How do you solve Max Value of Equation on LeetCode?

Max Value of Equation (LeetCode #1499) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a sliding window approach helps efficiently manage the constraints of the problem.

What is the brute force solution for Max Value of Equation?

The brute force approach for Max Value of Equation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible pair of points to see if they satisfy the condition |xi - xj| <= k. For each valid pair, we compute the value of the equation and keep track of the maximum found. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Max Value of Equation?

Max Value of Equation uses the following concepts: Array, Queue, Sliding Window, Heap (Priority Queue), Monotonic Queue. The recommended patterns to study are: Sliding Window, Priority Queue.

What are the interview tips for Max Value of Equation?

Always clarify the constraints and edge cases before diving into coding. Think about the data structures that can optimize your solution, like heaps or queues. Practice explaining your thought process as you code to simulate the interview environment.

What are common mistakes when solving Max Value of Equation?

Not properly managing the sliding window, leading to incorrect pairs being considered. Failing to understand how to use a priority queue effectively, which can lead to suboptimal performance.

#1499

Max Value of Equation

Hard

ArrayQueueSliding WindowHeap (Priority Queue)Monotonic QueueSliding WindowPriority Queue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses a sliding window approach with a priority queue to efficiently keep track of the maximum value of yi - xi as we iterate through the points. This allows us to quickly find the best candidate for the equation without checking all pairs.

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Algorithm

5 steps

1Step 1: Initialize a max heap (priority queue) to store tuples of (yi - xi, xi).
2Step 2: Iterate through each point (xj, yj) in points.
3Step 3: While the top of the heap has xi such that xj - xi > k, pop it from the heap.
4Step 4: If the heap is not empty, compute the value using the top of the heap and update max_value.
5Step 5: Push (yj - xj, xj) onto the heap.

solution.py12 lines

1import heapq
2
3def maxValue(points, k):
4    max_value = float('-inf')
5    heap = []
6    for xj, yj in points:
7        while heap and xj - heap[0][1] > k:
8            heapq.heappop(heap)
9        if heap:
10            max_value = max(max_value, yj + xj + -heap[0][0])
11        heapq.heappush(heap, (yj - xj, xj))
12    return max_value

ℹ

Complexity note: The time complexity is O(n log n) due to the use of a priority queue for maintaining the maximum value, where n is the number of points. The space complexity is O(n) because we store at most n elements in the heap.

1Using a sliding window approach helps efficiently manage the constraints of the problem.
2A priority queue allows us to quickly access the maximum value needed for our calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.