1499. Max Value of Equation

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.

Solution

• Brute force

  • iterate all possible combinations and find a max among them

  • Time Complexity: O(N^2), N is the number of input points

• Enhancement: Remove duplicated computation -> checking point (i, j) is equal to checking points (j, i)

  • Time Complexity: O(N^2/2)

• Enhacement2:

  • The input points are sorted, so For all (i, j) in [0....N-1], if j > i, then Xj >= Xi

    • yi + yj + |xi - xj| = yi + yj + xj - xi = (xj+yj) + (yi - xi)

    • -> If we can find the i, which (yi - xi) is the max, then we found it!

  • Ticky part: Becuase we only want to find the pair, so we can fixed at point j, and search for i,

    • where yi - xi is max, and xj - xi is <= k,

    • And, we don't need to checking all the combination when we fixed at i, we can check elements before i only([0...i-1], i), and the other combination (i, [i+1...N-1]) can be checked in the next fixed elements.

    • How to find the element which can result in the max or min result? Heap

public int findMax(int[][] points, int k){
    int ret = Integer.MIN_VALUE;
    PriorityQueue<int[]> maxHeap = new PriorityQueue<>((a,b) -> Integer.compare(b[1]-b[0], a[1]-a[0]));
    for(int[] point : points){
        while(!maxHeap.isEmpty() && point[0] - maxHeap.peek()[0] > k){
            maxHeap.poll();
        }
        if(!maxHeap.isEmpty()){
            ret = Math.max(ret, point[0] + point[1] + maxHeap.peek()[1] - maxHeap.peek()[0]);
        }
        
        maxHeap.add(point);
    }    
    return ret;
}