1499. Max Value of Equation LeetCode Solution

In this guide, you will get 1499. Max Value of Equation LeetCode Solution with the best time and space complexity. The solution to Max Value of Equation problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Max Value of Equation solution in C++
4. Max Value of Equation solution in Java
5. Max Value of Equation solution in Python
6. Additional Resources

Problem Statement of Max Value of Equation

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.

Complexity Analysis

• Time Complexity: O(n\log n)
• Space Complexity: O(n)

1499. Max Value of Equation LeetCode Solution in C++

class Solution {
 public:
  int findMaxValueOfEquation(vector<vector<int>>& points, int k) {
    int ans = INT_MIN;
    priority_queue<pair<int, int>> maxHeap;  // (y - x, x)

    for (const vector<int>& p : points) {
      const int x = p[0];
      const int y = p[1];
      while (!maxHeap.empty() && x - maxHeap.top().second > k)
        maxHeap.pop();
      if (!maxHeap.empty())
        ans = max(ans, x + y + maxHeap.top().first);
      maxHeap.emplace(y - x, x);
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

1499. Max Value of Equation LeetCode Solution in Java

class Solution {
  public int findMaxValueOfEquation(int[][] points, int k) {
    int ans = Integer.MIN_VALUE;
    // (y - x, x)
    Queue<Pair<Integer, Integer>> maxHeap = new PriorityQueue<>(
        Comparator.comparing(Pair<Integer, Integer>::getKey, Comparator.reverseOrder())
            .thenComparing(Pair<Integer, Integer>::getValue, Comparator.reverseOrder()));

    for (int[] p : points) {
      final int x = p[0];
      final int y = p[1];
      while (!maxHeap.isEmpty() && x - maxHeap.peek().getValue() > k)
        maxHeap.poll();
      if (!maxHeap.isEmpty())
        ans = Math.max(ans, x + y + maxHeap.peek().getKey());
      maxHeap.offer(new Pair<>(y - x, x));
    }

    return ans;
  }
}
// code provided by PROGIEZ

1499. Max Value of Equation LeetCode Solution in Python

class Solution:
  def findMaxValueOfEquation(self, points: list[list[int]], k: int) -> int:
    ans = -math.inf
    maxHeap = []  # (y - x, x)

    for x, y in points:
      while maxHeap and x + maxHeap[0][1] > k:
        heapq.heappop(maxHeap)
      if maxHeap:
        ans = max(ans, x + y - maxHeap[0][0])
      heapq.heappush(maxHeap, (x - y, -x))

    return ans
 # code by PROGIEZ

Additional Resources

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• Explore all problems on LeetCode website here

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