = deque()
    for x, y in points:
      while q and x - q[0][0] > k:
        q.popleft()
      if q:
        ans = max(ans, x + y + q[0][1] - q[0][0])
      while q and y - x >= q[-1][1] - q[-1][0]:
        q.pop()
      q.append((x, y))
    return ans

class Solution {
  public int findMaxValueOfEquation(int[][] points, int k) {
    int ans = -(1 << 30);
    Deque<int[]> q = new ArrayDeque<>();
    for (var p : points) {
      int x = p[0], y = p[1];
      while (!q.isEmpty() && x - q.peekFirst()[0] > k) {
        q.pollFirst();
      }
      if (!q.isEmpty()) {
        ans = Math.max(ans, x + y + q.peekFirst()[1] - q.peekFirst()[0]);
      }
      while (!q.isEmpty() && y - x >= q.peekLast()[1] - q.peekLast()[0]) {
        q.pollLast();
      }
      q.offerLast(p);
    }
    return ans;
  }
}

class Solution {
public:
  int findMaxValueOfEquation(vector<vector<int>>& points, int k) {
    int ans = -(1 << 30);
    deque<pair<int, int>> q;
    for (auto& p : points) {
      int x = p[0], y = p[1];
      while (!q.empty() && x - q.front().first > k) {
        q.pop_front();
      }
      if (!q.empty()) {
        ans = max(ans, x + y + q.front().second - q.front().first);
      }
      while (!q.empty() && y - x >= q.back().second - q.back().first) {
        q.pop_back();
      }
      q.emplace_back(x, y);
    }
    return ans;
  }
};

func findMaxValueOfEquation(points [][]int, k int) int {
  ans := -(1 << 30)
  q := [][2]int{}
  for _, p := range points {
    x, y := p[0], p[1]
    for len(q) > 0 && x-q[0][0] > k {
      q = q[1:]
    }
    if len(q) > 0 {
      ans = max(ans, x+y+q[0][1]-q[0][0])
    }
    for len(q) > 0 && y-x >= q[len(q)-1][1]-q[len(q)-1][0] {
      q = q[:len(q)-1]
    }
    q = append(q, [2]int{x, y})
  }
  return ans
}

function findMaxValueOfEquation(points: number[][], k: number): number {
  let ans = -(1 << 30);
  const q: number[][] = [];
  for (const [x, y] of points) {
    while (q.length > 0 && x - q[0][0] > k) {
      q.shift();
    }
    if (q.length > 0) {
      ans = Math.max(ans, x + y + q[0][1] - q[0][0]);
    }
    while (q.length > 0 && y - x > q[q.length - 1][1] - q[q.length - 1][0]) {
      q.pop();
    }
    q.push([x, y]);
  }
  return ans;
}