LeetCode 1499: Max Value of Equation Solution

Master LeetCode problem 1499 (Max Value of Equation), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1499. Max Value of Equation

Hard

Array Queue Sliding Window Heap (Priority Queue)Monotonic Queue

Problem Explanation

Explanation:

We can iterate through the points array, keeping track of the maximum value of yi + yj + |xi - xj| that we have encountered so far.
To maximize the equation yi + yj + |xi - xj|, we can rewrite it as (yi - xi) + (yj + xj). This way, for each point j, we only need to find the maximum value of (yi - xi) for all previous points i.
We can use a deque to store the indices of points where the difference (yi - xi) is maximum, within a range of k.
By maintaining a decreasing order deque based on the value (yi - xi), we can easily find the maximum value within the range k.
We iterate through the points array, updating the deque and calculating the maximum value of the equation.

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

Solution Code

class Solution {
    public int findMaxValueOfEquation(int[][] points, int k) {
        Deque<Integer> deque = new LinkedList<>();
        int maxVal = Integer.MIN_VALUE;
        
        for (int[] point : points) {
            while (!deque.isEmpty() && point[0] - points[deque.peekFirst()][0] > k) {
                deque.pollFirst();
            }
            if (!deque.isEmpty()) {
                maxVal = Math.max(maxVal, point[1] + points[deque.peekFirst()][1] + point[0] - points[deque.peekFirst()][0]);
            }
            while (!deque.isEmpty() && point[1] - point[0] > points[deque.peekLast()][1] - points[deque.peekLast()][0]) {
                deque.pollLast();
            }
            deque.offerLast(points.indexOf(point));
        }
        
        return maxVal;
    }
}

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