Max Value of Equation
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.
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.
Solve it on LeetCode
Approaches
1. Brute Force Approach
The brute force approach involves iterating through each pair of points and calculating the possible equation values directly. This results in a time-consuming solution, but it's a good starting point to understand the problem.
Intuition:
- Iterate over every possible pair of points
(i, j)wherei < j. - Check if the condition
|xi - xj| <= kholds. - If it holds, compute the equation value
(yi + yj + |xi - xj|). - Track the maximum equation value found.
Code:
- Time Complexity: O(n^2), where n is the number of points, as we are checking every pair.
- Space Complexity: O(1), no extra space is used apart from variables.
2. Optimized Approach using Priority Queue
To improve on the brute force approach, we can use a priority queue to help us find pairs that maximize the expression efficiently.
Intuition:
- Use a max heap (priority queue) to maintain the maximum values of
yi - xipairs. - Iterate over points while maintaining only valid points in the heap (those satisfying
xj - xi <= k). - For each point, calculate the potential value using the max in the priority queue.
- Update the priority queue with the current point.
Code:
- Time Complexity: O(n log n), because each point is inserted and removed from the heap.
- Space Complexity: O(n), due to the space needed for the heap.
3. Optimal Approach using Deque
To achieve even better time efficiency, we can use a deque data structure that allows insertion and deletions at both ends efficiently.
Intuition:
- Use a deque to maintain indices of elements such that the potential maximum calculation
(yi - xi)remains at the front. - For each point, first, clean the deque of elements that do not satisfy
|xi - xj| <= k. - Calculate maximum potential value using the front of the deque.
- Maintain the deque such that it stores only necessary indices for upcoming iterations.
Code:
- Time Complexity: O(n), since each element is inserted and removed from the deque at most once.
- Space Complexity: O(n), space required for the deque.