Formatted question description: https://leetcode.ca/all/1499.html

# 1499. Max Value of Equation (Hard)

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.

Find 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 <= 10^5
• points[i].length == 2
• -10^8 <= points[i], points[i] <= 10^8
• 0 <= k <= 2 * 10^8
• points[i] < points[j] for all 1 <= i < j <= points.length
• xi form a strictly increasing sequence.

Related Topics:
Array, Sliding Window

## Solution 1. Multiset

For the equation yi + yj + |xi - xj|, since j > i, so xj must be greater than xi, so the equation is the same as yi + yj + xj - xi = xj + yj - xi + yi. For a given i, -xi + yi is a constant, so we just need to find the maximum xj + yj satisfying the k constraint.

Keep a sliding window [i, j). The elements in the window satisfy the k constraint. Use a multiset<int> s to keep the x + y values in the window except for that for the A[i].

For this A[i], the maximum value we can get is A[i] - A[i] plus the maximum value in the multiset.

// OJ: https://leetcode.com/problems/max-value-of-equation/

// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
int findMaxValueOfEquation(vector<vector<int>>& A, int k) {
int i = 0, j = 0, N = A.size(), ans = INT_MIN;
multiset<int> s;
for (; i < N; ++i) {
for (; j < N && A[j] - A[i] <= k; ++j) s.insert(A[j] + A[j]);
s.erase(s.find(A[i] + A[i]));
if (s.size()) ans = max(ans, A[i] - A[i] + *s.rbegin());
}
return ans;
}
};


## Solution 2. Monoqueue

Since we only care about the maximum value in a sliding window, we can use a descending monoqueue to keep track of the maximum value.

// OJ: https://leetcode.com/problems/max-value-of-equation/

// Time: O(N)
// Space: O(N)
class Solution {
public:
int findMaxValueOfEquation(vector<vector<int>>& A, int k) {
int i = 0, j = 0, N = A.size(), ans = INT_MIN;
deque<int> q; // descending monoqueue
for (; i < N; ++i) {
for (; j < N && A[j] - A[i] <= k; ++j) {
int sum = A[j] + A[j];
while (q.size() && q.back() < sum) q.pop_back();
q.push_back(sum);
}
if (q.size() && q.front() == A[i] + A[i]) q.pop_front();
if (q.size()) ans = max(ans, A[i] - A[i] + q.front());
}
return ans;
}
};


Java

class Solution {
public int findMaxValueOfEquation(int[][] points, int k) {
TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
int length = points.length;
for (int i = 0; i < length; i++) {
int[] point = points[i];
map.put(point, i);
}
int[] differences = new int[length];
for (int i = 0; i < length; i++)
differences[i] = points[i] - points[i];
Deque<Integer> deque = new ArrayDeque<Integer>();
deque.offer(0);
int maxSum = Integer.MIN_VALUE;
for (int i = 1; i < length; i++) {
int[] point = points[i];
int x = point, y = point;
int sum = x + y;
while (!deque.isEmpty() && x - points[deque.peekFirst()] > k)
deque.pollFirst();
if (deque.isEmpty())
deque.offerLast(i);
else {
int prevIndex = deque.peekLast();
for (int j = prevIndex + 1; j < i; j++) {
while (!deque.isEmpty() && differences[deque.peekLast()] <= differences[j])
deque.pollLast();
deque.offerLast(j);
}
int prevMax = differences[deque.peekFirst()];
int curSum = prevMax + sum;
maxSum = Math.max(maxSum, curSum);
while (!deque.isEmpty() && differences[deque.peekLast()] <= differences[i])
deque.pollLast();
deque.offerLast(i);
}
}
return maxSum;
}
}