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Commit 520728e

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feat: add solutions to lc problem: No.3652 (doocs#4728)
1 parent eb6d08f commit 520728e

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‎solution/3600-3699/3652.Best Time to Buy and Sell Stock using Strategy/README.md‎

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<!-- solution:start -->
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### 方法一
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### 方法一:前缀和 + 枚举
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我们用一个数组 $\textit{s}$ 来表示前缀和,其中 $\textit{s}[i]$ 表示前 $i$ 天的利润和,即 $\textit{s}[i] = \sum_{j=0}^{i-1} \textit{prices}[j] \times \textit{strategy}[j]$。我们还用一个数组 $\textit{t}$ 来表示前缀和,其中 $\textit{t}[i]$ 表示前 $i$ 天的股票价格和,即 $\textit{t}[i] = \sum_{j=0}^{i-1} \textit{prices}[j]$。
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初始时,最大利润为 $\textit{s}[n]$。我们枚举修改的子数组的右端点 $i,ドル则左端点为 $i-k$。修改后,子数组内前 $k/2$ 天的策略变为 0ドル,ドル后 $k/2$ 天的策略变为 1ドル,ドル因此利润变化为:
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$$\Delta = -(\textit{s}[i] - \textit{s}[i-k]) + (\textit{t}[i] - \textit{t}[i-k/2])$$
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因此,我们可以通过枚举所有可能的 $i$ 来更新最大利润。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组的长度。
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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 maxProfit(self, prices: List[int], strategy: List[int], k: int) -> int:
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n = len(prices)
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s = [0] * (n + 1)
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t = [0] * (n + 1)
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for i, (a, b) in enumerate(zip(prices, strategy), 1):
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s[i] = s[i - 1] + a * b
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t[i] = t[i - 1] + a
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ans = s[n]
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for i in range(k, n + 1):
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ans = max(ans, s[n] - (s[i] - s[i - k]) + t[i] - t[i - k // 2])
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return ans
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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 maxProfit(int[] prices, int[] strategy, int k) {
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int n = prices.length;
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long[] s = new long[n + 1];
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long[] t = new long[n + 1];
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for (int i = 1; i <= n; i++) {
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int a = prices[i - 1];
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int b = strategy[i - 1];
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s[i] = s[i - 1] + a * b;
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t[i] = t[i - 1] + a;
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}
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long ans = s[n];
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for (int i = k; i <= n; i++) {
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ans = Math.max(ans, s[n] - (s[i] - s[i - k]) + (t[i] - t[i - k / 2]));
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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 maxProfit(vector<int>& prices, vector<int>& strategy, int k) {
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int n = prices.size();
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vector<long long> s(n + 1), t(n + 1);
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for (int i = 1; i <= n; i++) {
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int a = prices[i - 1];
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int b = strategy[i - 1];
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s[i] = s[i - 1] + a * b;
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t[i] = t[i - 1] + a;
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}
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long long ans = s[n];
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for (int i = k; i <= n; i++) {
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ans = max(ans, s[n] - (s[i] - s[i - k]) + (t[i] - t[i - k / 2]));
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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 maxProfit(prices []int, strategy []int, k int) int64 {
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n := len(prices)
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s := make([]int64, n+1)
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t := make([]int64, n+1)
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for i := 1; i <= n; i++ {
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a := prices[i-1]
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b := strategy[i-1]
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s[i] = s[i-1] + int64(a*b)
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t[i] = t[i-1] + int64(a)
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}
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ans := s[n]
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for i := k; i <= n; i++ {
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ans = max(ans, s[n]-(s[i]-s[i-k])+(t[i]-t[i-k/2]))
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}
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return ans
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}
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```
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#### TypeScript
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```ts
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function maxProfit(prices: number[], strategy: number[], k: number): number {
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const n = prices.length;
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const s: number[] = Array(n + 1).fill(0);
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const t: number[] = Array(n + 1).fill(0);
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for (let i = 1; i <= n; i++) {
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const a = prices[i - 1];
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const b = strategy[i - 1];
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s[i] = s[i - 1] + a * b;
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t[i] = t[i - 1] + a;
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}
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let ans = s[n];
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for (let i = k; i <= n; i++) {
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const val = s[n] - (s[i] - s[i - k]) + (t[i] - t[i - Math.floor(k / 2)]);
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ans = Math.max(ans, val);
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}
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return ans;
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}
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```
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<!-- tabs:end -->

‎solution/3600-3699/3652.Best Time to Buy and Sell Stock using Strategy/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Prefix Sum + Enumeration
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We use an array $\textit{s}$ to represent the prefix sum, where $\textit{s}[i]$ is the total profit for the first $i$ days, i.e., $\textit{s}[i] = \sum_{j=0}^{i-1} \textit{prices}[j] \times \textit{strategy}[j]$. We also use an array $\textit{t}$ to represent the prefix sum of stock prices, where $\textit{t}[i] = \sum_{j=0}^{i-1} \textit{prices}[j]$.
