+ problem 1499

Author: fruit-in

Problemset/1499-Max Value of Equation/README.md

@@ -0,0 +1,67 @@
+# 1499. 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 <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> such that <code>x<sub>i</sub> < x<sub>j</sub></code> for all `1 <= i < j <= points.length`. You are also given an integer `k`.
+
+Return *the maximum value of the equation* <code>y<sub>i</sub> + y<sub>j</sub> + |x<sub>i</sub> - x<sub>j</sub>|</code> where <code>|x<sub>i</sub> - x<sub>j</sub>| <= k</code> and `1 <= i < j <= points.length`.
+
+It is guaranteed that there exists at least one pair of points that satisfy the constraint <code>|x<sub>i</sub> - x<sub>j</sub>| <= k</code>.
+
+#### Example 1:
+<pre>
+<strong>Input:</strong> points = [[1,3],[2,0],[5,10],[6,-10]], k = 1
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> 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.
+</pre>
+
+#### Example 2:
+<pre>
+<strong>Input:</strong> points = [[0,0],[3,0],[9,2]], k = 3
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> 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.
+</pre>
+
+#### Constraints:
+* <code>2 <= points.length <= 10<sup>5</sup></code>
+* `points[i].length == 2`
+* <code>-10<sup>8</sup> <= x<sub>i</sub>, y<sub>i</sub> <= 10<sup>8</sup></code>
+* <code>0 <= k <= 2 * 10<sup>8</sup></code>
+* <code>x<sub>i</sub> < x<sub>j</sub></code> for all <code>1 <= i < j <= points.length</code>
+* <code>x<sub>i</sub></code> form a strictly increasing sequence.
+
+## Solutions (Rust)
+
+### 1. Solution
+```Rust
+use std::collections::VecDeque;
+
+impl Solution {
+    pub fn find_max_value_of_equation(points: Vec<Vec<i32>>, k: i32) -> i32 {
+        let mut deque = VecDeque::new();
+        let mut ret = i32::MIN;
+
+        for j in 0..points.len() {
+            let (xj, yj) = (points[j][0], points[j][1]);
+
+            while xj - deque.front().unwrap_or(&(xj, 0)).0 > k {
+                deque.pop_front();
+            }
+
+            if let Some(&(xi, yi)) = deque.front() {
+                ret = ret.max(yi + yj + xj - xi);
+            }
+
+            while let Some(&(xi, yi)) = deque.back() {
+                if yi - xi <= yj - xj {
+                    deque.pop_back();
+                } else {
+                    break;
+                }
+            }
+
+            deque.push_back((xj, yj));
+        }
+
+        ret
+    }
+}
+```

