Skip to content

Latest commit

 

History

History
146 lines (120 loc) · 4.41 KB

README_EN.md

File metadata and controls

146 lines (120 loc) · 4.41 KB
Raw

1499. Max Value of Equation

中文文档

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

Python3

class Solution:
    def findMaxValueOfEquation(self, points: List[List[int]], k: int) -> int:
        q = deque([points[0]])
        ans = -inf
        for x, y in points[1:]:
            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

Java

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

C++

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

Go

func findMaxValueOfEquation(points [][]int, k int) int {
	q := [][]int{}
	ans := math.MinInt32
	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, y+x+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, p)
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

...