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feat: add solutions to lc problem: No.2898 (#1792)
No.2898.Maximum Linear Stock Score
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# [2898. Maximum Linear Stock Score](https://leetcode.cn/problems/maximum-linear-stock-score)
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[English Version](/solution/2800-2899/2898.Maximum%20Linear%20Stock%20Score/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<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>
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<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>
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<ul>
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<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>
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</ul>
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<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>
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<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>
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<p>Return <em>the <strong>maximum</strong> <strong>score</strong> that a linear selection can have</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> prices = [1,5,3,7,8]
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<strong>Output:</strong> 20
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<strong>Explanation:</strong> We can select the indexes [2,4,5]. We show that our selection is linear:
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For j = 2, we have:
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indexes[2] - indexes[1] = 4 - 2 = 2.
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prices[4] - prices[2] = 7 - 5 = 2.
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For j = 3, we have:
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indexes[3] - indexes[2] = 5 - 4 = 1.
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prices[5] - prices[4] = 8 - 7 = 1.
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The sum of the elements is: prices[2] + prices[4] + prices[5] = 20.
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It can be shown that the maximum sum a linear selection can have is 20.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> prices = [5,6,7,8,9]
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<strong>Output:</strong> 35
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<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.
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The sum of all the elements is 35 which is the maximum possible some out of every selection.</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= prices.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= prices[i] &lt;= 10<sup>9</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:哈希表**
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我们可以将式子进行变换,得到:
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$$
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prices[i] - i = prices[j] - j
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$$
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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。
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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。
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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。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maxScore(self, prices: List[int]) -> int:
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cnt = Counter()
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for i, x in enumerate(prices):
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cnt[x - i] += x
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return max(cnt.values())
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public long maxScore(int[] prices) {
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Map<Integer, Long> cnt = new HashMap<>();
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for (int i = 0; i < prices.length; ++i) {
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cnt.merge(prices[i] - i, (long) prices[i], Long::sum);
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}
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long ans = 0;
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for (long v : cnt.values()) {
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ans = Math.max(ans, v);
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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long long maxScore(vector<int>& prices) {
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unordered_map<int, long long> cnt;
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for (int i = 0; i < prices.size(); ++i) {
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cnt[prices[i] - i] += prices[i];
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}
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long long ans = 0;
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for (auto& [_, v] : cnt) {
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ans = max(ans, v);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxScore(prices []int) (ans int64) {
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cnt := map[int]int{}
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for i, x := range prices {
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cnt[x-i] += x
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}
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for _, v := range cnt {
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ans = max(ans, int64(v))
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}
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return
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}
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func max(a, b int64) int64 {
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if a > b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function maxScore(prices: number[]): number {
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const cnt: Map<number, number> = new Map();
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for (let i = 0; i < prices.length; ++i) {
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const j = prices[i] - i;
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cnt.set(j, (cnt.get(j) || 0) + prices[i]);
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}
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return Math.max(...cnt.values());
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2898. Maximum Linear Stock Score](https://leetcode.com/problems/maximum-linear-stock-score)
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[中文文档](/solution/2800-2899/2898.Maximum%20Linear%20Stock%20Score/README.md)
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## Description
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<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>
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<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>
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<ul>
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<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>
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</ul>
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<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>
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<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>
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<p>Return <em>the <strong>maximum</strong> <strong>score</strong> that a linear selection can have</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> prices = [1,5,3,7,8]
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<strong>Output:</strong> 20
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<strong>Explanation:</strong> We can select the indexes [2,4,5]. We show that our selection is linear:
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For j = 2, we have:
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indexes[2] - indexes[1] = 4 - 2 = 2.
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prices[4] - prices[2] = 7 - 5 = 2.
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For j = 3, we have:
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indexes[3] - indexes[2] = 5 - 4 = 1.
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prices[5] - prices[4] = 8 - 7 = 1.
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The sum of the elements is: prices[2] + prices[4] + prices[5] = 20.
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It can be shown that the maximum sum a linear selection can have is 20.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> prices = [5,6,7,8,9]
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<strong>Output:</strong> 35
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<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.
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The sum of all the elements is 35 which is the maximum possible some out of every selection.</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= prices.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= prices[i] &lt;= 10<sup>9</sup></code></li>
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</ul>
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## Solutions
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**Solution 1: Hash Table**
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We can transform the equation as follows:
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$$
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prices[i] - i = prices[j] - j
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$$
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In fact, the problem is to find the maximum sum of all $prices[i]$ under the same $prices[i] - i$.
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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.
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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.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def maxScore(self, prices: List[int]) -> int:
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cnt = Counter()
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for i, x in enumerate(prices):
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cnt[x - i] += x
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return max(cnt.values())
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```
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### **Java**
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```java
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class Solution {
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public long maxScore(int[] prices) {
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Map<Integer, Long> cnt = new HashMap<>();
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for (int i = 0; i < prices.length; ++i) {
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cnt.merge(prices[i] - i, (long) prices[i], Long::sum);
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}
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long ans = 0;
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for (long v : cnt.values()) {
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ans = Math.max(ans, v);
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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long long maxScore(vector<int>& prices) {
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unordered_map<int, long long> cnt;
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for (int i = 0; i < prices.size(); ++i) {
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cnt[prices[i] - i] += prices[i];
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}
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long long ans = 0;
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for (auto& [_, v] : cnt) {
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ans = max(ans, v);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxScore(prices []int) (ans int64) {
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cnt := map[int]int{}
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for i, x := range prices {
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cnt[x-i] += x
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}
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for _, v := range cnt {
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ans = max(ans, int64(v))
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}
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return
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}
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func max(a, b int64) int64 {
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if a > b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function maxScore(prices: number[]): number {
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const cnt: Map<number, number> = new Map();
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for (let i = 0; i < prices.length; ++i) {
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const j = prices[i] - i;
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cnt.set(j, (cnt.get(j) || 0) + prices[i]);
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}
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return Math.max(...cnt.values());
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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class Solution {
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public:
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long long maxScore(vector<int>& prices) {
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unordered_map<int, long long> cnt;
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for (int i = 0; i < prices.size(); ++i) {
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cnt[prices[i] - i] += prices[i];
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}
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long long ans = 0;
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for (auto& [_, v] : cnt) {
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ans = max(ans, v);
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}
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return ans;
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}
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};
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func maxScore(prices []int) (ans int64) {
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cnt := map[int]int{}
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for i, x := range prices {
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cnt[x-i] += x
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}
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for _, v := range cnt {
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ans = max(ans, int64(v))
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}
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return
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}
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func max(a, b int64) int64 {
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if a > b {
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return a
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}
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return b
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}
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class Solution {
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public long maxScore(int[] prices) {
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Map<Integer, Long> cnt = new HashMap<>();
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for (int i = 0; i < prices.length; ++i) {
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cnt.merge(prices[i] - i, (long) prices[i], Long::sum);
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}
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long ans = 0;
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for (long v : cnt.values()) {
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ans = Math.max(ans, v);
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}
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return ans;
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}
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}
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class Solution:
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def maxScore(self, prices: List[int]) -> int:
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cnt = Counter()
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for i, x in enumerate(prices):
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cnt[x - i] += x
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return max(cnt.values())
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function maxScore(prices: number[]): number {
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const cnt: Map<number, number> = new Map();
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for (let i = 0; i < prices.length; ++i) {
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const j = prices[i] - i;
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cnt.set(j, (cnt.get(j) || 0) + prices[i]);
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}
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return Math.max(...cnt.values());
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}

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