#1499

Hard

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Max Value of Equation

Max Value of Equation asks to find the maximum value of a specific equation on a set of 2D points using sliding window techniques.

array queue sliding window heap priority queue monotonic queue

Problem Statement

You are given an array of points, where each point is represented by its x and y coordinates, sorted by x-values. The task is to find the maximum value of the equation yi + yj + |xi - xj| for any valid pair of points, where |xi - xj| <= k and 1 <= i < j <= points.length.

The equation is subject to the constraint that |xi - xj| must be less than or equal to k. Your solution should return the maximum possible value of the equation for any such pair of points in the array.

Examples

Example 1

Input: points = [[1,3],[2,0],[5,10],[6,-10]], k = 1

Output: 4

Example 2

Input: points = [[0,0],[3,0],[9,2]], k = 3

Output: 3

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 Approach

Sliding Window with Priority Queue

Use a sliding window to maintain valid pairs of points such that |xi - xj| <= k. A priority queue can be used to track the best yi - xi values in the window, ensuring we can calculate the equation efficiently.

Efficient Lookup

While iterating through the points, for each point, calculate the potential maximum value using the priority queue. Keep the window size within the constraint of k to avoid unnecessary computations.

Update with Running State

For every new point processed, update the priority queue with the new point's value and ensure the previous points that no longer satisfy the |xi - xj| <= k constraint are removed from the window.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n log n), where n is the number of points, due to the use of the priority queue. Space complexity is O(n) for storing the points and maintaining the priority queue.

What Interviewers Usually Probe

Look for an understanding of how to maintain a sliding window.
Check if the candidate correctly uses a priority queue for efficient lookups.
Evaluate if the candidate handles the constraint |xi - xj| <= k appropriately while iterating.

Common Pitfalls or Variants

Common pitfalls

Failing to properly manage the sliding window size could lead to incorrect results.
Not using the priority queue effectively might result in inefficient or incorrect computation of the maximum value.
Ignoring the constraint |xi - xj| <= k could cause invalid pairs to be considered.

Follow-up variants

Increase the problem's complexity by adding additional constraints on the y-values of points.
Consider modifying the equation to include additional terms or weights for each point.
Allow for dynamic updates to the points array and re-calculate the maximum value accordingly.

FAQ

What is the main pattern used in the Max Value of Equation problem?

The problem uses the sliding window technique combined with a priority queue to efficiently find the maximum value of the given equation.

How do I handle the constraint |xi - xj| <= k in Max Value of Equation?

By maintaining a sliding window that ensures only valid pairs of points with |xi - xj| <= k are considered.

What data structure is best for solving Max Value of Equation?

A priority queue is ideal for maintaining the best values of yi - xi within the sliding window.

What is the time complexity of the Max Value of Equation solution?

The time complexity is O(n log n) due to the use of the priority queue for efficient lookups and updates.

How can GhostInterview help with Max Value of Equation?

GhostInterview provides tailored guidance, helping you implement the sliding window and priority queue efficiently, ensuring a clean solution.

terminal

Solution

Solution 1

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class Solution:
    def findMaxValueOfEquation(self, points: List[List[int]], k: int) -> int:
        ans = -inf
        pq = []
        for x, y in points:
            while pq and x - pq[0][1] > k:
                heappop(pq)
            if pq:
                ans = max(ans, x + y - pq[0][0])
            heappush(pq, (x - y, x))
        return ans

Solution 2

#### Python3

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class Solution:
    def findMaxValueOfEquation(self, points: List[List[int]], k: int) -> int:
        ans = -inf
        pq = []
        for x, y in points:
            while pq and x - pq[0][1] > k:
                heappop(pq)
            if pq:
                ans = max(ans, x + y - pq[0][0])
            heappush(pq, (x - y, x))
        return ans

Continue Practicing

#1438 longest continuous subarray with absolute diff less than or equal to limit #1425 constrained subsequence sum #862 shortest subarray with sum at least k