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Max Value of Equation

Difficulty: Hard

Problem Description

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.

Key Insights

  • The problem requires finding pairs of points where the x-coordinates have a difference less than or equal to k.
  • The equation to maximize can be rearranged to focus on the difference between y-coordinates and x-coordinates.
  • A sliding window or monotonic queue approach can be effective to maintain potential candidates for the equation.
  • The points are already sorted by x-values, which can help in efficiently managing the indices.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(n)

Solution

To solve the problem, we can use a sliding window technique with a deque (double-ended queue) to maintain candidates for the maximum value of the equation. As we iterate through the points, we will keep adding points to the deque while ensuring that the x-coordinate difference remains within k. We'll calculate the potential maximum value using the current y-coordinate and the best candidate from the deque.

  1. Initialize a variable to store the maximum value.
  2. Use a deque to store indices of the points.
  3. For each point:
    • Remove points from the front of the deque if they are out of the k range.
    • Compute the equation using the front of the deque, which gives the maximum possible value for the current point.
    • Add the current point to the deque, maintaining the order based on the potential value of the equation.

Code Solutions

from collections import deque

def findMaxValueOfEquation(points, k):
    max_value = float('-inf')
    dq = deque()
    
    for x, y in points:
        # Remove points that are out of the k range
        while dq and x - dq[0][0] > k:
            dq.popleft()
        
        # If the deque is not empty, calculate the max value
        if dq:
            max_value = max(max_value, y + x + dq[0][1] - dq[0][0])
        
        # Maintain the deque in decreasing order of (y - x)
        while dq and dq[-1][1] - dq[-1][0] < y - x:
            dq.pop()
        
        # Add the current point to the deque
        dq.append((x, y))
    
    return max_value
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