Problemset/1499-Max Value of Equation/README_CN.md

@@ -0,0 +1,66 @@
+# 1499. 满足不等式的最大值
+给你一个数组 `points` 和一个整数 `k` 。数组中每个元素都表示二维平面上的点的坐标，并按照横坐标 x 的值从小到大排序。也就是说 <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> ，并且在 `1 <= i < j <= points.length` 的前提下， <code>x<sub>i</sub> < x<sub>j</sub></code> 总成立。
+
+请你找出 <code>y<sub>i</sub> + y<sub>j</sub> + |x<sub>i</sub> - x<sub>j</sub>|</code> 的 **最大值**，其中 <code>|x<sub>i</sub> - x<sub>j</sub>| <= k</code> 且 `1 <= i < j <= points.length`。
+
+题目测试数据保证至少存在一对能够满足 <code>|x<sub>i</sub> - x<sub>j</sub>| <= k</code> 的点。
+
+#### 示例 1:
+<pre>
+<strong>输入:</strong> points = [[1,3],[2,0],[5,10],[6,-10]], k = 1
+<strong>输出:</strong> 4
+<strong>解释:</strong> 前两个点满足 |xi - xj| <= 1 ，代入方程计算，则得到值 3 + 0 + |1 - 2| = 4 。第三个和第四个点也满足条件，得到值 10 + -10 + |5 - 6| = 1 。
+没有其他满足条件的点，所以返回 4 和 1 中最大的那个。
+</pre>
+
+#### 示例 2:
+<pre>
+<strong>输入:</strong> points = [[0,0],[3,0],[9,2]], k = 3
+<strong>输出:</strong> 3
+<strong>解释:</strong> 只有前两个点满足 |xi - xj| <= 3 ，代入方程后得到值 0 + 0 + |0 - 3| = 3 。
+</pre>
+
+#### 提示:
+* <code>2 <= points.length <= 10<sup>5</sup></code>
+* `points[i].length == 2`
+* <code>-10<sup>8</sup> <= x<sub>i</sub>, y<sub>i</sub> <= 10<sup>8</sup></code>
+* <code>0 <= k <= 2 * 10<sup>8</sup></code>
+* 对于所有的`1 <= i < j <= points.length` ，`points[i][0] < points[j][0]` 都成立。也就是说，<code>x<sub>i</sub></code> 是严格递增的。
+
+## 题解 (Rust)
+
+### 1. 题解
+```Rust
+use std::collections::VecDeque;
+
+impl Solution {
+    pub fn find_max_value_of_equation(points: Vec<Vec<i32>>, k: i32) -> i32 {
+        let mut deque = VecDeque::new();
+        let mut ret = i32::MIN;
+
+        for j in 0..points.len() {
+            let (xj, yj) = (points[j][0], points[j][1]);
+
+            while xj - deque.front().unwrap_or(&(xj, 0)).0 > k {
+                deque.pop_front();
+            }
+
+            if let Some(&(xi, yi)) = deque.front() {
+                ret = ret.max(yi + yj + xj - xi);
+            }
+
+            while let Some(&(xi, yi)) = deque.back() {
+                if yi - xi <= yj - xj {
+                    deque.pop_back();
+                } else {
+                    break;
+                }
+            }
+
+            deque.push_back((xj, yj));
+        }
+
+        ret
+    }
+}
+```

Problemset/1499-Max Value of Equation/Solution.rs

@@ -0,0 +1,32 @@
+use std::collections::VecDeque;
+
+impl Solution {
+    pub fn find_max_value_of_equation(points: Vec<Vec<i32>>, k: i32) -> i32 {
+        let mut deque = VecDeque::new();
+        let mut ret = i32::MIN;
+
+        for j in 0..points.len() {
+            let (xj, yj) = (points[j][0], points[j][1]);
+
+            while xj - deque.front().unwrap_or(&(xj, 0)).0 > k {
+                deque.pop_front();
+            }
+
+            if let Some(&(xi, yi)) = deque.front() {
+                ret = ret.max(yi + yj + xj - xi);
+            }
+
+            while let Some(&(xi, yi)) = deque.back() {
+                if yi - xi <= yj - xj {
+                    deque.pop_back();
+                } else {
+                    break;
+                }
+            }
+
+            deque.push_back((xj, yj));
+        }
+
+        ret
+    }
+}