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Initially, the maximum profit is $\textit{s}[n]$. We enumerate the right endpoint $i$ of the subarray to be modified, with the left endpoint being $i-k$. After modification, the first $k/2$ days of the subarray have strategy 0ドル,ドル and the last $k/2$ days have strategy 1ドル,ドル so the profit change is:
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$$\Delta = -(\textit{s}[i] - \textit{s}[i-k]) + (\textit{t}[i] - \textit{t}[i-k/2])$$
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Therefore, we can update the maximum profit by enumerating all possible $i$.
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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 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 maxProfit(self, prices: List[int], strategy: List[int], k: int) -> int:
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n = len(prices)
176+
s = [0] * (n + 1)
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t = [0] * (n + 1)
178+
for i, (a, b) in enumerate(zip(prices, strategy), 1):
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s[i] = s[i - 1] + a * b
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t[i] = t[i - 1] + a
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ans = s[n]
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for i in range(k, n + 1):
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ans = max(ans, s[n] - (s[i] - s[i - k]) + t[i] - t[i - k // 2])
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return ans
164185
```
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166187
#### Java
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168189
```java
169-
190+
class Solution {
191+
public long maxProfit(int[] prices, int[] strategy, int k) {
192+
int n = prices.length;
193+
long[] s = new long[n + 1];
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long[] t = new long[n + 1];
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for (int i = 1; i <= n; i++) {
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int a = prices[i - 1];
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int b = strategy[i - 1];
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s[i] = s[i - 1] + a * b;
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t[i] = t[i - 1] + a;
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}
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long ans = s[n];
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for (int i = k; i <= n; i++) {
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ans = Math.max(ans, s[n] - (s[i] - s[i - k]) + (t[i] - t[i - k / 2]));
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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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174212
```cpp
175-
213+
class Solution {
214+
public:
215+
long long maxProfit(vector<int>& prices, vector<int>& strategy, int k) {
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int n = prices.size();
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vector<long long> s(n + 1), t(n + 1);
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for (int i = 1; i <= n; i++) {
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int a = prices[i - 1];
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int b = strategy[i - 1];
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s[i] = s[i - 1] + a * b;
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t[i] = t[i - 1] + a;
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}
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long long ans = s[n];
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for (int i = k; i <= n; i++) {
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ans = max(ans, s[n] - (s[i] - s[i - k]) + (t[i] - t[i - k / 2]));
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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 maxProfit(prices []int, strategy []int, k int) int64 {
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n := len(prices)
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s := make([]int64, n+1)
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t := make([]int64, n+1)
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for i := 1; i <= n; i++ {
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a := prices[i-1]
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b := strategy[i-1]
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s[i] = s[i-1] + int64(a*b)
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t[i] = t[i-1] + int64(a)
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}
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ans := s[n]
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for i := k; i <= n; i++ {
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ans = max(ans, s[n]-(s[i]-s[i-k])+(t[i]-t[i-k/2]))
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}
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return ans
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}
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```
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#### TypeScript
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```ts
259+
function maxProfit(prices: number[], strategy: number[], k: number): number {
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const n = prices.length;
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const s: number[] = Array(n + 1).fill(0);
