Here, $$$|x_i - x_j| = x_j - x_i$$$, so you can re-write the expression as $$$(y_i - x_i) + (y_j + x_j)$$$. You can iterate through points, adding values of $$$y_i - x_i$$$ to multiset or priority queue, and removing the ones for which $$$x_j - x_i \gt k$$$. Each time, just update the answer if $$$mx + y_j + x_j$$$ is bigger than the biggest answer so far.

multiset st
ans = 0
i = 0
for j in range(n):
  while points[i].x + k < points[j].x:
    st.erase(points[i].y - points[i].x)
    increment i
  if st not empty:
    ans = max(ans, points[j].x + points[j].y + st.max)
  st.insert(points[j].y - points[j].x)
print ans

To maximise yi+yj + |xi-xj|. Since the points are sorted by x-coordinates, we can rewrite this as (yj+xj) + (yi-xi). As we iterate through each point j, our goal is to find the maximum value of (yi-xi) for all previous points i that satisfy the condition xj-xi <= k. A sliding window approach is effective in this case.

import heapq
class Solution:
    def findMaxValueOfEquation(self, p: list[list[int]], k: int) -> int:
        q = []; r = -float('inf')
        for x, y in p:
            while q and x - q[0][1] > k: heapq.heappop(q)
            if q: r = max(r, x + y - q[0][0])
            heapq.heappush(q, (x - y, x))
        return r