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feat: add solutions to lc problem: No.3259 (#3433)
No.3259.Maximum Energy Boost From Two Drinks
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‎solution/3200-3299/3259.Maximum Energy Boost From Two Drinks/README.md

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<!-- solution:start -->
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### 方法一
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### 方法一：动态规划
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我们定义 $f[i][0]$ 表示在第 $i$ 小时选择能量饮料 A 获得的最大强化能量，定义 $f[i][1]$ 表示在第 $i$ 小时选择能量饮料 B 获得的最大强化能量。初始时 $f[0][0] = \textit{energyDrinkA}[0]$, $f[0][1] = \textit{energyDrinkB}[0]$。答案为 $\max(f[n - 1][0], f[n - 1][1])$。
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对于 $i > 0$，我们有以下状态转移方程：
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$$
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\begin{aligned}
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f[i][0] & = \max(f[i - 1][0] + \textit{energyDrinkA}[i], f[i - 1][1]) \\
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f[i][1] & = \max(f[i - 1][1] + \textit{energyDrinkB}[i], f[i - 1][0])
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\end{aligned}
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$$
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最后返回 $\max(f[n - 1][0], f[n - 1][1])$ 即可。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组的长度。
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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 maxEnergyBoost(self, energyDrinkA: List[int], energyDrinkB: List[int]) -> int:
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n = len(energyDrinkA)
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f = [[0] * 2 for _ in range(n)]
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f[0][0] = energyDrinkA[0]
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f[0][1] = energyDrinkB[0]
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for i in range(1, n):
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f[i][0] = max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1])
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f[i][1] = max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0])
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return max(f[n - 1])
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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 maxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {
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int n = energyDrinkA.length;
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long[][] f = new long[n][2];
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f[0][0] = energyDrinkA[0];
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f[0][1] = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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f[i][0] = Math.max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
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f[i][1] = Math.max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
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}
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return Math.max(f[n - 1][0], f[n - 1][1]);
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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 maxEnergyBoost(vector<int>& energyDrinkA, vector<int>& energyDrinkB) {
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int n = energyDrinkA.size();
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vector<vector<long long>> f(n, vector<long long>(2));
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f[0][0] = energyDrinkA[0];
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f[0][1] = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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f[i][0] = max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
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f[i][1] = max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
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}
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return max(f[n - 1][0], f[n - 1][1]);
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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 maxEnergyBoost(energyDrinkA []int, energyDrinkB []int) int64 {
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n := len(energyDrinkA)
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f := make([][2]int64, n)
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f[0][0] = int64(energyDrinkA[0])
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f[0][1] = int64(energyDrinkB[0])
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for i := 1; i < n; i++ {
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f[i][0] = max(f[i-1][0]+int64(energyDrinkA[i]), f[i-1][1])
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f[i][1] = max(f[i-1][1]+int64(energyDrinkB[i]), f[i-1][0])
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}
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return max(f[n-1][0], f[n-1][1])
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}
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```
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#### TypeScript
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```ts
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function maxEnergyBoost(energyDrinkA: number[], energyDrinkB: number[]): number {
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const n = energyDrinkA.length;
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const f: number[][] = Array.from({ length: n }, () => [0, 0]);
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f[0][0] = energyDrinkA[0];
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f[0][1] = energyDrinkB[0];
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for (let i = 1; i < n; i++) {
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f[i][0] = Math.max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
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f[i][1] = Math.max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
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}
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return Math.max(...f[n - 1]!);
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}
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```
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<!-- tabs:end -->

