Q25_max_value_of_equation.cpp

// 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.

#include<bits/stdc++.h>
using namespace std;

class Solution {
public:
    int findMaxValueOfEquation(vector<vector<int>>& points, int k) {
        // yi + yj -(xi - xj) = yi + yj - xi + xj = (xj + yj) + (yi - xi)
        int n = points.size();
        int ans = INT_MIN;
        priority_queue<pair<int, int>> pq;

        pq.push({points[0][1] - points[0][0], points[0][0]});

        for(int i = 1; i < n; i++) {
            int curr = points[i][0] + points[i][1];
            while(!pq.empty() && points[i][0] - pq.top().second > k) pq.pop();
            if(!pq.empty()) ans = max(ans, curr + pq.top().first);

            pq.push({points[i][1] - points[i][0], points[i][0]});
        }
        return ans;
    }
};

// Simplification of the function
// yi + yj + |xi - xj| = yi + yj - (xi - xj) [because the x coordinates are sorted]
// so |xi - xj| = -(xi - xj), as |x| = -x when x < 0 [here xi < xj]
// yi + yj + |xi - xj| = yi + yj - xi + xj = (yj + xj) + (yi - xi)

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