LeetCode-in-Java

1499. Max Value of Equation

Hard

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 |x_i - x_j| <= 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.

Solution

public class Solution {
    public int findMaxValueOfEquation(int[][] points, int k) {
        int res = Integer.MIN_VALUE;
        int max = Integer.MIN_VALUE;
        int r = 0;
        int rMax = 0;
        for (int l = 0; l < points.length - 1; l++) {
            if (rMax == l) {
                max = Integer.MIN_VALUE;
                r = l + 1;
                rMax = r;
            }
            while (r < points.length && points[r][0] - points[l][0] <= k) {
                int v = points[r][0] + points[r][1];
                if (max < v) {
                    max = v;
                    rMax = r;
                }
                r++;
            }
            if (points[rMax][0] - points[l][0] <= k) {
                res =
                        Math.max(
                                res,
                                points[rMax][0] - points[l][0] + points[rMax][1] + points[l][1]);
            }
        }
        return res;
    }
}

This site is open source. Improve this page.