Max Value of Equation

Hard

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ArraysSliding WindowsStacksGreedy Algorithms

You are given an array points containing the coordinates of points on a 2D plane, sorted by the x-values, where points[i] = [x_i, y_i] such that x_i < x_j for all 1 <= i < j <= points.length. You are also given an integer k.

Return the maximum value of the equation y_i + y_j + |x_i - x_j| where |x_i - x_j| <= k and 1 <= i < j <= points.length.

It is guaranteed that there exists at least one pair of points that satisfy the constraint |x_i - x_j| <= 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 |x_i - x_j| <= 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 <= x_i, y_i <= 10⁸
0 <= k <= 2 * 10⁸
x_i < x_j for all 1 <= i < j <= points.length
x_i form a strictly increasing sequence.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the values of x and y coordinates in the points array? Can they be negative, zero, or floating-point numbers?
What is the maximum size of the `points` array? Is there a concern for exceeding memory limits with larger inputs?
If no pair of points satisfies the condition `|xi - xj| <= k`, should I return a specific value, like negative infinity or null?
Is the input `points` array guaranteed to be sorted by the x-coordinate, or do I need to handle unsorted input?
What data type should I use to represent the result of the equation, considering the potential for large coordinate values (e.g., `long` or `double`)?

Brute Force Solution

Approach

The goal is to find the maximum value that can be obtained by combining two points from a given set based on a specific formula. A brute force approach simply tries every possible pair of points to see which one results in the highest value.

Here's how the algorithm would work step-by-step:

Consider each possible pair of points from the set.
For each pair of points, calculate the value based on the given formula involving their positions.
Compare this calculated value with the current maximum value found so far.
If the calculated value is greater than the current maximum value, update the maximum value.
After checking all possible pairs of points, the final maximum value will be the answer.

Code Implementation

def find_max_value_equation_brute_force(points):
    max_value = float('-inf')

    # Iterate through all possible pairs of points
    for first_point_index in range(len(points)):
        for second_point_index in range(first_point_index + 1, len(points)):

            first_point_x = points[first_point_index][0]
            first_point_y = points[first_point_index][1]
            second_point_x = points[second_point_index][0]
            second_point_y = points[second_point_index][1]

            #This check is required by constraints
            if abs(first_point_x - second_point_x) <= abs(first_point_y - second_point_y):

                equation_value = first_point_y + second_point_y + abs(first_point_x - second_point_x)

                #Update max value, if equation is greater
                if equation_value > max_value:
                    max_value = equation_value

    # Return the largest value found
    return max_value

Big(O) Analysis

Time Complexity

O(n²) – The provided solution iterates through all possible pairs of points in the input array of size n. For each point, it compares it with every other point. This involves a nested loop structure where the outer loop iterates n times and the inner loop (in the worst case) also iterates close to n times. Therefore, the total number of operations grows proportionally to n multiplied by n, approximating n * n/2. This simplifies to a time complexity of O(n²).

Space Complexity

O(1) – The algorithm iterates through pairs of points from the input array, but it does not use any auxiliary data structures that scale with the input size N (the number of points). It only uses a constant amount of extra space to store variables like the current maximum value and loop counters, regardless of how many points are in the input. Therefore, the space complexity is O(1), indicating constant auxiliary space usage.

Optimal Solution

Approach

The key to solving this problem efficiently is to avoid checking every possible pair of points. We use a data structure to keep track of the best potential candidates for maximizing our equation as we move through the points, allowing us to quickly find the optimal pair.

Here's how the algorithm would work step-by-step:

Imagine you're walking through the points from left to right.
As you go, keep track of the 'best' points you've seen so far. 'Best' means the points that would give you the highest score if combined with points further down the line.
Use a special list to remember these best points. This list is organized so the 'best' one is always easy to access.
For each point you encounter, compare it with the 'best' points on your list to see if they meet the distance requirement and calculate the score.
If you find a pair that works, keep track of the highest score you've seen so far.
Also, before moving to the next point, update your list of 'best' points by adding the current point and removing any points that are worse than the current point based on our equation's characteristics.
By only looking at the 'best' points, you significantly reduce the number of comparisons you need to make, making the solution much faster.

Code Implementation

from collections import deque

def find_max_value_of_equation(points, max_allowed_difference):
    maximum_value = float('-inf')
    queue = deque()

    for x_coordinate, y_coordinate in points:
        # Remove points outside the allowed distance.
        while queue and x_coordinate - queue[0][1] > max_allowed_difference:
            queue.popleft()

        # Calculate the equation value with the best point so far.
        if queue:
            maximum_value = max(maximum_value, x_coordinate + y_coordinate + queue[0][0])

        # Maintain the queue to have the best candidates.
        # Only keep points with a higher potential score.
        while queue and y_coordinate - x_coordinate >= queue[-1][0]:
            queue.pop()
        
        queue.append((y_coordinate - x_coordinate, x_coordinate))

    return maximum_value

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each of the n points in the input array once. Inside the loop, it performs constant-time operations: accessing the 'best' points (which is an optimized list), calculating scores, and updating the 'best' points list. The key optimization is that the 'best' points list maintains a sorted structure such that finding suitable pairs is done in constant time per point, and adding/removing 'best' points is also amortized constant time. Therefore, the time complexity is dominated by the single pass through the n points, resulting in O(n) time.

Space Complexity

O(N) – The algorithm uses a list (or queue) to store potential 'best' points encountered so far. In the worst-case scenario, this list could contain all N points if they are all considered good candidates initially. Therefore, the auxiliary space required grows linearly with the number of input points. This results in a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty input array	Return negative infinity or a suitably small value to indicate no valid equation exists.
Input array with only one or zero points	Return negative infinity since we need at least two points to form an equation.
All points have the same x-coordinate	The priority queue should efficiently handle all x-coordinates being the same, but verify no division by zero occurs if used in the calculation.
Points are sorted in descending order of x	The algorithm should still function correctly as it considers all pairs within k distance, regardless of the order.
k is zero	Return negative infinity, because no pairs can possibly exist.
Large values for x and y coordinates leading to potential integer overflow	Use 64-bit integers (long) to prevent potential integer overflow when calculating the equation's value.
Large k value that encompasses almost all points	The time complexity is still O(n log n) but ensure the priority queue does not exhaust available memory.
Negative or very small k value	k represents a distance and should be non-negative; either throw an exception or interpret it as 0.

Empty input array

How to Handle:

Return negative infinity or a suitably small value to indicate no valid equation exists.

Input array with only one or zero points

How to Handle:

Return negative infinity since we need at least two points to form an equation.

All points have the same x-coordinate

How to Handle:

The priority queue should efficiently handle all x-coordinates being the same, but verify no division by zero occurs if used in the calculation.

Points are sorted in descending order of x

How to Handle:

The algorithm should still function correctly as it considers all pairs within k distance, regardless of the order.

k is zero

How to Handle:

Return negative infinity, because no pairs can possibly exist.

Large values for x and y coordinates leading to potential integer overflow

How to Handle:

Use 64-bit integers (long) to prevent potential integer overflow when calculating the equation's value.

Large k value that encompasses almost all points

How to Handle:

The time complexity is still O(n log n) but ensure the priority queue does not exhaust available memory.

Negative or very small k value

How to Handle:

k represents a distance and should be non-negative; either throw an exception or interpret it as 0.