feat: add solutions to lc problem: No.2898 (doocs#1792)

No.2898.Maximum Linear Stock Score

Author: yanglbme

solution/2800-2899/2898.Maximum Linear Stock Score/README.md

@@ -0,0 +1,169 @@
+# [2898. Maximum Linear Stock Score](https://leetcode.cn/problems/maximum-linear-stock-score)
+
+[English Version](/solution/2800-2899/2898.Maximum%20Linear%20Stock%20Score/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>Given a <strong>1-indexed</strong> integer array <code>prices</code>, where <code>prices[i]</code> is the price of a particular stock on the <code>i<sup>th</sup></code> day, your task is to select some of the elements of <code>prices</code> such that your selection is <strong>linear</strong>.</p>
+
+<p>A selection <code>indexes</code>, where <code>indexes</code> is a <strong>1-indexed</strong> integer array of length <code>k</code> which is a subsequence of the array <code>[1, 2, ..., n]</code>, is <strong>linear</strong> if:</p>
+
+<ul>
+	<li>For every <code>1 &lt; j &lt;= k</code>, <code>prices[indexes[j]] - prices[indexes[j - 1]] == indexes[j] - indexes[j - 1]</code>.</li>
+</ul>
+
+<p>A <b>subsequence</b> is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.</p>
+
+<p>The <strong>score</strong> of a selection <code>indexes</code>, is equal to the sum of the following array: <code>[prices[indexes[1]], prices[indexes[2]], ..., prices[indexes[k]]</code>.</p>
+
+<p>Return <em>the <strong>maximum</strong> <strong>score</strong> that a linear selection can have</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [1,5,3,7,8]
+<strong>Output:</strong> 20
+<strong>Explanation:</strong> We can select the indexes [2,4,5]. We show that our selection is linear:
+For j = 2, we have:
+indexes[2] - indexes[1] = 4 - 2 = 2.
+prices[4] - prices[2] = 7 - 5 = 2.
+For j = 3, we have:
+indexes[3] - indexes[2] = 5 - 4 = 1.
+prices[5] - prices[4] = 8 - 7 = 1.
+The sum of the elements is: prices[2] + prices[4] + prices[5] = 20.
+It can be shown that the maximum sum a linear selection can have is 20.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [5,6,7,8,9]
+<strong>Output:</strong> 35
+<strong>Explanation:</strong> We can select all of the indexes [1,2,3,4,5]. Since each element has a difference of exactly 1 from its previous element, our selection is linear.
+The sum of all the elements is 35 which is the maximum possible some out of every selection.</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= prices.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= prices[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：哈希表**
+
+我们可以将式子进行变换，得到：
+
+$$
+prices[i] - i = prices[j] - j
+$$
+
+题目实际上求的是相同的 $prices[i] - i$ 下，所有 $prices[i]$ 的和的最大值和。
+
+因此，我们可以用一个哈希表 $cnt$ 来存储 $prices[i] - i$ 下，所有 $prices[i]$ 的和，最后取哈希表中的最大值即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $prices$ 的长度。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def maxScore(self, prices: List[int]) -> int:
+        cnt = Counter()
+        for i, x in enumerate(prices):
+            cnt[x - i] += x
+        return max(cnt.values())
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public long maxScore(int[] prices) {
+        Map<Integer, Long> cnt = new HashMap<>();
+        for (int i = 0; i < prices.length; ++i) {
+            cnt.merge(prices[i] - i, (long) prices[i], Long::sum);
+        }
+        long ans = 0;
+        for (long v : cnt.values()) {
+            ans = Math.max(ans, v);
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    long long maxScore(vector<int>& prices) {
+        unordered_map<int, long long> cnt;
+        for (int i = 0; i < prices.size(); ++i) {
+            cnt[prices[i] - i] += prices[i];
+        }
+        long long ans = 0;
+        for (auto& [_, v] : cnt) {
+            ans = max(ans, v);
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func maxScore(prices []int) (ans int64) {
+	cnt := map[int]int{}
+	for i, x := range prices {
+		cnt[x-i] += x
+	}
+	for _, v := range cnt {
+		ans = max(ans, int64(v))
+	}
+	return
+}
+
+func max(a, b int64) int64 {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+### **TypeScript**
+
+```ts
+function maxScore(prices: number[]): number {
+    const cnt: Map<number, number> = new Map();
+    for (let i = 0; i < prices.length; ++i) {
+        const j = prices[i] - i;
+        cnt.set(j, (cnt.get(j) || 0) + prices[i]);
+    }
+    return Math.max(...cnt.values());
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2800-2899/2898.Maximum Linear Stock Score/README_EN.md

