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][0], points[i][1] <= 10^8
  • 0 <= k <= 2 * 10^8
  • points[i][0] < points[j][0] 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][1] - A[i][0] 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][0] - A[i][0] <= k; ++j) s.insert(A[j][0] + A[j][1]);
            s.erase(s.find(A[i][0] + A[i][1]));
            if (s.size()) ans = max(ans, A[i][1] - A[i][0] + *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][0] - A[i][0] <= k; ++j) {
                int sum = A[j][0] + A[j][1];
                while (q.size() && q.back() < sum) q.pop_back();
                q.push_back(sum);
            }
            if (q.size() && q.front() == A[i][0] + A[i][1]) q.pop_front();
            if (q.size()) ans = max(ans, A[i][1] - A[i][0] + 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[0], i);
        }
        int[] differences = new int[length];
        for (int i = 0; i < length; i++)
            differences[i] = points[i][1] - points[i][0];
        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[0], y = point[1];
            int sum = x + y;
            while (!deque.isEmpty() && x - points[deque.peekFirst()][0] > 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;
    }
}

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