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

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169 changes: 169 additions & 0 deletions solution/2800-2899/2898.Maximum Linear Stock Score/README.md
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# [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),ドル空间复杂度 $O(n)$。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 -->
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# [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,ドル and finally take the maximum value in the hash table.

The time complexity is $O(n),ドル and the space complexity is $O(n),ドル 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 -->
14 changes: 14 additions & 0 deletions solution/2800-2899/2898.Maximum Linear Stock Score/Solution.cpp
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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;
}
};
17 changes: 17 additions & 0 deletions solution/2800-2899/2898.Maximum Linear Stock Score/Solution.go
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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
}
13 changes: 13 additions & 0 deletions solution/2800-2899/2898.Maximum Linear Stock Score/Solution.java
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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;
}
}
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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())
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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());
}
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