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const t: number[] = Array(n + 1).fill(0);
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for (let i = 1; i <= n; i++) {
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const a = prices[i - 1];
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const b = strategy[i - 1];
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s[i] = s[i - 1] + a * b;
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t[i] = t[i - 1] + a;
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}
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let ans = s[n];
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for (let i = k; i <= n; i++) {
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const val = s[n] - (s[i] - s[i - k]) + (t[i] - t[i - Math.floor(k / 2)]);
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ans = Math.max(ans, val);
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}
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return ans;
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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 maxProfit(vector<int>& prices, vector<int>& strategy, int k) {
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int n = prices.size();
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vector<long long> s(n + 1), t(n + 1);
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for (int i = 1; i <= n; i++) {
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int a = prices[i - 1];
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int b = strategy[i - 1];
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s[i] = s[i - 1] + a * b;
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t[i] = t[i - 1] + a;
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}
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long long ans = s[n];
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for (int i = k; i <= n; i++) {
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ans = max(ans, s[n] - (s[i] - s[i - k]) + (t[i] - t[i - k / 2]));
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}
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return ans;
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}
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};
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func maxProfit(prices []int, strategy []int, k int) int64 {
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n := len(prices)
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s := make([]int64, n+1)
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t := make([]int64, n+1)
5+
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for i := 1; i <= n; i++ {
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a := prices[i-1]
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b := strategy[i-1]
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s[i] = s[i-1] + int64(a*b)
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t[i] = t[i-1] + int64(a)
11+
}
12+
13+
ans := s[n]
14+
for i := k; i <= n; i++ {
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ans = max(ans, s[n]-(s[i]-s[i-k])+(t[i]-t[i-k/2]))
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}
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return ans
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}
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class Solution {
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public long maxProfit(int[] prices, int[] strategy, int k) {
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int n = prices.length;
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long[] s = new long[n + 1];
5+
long[] t = new long[n + 1];
6+
for (int i = 1; i <= n; i++) {
7+
int a = prices[i - 1];
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int b = strategy[i - 1];
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s[i] = s[i - 1] + a * b;
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t[i] = t[i - 1] + a;
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}
12+
long ans = s[n];
13+
for (int i = k; i <= n; i++) {
14+
ans = Math.max(ans, s[n] - (s[i] - s[i - k]) + (t[i] - t[i - k / 2]));
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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 maxProfit(self, prices: List[int], strategy: List[int], k: int) -> int:
3+
n = len(prices)
4+
s = [0] * (n + 1)
5+
t = [0] * (n + 1)
6+
for i, (a, b) in enumerate(zip(prices, strategy), 1):
7+
s[i] = s[i - 1] + a * b
8+
t[i] = t[i - 1] + a
9+
ans = s[n]
10+
for i in range(k, n + 1):
11+
ans = max(ans, s[n] - (s[i] - s[i - k]) + t[i] - t[i - k // 2])
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return ans
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function maxProfit(prices: number[], strategy: number[], k: number): number {
2+
const n = prices.length;
3+
const s: number[] = Array(n + 1).fill(0);
4+
const t: number[] = Array(n + 1).fill(0);
5+
6+
for (let i = 1; i <= n; i++) {
7+
const a = prices[i - 1];
8+
const b = strategy[i - 1];
9+
s[i] = s[i - 1] + a * b;
10+
t[i] = t[i - 1] + a;
11+
}
12+
13+
let ans = s[n];
14+
for (let i = k; i <= n; i++) {
15+
const val = s[n] - (s[i] - s[i - k]) + (t[i] - t[i - Math.floor(k / 2)]);
16+
ans = Math.max(ans, val);
17+
}
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return ans;
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}

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