@@ -0,0 +1,161 @@
+# [2898. Maximum Linear Stock Score](https://leetcode.com/problems/maximum-linear-stock-score)
+
+[中文文档](/solution/2800-2899/2898.Maximum%20Linear%20Stock%20Score/README.md)
+
+## Description
+
+<p>Given a <strong>1-indexed</strong> integer array <code>prices</code>, where <code>prices[i]</code> is the price of a particular stock on the <code>i<sup>th</sup></code> day, your task is to select some of the elements of <code>prices</code> such that your selection is <strong>linear</strong>.</p>
+
+<p>A selection <code>indexes</code>, where <code>indexes</code> is a <strong>1-indexed</strong> integer array of length <code>k</code> which is a subsequence of the array <code>[1, 2, ..., n]</code>, is <strong>linear</strong> if:</p>
+
+<ul>
+	<li>For every <code>1 &lt; j &lt;= k</code>, <code>prices[indexes[j]] - prices[indexes[j - 1]] == indexes[j] - indexes[j - 1]</code>.</li>
+</ul>
+
+<p>A <b>subsequence</b> is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.</p>
+
+<p>The <strong>score</strong> of a selection <code>indexes</code>, is equal to the sum of the following array: <code>[prices[indexes[1]], prices[indexes[2]], ..., prices[indexes[k]]</code>.</p>
+
+<p>Return <em>the <strong>maximum</strong> <strong>score</strong> that a linear selection can have</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [1,5,3,7,8]
+<strong>Output:</strong> 20
+<strong>Explanation:</strong> We can select the indexes [2,4,5]. We show that our selection is linear:
+For j = 2, we have:
+indexes[2] - indexes[1] = 4 - 2 = 2.
+prices[4] - prices[2] = 7 - 5 = 2.
+For j = 3, we have:
+indexes[3] - indexes[2] = 5 - 4 = 1.
+prices[5] - prices[4] = 8 - 7 = 1.
+The sum of the elements is: prices[2] + prices[4] + prices[5] = 20.
+It can be shown that the maximum sum a linear selection can have is 20.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [5,6,7,8,9]
+<strong>Output:</strong> 35
+<strong>Explanation:</strong> We can select all of the indexes [1,2,3,4,5]. Since each element has a difference of exactly 1 from its previous element, our selection is linear.
+The sum of all the elements is 35 which is the maximum possible some out of every selection.</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= prices.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= prices[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+## Solutions
+
+**Solution 1: Hash Table**
+
+We can transform the equation as follows:
+
+$$
+prices[i] - i = prices[j] - j
+$$
+
+In fact, the problem is to find the maximum sum of all $prices[i]$ under the same $prices[i] - i$.
+
+Therefore, we can use a hash table $cnt$ to store the sum of all $prices[i]$ under the same $prices[i] - i$, and finally take the maximum value in the hash table.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the $prices$ array.
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def maxScore(self, prices: List[int]) -> int:
+        cnt = Counter()
+        for i, x in enumerate(prices):
+            cnt[x - i] += x
+        return max(cnt.values())
+```
+
+### **Java**
+
+```java
+class Solution {
+    public long maxScore(int[] prices) {
+        Map<Integer, Long> cnt = new HashMap<>();
+        for (int i = 0; i < prices.length; ++i) {
+            cnt.merge(prices[i] - i, (long) prices[i], Long::sum);
+        }
+        long ans = 0;
+        for (long v : cnt.values()) {
+            ans = Math.max(ans, v);
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    long long maxScore(vector<int>& prices) {
+        unordered_map<int, long long> cnt;
+        for (int i = 0; i < prices.size(); ++i) {
+            cnt[prices[i] - i] += prices[i];
+        }
+        long long ans = 0;
+        for (auto& [_, v] : cnt) {
+            ans = max(ans, v);
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func maxScore(prices []int) (ans int64) {
+	cnt := map[int]int{}
+	for i, x := range prices {
+		cnt[x-i] += x
+	}
+	for _, v := range cnt {
+		ans = max(ans, int64(v))
+	}
+	return
+}
+
+func max(a, b int64) int64 {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+### **TypeScript**
+
+```ts
+function maxScore(prices: number[]): number {
+    const cnt: Map<number, number> = new Map();
+    for (let i = 0; i < prices.length; ++i) {
+        const j = prices[i] - i;
+        cnt.set(j, (cnt.get(j) || 0) + prices[i]);
+    }
+    return Math.max(...cnt.values());
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2800-2899/2898.Maximum Linear Stock Score/Solution.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+    long long maxScore(vector<int>& prices) {
+        unordered_map<int, long long> cnt;
+        for (int i = 0; i < prices.size(); ++i) {
+            cnt[prices[i] - i] += prices[i];
+        }
+        long long ans = 0;
+        for (auto& [_, v] : cnt) {
+            ans = max(ans, v);
+        }
+        return ans;
+    }
+};

solution/2800-2899/2898.Maximum Linear Stock Score/Solution.go

@@ -0,0 +1,17 @@
+func maxScore(prices []int) (ans int64) {
+	cnt := map[int]int{}
+	for i, x := range prices {
+		cnt[x-i] += x
+	}
+	for _, v := range cnt {
+		ans = max(ans, int64(v))
+	}
+	return
+}
+
+func max(a, b int64) int64 {
+	if a > b {
+		return a
+	}
+	return b
+}

solution/2800-2899/2898.Maximum Linear Stock Score/Solution.java

@@ -0,0 +1,13 @@
+class Solution {
+    public long maxScore(int[] prices) {
+        Map<Integer, Long> cnt = new HashMap<>();
+        for (int i = 0; i < prices.length; ++i) {
+            cnt.merge(prices[i] - i, (long) prices[i], Long::sum);
+        }
+        long ans = 0;
+        for (long v : cnt.values()) {
+            ans = Math.max(ans, v);
+        }
+        return ans;
+    }
+}

solution/2800-2899/2898.Maximum Linear Stock Score/Solution.py

@@ -0,0 +1,6 @@
+class Solution:
+    def maxScore(self, prices: List[int]) -> int:
+        cnt = Counter()
+        for i, x in enumerate(prices):
+            cnt[x - i] += x
+        return max(cnt.values())

solution/2800-2899/2898.Maximum Linear Stock Score/Solution.ts

@@ -0,0 +1,8 @@
+function maxScore(prices: number[]): number {
+    const cnt: Map<number, number> = new Map();
+    for (let i = 0; i < prices.length; ++i) {
+        const j = prices[i] - i;
+        cnt.set(j, (cnt.get(j) || 0) + prices[i]);
+    }
+    return Math.max(...cnt.values());
+}