Max Value of Equation

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.

Find 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 <= 10^5

• points[i].length == 2

• -10^8 <= points[i][0], points[i][1] <= 10^8

• 0 <= k <= 2 * 10^8

• points[i][0] < points[j][0] for all 1 <= i < j <= points.length

• xi form a strictly increasing sequence.

class Solution {
    public int findMaxValueOfEquation(int[][] points, int k) {
        // Because xi < xj,
        // yi + yj + |xi - xj| = (yi - xi) + (yj + xj)
        // So we only need to find out the maximum yi - xi.
        Deque<int[]> dq = new LinkedList<>();
        // Maintaing an increasing deque from right to left
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < points.length; i++) {
            // Kicking out of range k points
            while (dq.size() != 0 && dq.peekFirst()[0] + k < points[i][0])
                dq.pollFirst();
            if (dq.size() != 0)
                max = Math.max(max, dq.peekFirst()[1] + points[i][1] + (points[i][0] - dq.peekFirst()[0]));
            // Maintaining decreasing dq
            while (dq.size() != 0 && dq.peekLast()[1] - dq.peekLast()[0] < points[i][1] - points[i][0])
                dq.pollLast();
            dq.add(points[i]);
        }
        return max;
    }
}