‎solution/3200-3299/3259.Maximum Energy Boost From Two Drinks/README_EN.md

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Dynamic Programming
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We define $f[i][0]$ to represent the maximum boost energy obtained by choosing energy drink A at the $i$-th hour, and $f[i][1]$ to represent the maximum boost energy obtained by choosing energy drink B at the $i$-th hour. Initially, $f[0][0] = \textit{energyDrinkA}[0]$, $f[0][1] = \textit{energyDrinkB}[0]$. The answer is $\max(f[n - 1][0], f[n - 1][1])$.
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For $i > 0$, we have the following state transition equations:
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$$
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\begin{aligned}
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f[i][0] & = \max(f[i - 1][0] + \textit{energyDrinkA}[i], f[i - 1][1]) \\
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f[i][1] & = \max(f[i - 1][1] + \textit{energyDrinkB}[i], f[i - 1][0])
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\end{aligned}
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$$
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Finally, return $\max(f[n - 1][0], f[n - 1][1])$.
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The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ 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 maxEnergyBoost(self, energyDrinkA: List[int], energyDrinkB: List[int]) -> int:
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n = len(energyDrinkA)
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f = [[0] * 2 for _ in range(n)]
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f[0][0] = energyDrinkA[0]
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f[0][1] = energyDrinkB[0]
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for i in range(1, n):
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f[i][0] = max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1])
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f[i][1] = max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0])
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return max(f[n - 1])
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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 maxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {
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int n = energyDrinkA.length;
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long[][] f = new long[n][2];
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f[0][0] = energyDrinkA[0];
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f[0][1] = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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f[i][0] = Math.max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
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f[i][1] = Math.max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
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}
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return Math.max(f[n - 1][0], f[n - 1][1]);
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}
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}
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```
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#### C++
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```cpp
90-
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class Solution {
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public:
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long long maxEnergyBoost(vector<int>& energyDrinkA, vector<int>& energyDrinkB) {
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int n = energyDrinkA.size();
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vector<vector<long long>> f(n, vector<long long>(2));
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f[0][0] = energyDrinkA[0];
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f[0][1] = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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f[i][0] = max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
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f[i][1] = max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
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}
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return max(f[n - 1][0], f[n - 1][1]);
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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 maxEnergyBoost(energyDrinkA []int, energyDrinkB []int) int64 {
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n := len(energyDrinkA)
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f := make([][2]int64, n)
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f[0][0] = int64(energyDrinkA[0])
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f[0][1] = int64(energyDrinkB[0])
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for i := 1; i < n; i++ {
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f[i][0] = max(f[i-1][0]+int64(energyDrinkA[i]), f[i-1][1])
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f[i][1] = max(f[i-1][1]+int64(energyDrinkB[i]), f[i-1][0])
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}
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return max(f[n-1][0], f[n-1][1])
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}
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```
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#### TypeScript
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```ts
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function maxEnergyBoost(energyDrinkA: number[], energyDrinkB: number[]): number {
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const n = energyDrinkA.length;
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const f: number[][] = Array.from({ length: n }, () => [0, 0]);
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f[0][0] = energyDrinkA[0];
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f[0][1] = energyDrinkB[0];
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for (let i = 1; i < n; i++) {
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f[i][0] = Math.max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
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f[i][1] = Math.max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
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}
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return Math.max(...f[n - 1]!);
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}
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```
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<!-- tabs:end -->

‎solution/3200-3299/3259.Maximum Energy Boost From Two Drinks/Solution.cpp

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class Solution {
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public:
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long long maxEnergyBoost(vector<int>& energyDrinkA, vector<int>& energyDrinkB) {
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int n = energyDrinkA.size();
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vector<vector<long long>> f(n, vector<long long>(2));
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f[0][0] = energyDrinkA[0];
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f[0][1] = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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f[i][0] = max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
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f[i][1] = max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
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}
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return max(f[n - 1][0], f[n - 1][1]);
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}
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};

‎solution/3200-3299/3259.Maximum Energy Boost From Two Drinks/Solution.go

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func maxEnergyBoost(energyDrinkA []int, energyDrinkB []int) int64 {
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n := len(energyDrinkA)
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f := make([][2]int64, n)
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f[0][0] = int64(energyDrinkA[0])
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f[0][1] = int64(energyDrinkB[0])
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for i := 1; i < n; i++ {
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f[i][0] = max(f[i-1][0]+int64(energyDrinkA[i]), f[i-1][1])
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f[i][1] = max(f[i-1][1]+int64(energyDrinkB[i]), f[i-1][0])
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}
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return max(f[n-1][0], f[n-1][1])
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}

‎solution/3200-3299/3259.Maximum Energy Boost From Two Drinks/Solution.java

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class Solution {
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public long maxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {
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int n = energyDrinkA.length;
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long[][] f = new long[n][2];
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f[0][0] = energyDrinkA[0];
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f[0][1] = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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f[i][0] = Math.max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
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f[i][1] = Math.max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
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}
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return Math.max(f[n - 1][0], f[n - 1][1]);
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}
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}

‎solution/3200-3299/3259.Maximum Energy Boost From Two Drinks/Solution.py

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class Solution:
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def maxEnergyBoost(self, energyDrinkA: List[int], energyDrinkB: List[int]) -> int:
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n = len(energyDrinkA)
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f = [[0] * 2 for _ in range(n)]
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f[0][0] = energyDrinkA[0]
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f[0][1] = energyDrinkB[0]
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for i in range(1, n):
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f[i][0] = max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1])
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f[i][1] = max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0])
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return max(f[n - 1])

‎solution/3200-3299/3259.Maximum Energy Boost From Two Drinks/Solution.ts

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function maxEnergyBoost(energyDrinkA: number[], energyDrinkB: number[]): number {
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const n = energyDrinkA.length;
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const f: number[][] = Array.from({ length: n }, () => [0, 0]);
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f[0][0] = energyDrinkA[0];
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f[0][1] = energyDrinkB[0];
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for (let i = 1; i < n; i++) {
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f[i][0] = Math.max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
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f[i][1] = Math.max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
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}
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return Math.max(...f[n - 1]!);
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}

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