README.md

@@ -876,6 +876,7 @@
 [1493][1493l]|[Longest Subarray of 1's After Deleting One Element][1493]                            |![rb]  ![rs]
 [1496][1496l]|[Path Crossing][1496]                                                                 |![rs]
 [1497][1497l]|[Check If Array Pairs Are Divisible by k][1497]                                       |![rs]
+[1499][1499l]|[Max Value of Equation][1499]                                                         |![rs]
 [1502][1502l]|[Can Make Arithmetic Progression From Sequence][1502]                                 |![rb]  ![rs]
 [1503][1503l]|[Last Moment Before All Ants Fall Out of a Plank][1503]                               |![rb]  ![rs]
 [1504][1504l]|[Count Submatrices With All Ones][1504]                                               |![rs]
@@ -2273,6 +2274,7 @@
 [1493]:Problemset/1493-Longest%20Subarray%20of%201's%20After%20Deleting%20One%20Element/README.md#1493-longest-subarray-of-1s-after-deleting-one-element
 [1496]:Problemset/1496-Path%20Crossing/README.md#1496-path-crossing
 [1497]:Problemset/1497-Check%20If%20Array%20Pairs%20Are%20Divisible%20by%20k/README.md#1497-check-if-array-pairs-are-divisible-by-k
+[1499]:Problemset/1499-Max%20Value%20of%20Equation/README.md#1499-max-value-of-equation
 [1502]:Problemset/1502-Can%20Make%20Arithmetic%20Progression%20From%20Sequence/README.md#1502-can-make-arithmetic-progression-from-sequence
 [1503]:Problemset/1503-Last%20Moment%20Before%20All%20Ants%20Fall%20Out%20of%20a%20Plank/README.md#1503-last-moment-before-all-ants-fall-out-of-a-plank
 [1504]:Problemset/1504-Count%20Submatrices%20With%20All%20Ones/README.md#1504-count-submatrices-with-all-ones
@@ -3669,6 +3671,7 @@
 [1493l]:https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/
 [1496l]:https://leetcode.com/problems/path-crossing/
 [1497l]:https://leetcode.com/problems/check-if-array-pairs-are-divisible-by-k/
+[1499l]:https://leetcode.com/problems/max-value-of-equation/
 [1502l]:https://leetcode.com/problems/can-make-arithmetic-progression-from-sequence/
 [1503l]:https://leetcode.com/problems/last-moment-before-all-ants-fall-out-of-a-plank/
 [1504l]:https://leetcode.com/problems/count-submatrices-with-all-ones/

README_CN.md

@@ -876,6 +876,7 @@
 [1493][1493l]|[删掉一个元素以后全为 1 的最长子数组][1493]               |![rb]  ![rs]
 [1496][1496l]|[判断路径是否相交][1496]                                  |![rs]
 [1497][1497l]|[检查数组对是否可以被 k 整除][1497]                       |![rs]
+[1499][1499l]|[满足不等式的最大值][1499]                                |![rs]
 [1502][1502l]|[判断能否形成等差数列][1502]                              |![rb]  ![rs]
 [1503][1503l]|[所有蚂蚁掉下来前的最后一刻][1503]                        |![rb]  ![rs]
 [1504][1504l]|[统计全 1 子矩形][1504]                                   |![rs]
@@ -2273,6 +2274,7 @@
 [1493]:Problemset/1493-Longest%20Subarray%20of%201's%20After%20Deleting%20One%20Element/README_CN.md#1493-删掉一个元素以后全为-1-的最长子数组
 [1496]:Problemset/1496-Path%20Crossing/README_CN.md#1496-判断路径是否相交
 [1497]:Problemset/1497-Check%20If%20Array%20Pairs%20Are%20Divisible%20by%20k/README_CN.md#1497-检查数组对是否可以被-k-整除
+[1499]:Problemset/1499-Max%20Value%20of%20Equation/README_CN.md#1499-满足不等式的最大值
 [1502]:Problemset/1502-Can%20Make%20Arithmetic%20Progression%20From%20Sequence/README_CN.md#1502-判断能否形成等差数列
 [1503]:Problemset/1503-Last%20Moment%20Before%20All%20Ants%20Fall%20Out%20of%20a%20Plank/README_CN.md#1503-所有蚂蚁掉下来前的最后一刻
 [1504]:Problemset/1504-Count%20Submatrices%20With%20All%20Ones/README_CN.md#1504-统计全-1-子矩形
@@ -3669,6 +3671,7 @@
 [1493l]:https://leetcode.cn/problems/longest-subarray-of-1s-after-deleting-one-element/
 [1496l]:https://leetcode.cn/problems/path-crossing/
 [1497l]:https://leetcode.cn/problems/check-if-array-pairs-are-divisible-by-k/
+[1499l]:https://leetcode.cn/problems/max-value-of-equation/
 [1502l]:https://leetcode.cn/problems/can-make-arithmetic-progression-from-sequence/
 [1503l]:https://leetcode.cn/problems/last-moment-before-all-ants-fall-out-of-a-plank/
 [1504l]:https://leetcode.cn/problems/count-submatrices-with-all-ones/