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g1401_1500.s1499_max_value_of_equation.Solution Maven / Gradle / Ivy

package g1401_1500.s1499_max_value_of_equation;

// #Hard #Array #Heap_Priority_Queue #Sliding_Window #Queue #Monotonic_Queue
// #2022_03_21_Time_7_ms_(98.61%)_Space_105.2_MB_(79.40%)

/**
 * 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 |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.
**/
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;
    }
}