LeetCode [1499] Max Value of Equation

 1499. Max Value of Equation

Hard

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.

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class Solution { public int findMaxValueOfEquation(int[][] points, int k) { int maxR = Integer.MIN_VALUE; //<yi-xi, xi> PriorityQueue<int[]> pq = new PriorityQueue<>((a,b) -> { if(a[0]==b[0]) return a[1]-b[1]; else return b[0]-a[0]; } ); for(int i=0; i<points.length; ++i){ while(!pq.isEmpty() && points[i][0]-pq.peek()[1]>k){ pq.poll(); } if(!pq.isEmpty()){ int r = points[i][0]+points[i][1]+pq.peek()[0]; maxR = Math.max(r, maxR); } pq.add(new int[]{points[i][1]-points[i][0], points[i][0]}); } return maxR; } }