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Commit faa39e9

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feat: add solutions to lc problem: No.3259 (doocs#3699)
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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@@ -180,4 +180,90 @@ function maxEnergyBoost(energyDrinkA: number[], energyDrinkB: number[]): number
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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二:动态规划(空间优化)
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我们注意到,状态 $f[i]$ 至于 $f[i - 1]$ 有关,而与 $f[i - 2]$ 无关。因此我们可以只使用两个变量 $f$ 和 $g$ 来维护状态,从而将空间复杂度优化到 $O(1)$。
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时间复杂度 $O(n),ドル其中 $n$ 为数组的长度。空间复杂度 $O(1)$。
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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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f, g = energyDrinkA[0], energyDrinkB[0]
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for a, b in zip(energyDrinkA[1:], energyDrinkB[1:]):
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f, g = max(f + a, g), max(g + b, f)
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return max(f, g)
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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 = energyDrinkA[0], g = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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long ff = Math.max(f + energyDrinkA[i], g);
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g = Math.max(g + energyDrinkB[i], f);
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f = ff;
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}
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return Math.max(f, g);
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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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long long f = energyDrinkA[0], g = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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long long ff = max(f + energyDrinkA[i], g);
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g = max(g + energyDrinkB[i], f);
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f = ff;
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}
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return max(f, g);
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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, g := energyDrinkA[0], energyDrinkB[0]
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for i := 1; i < n; i++ {
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f, g = max(f+energyDrinkA[i], g), max(g+energyDrinkB[i], f)
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}
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return int64(max(f, g))
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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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let [f, g] = [energyDrinkA[0], energyDrinkB[0]];
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for (let i = 1; i < n; ++i) {
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[f, g] = [Math.max(f + energyDrinkA[i], g), Math.max(g + energyDrinkB[i], f)];
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}
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return Math.max(f, g);
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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

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<!-- solution:end -->
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<!-- solution:start -->
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### Solution 2: Dynamic Programming (Space Optimization)
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We notice that the state $f[i]$ is only related to $f[i - 1]$ and not to $f[i - 2]$. Therefore, we can use only two variables $f$ and $g$ to maintain the state, thus optimizing the space complexity to $O(1)$.
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The time complexity is $O(n),ドル where $n$ is the length of the array. The space complexity is $O(1)$.
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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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f, g = energyDrinkA[0], energyDrinkB[0]
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for a, b in zip(energyDrinkA[1:], energyDrinkB[1:]):
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f, g = max(f + a, g), max(g + b, f)
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return max(f, g)
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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 = energyDrinkA[0], g = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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long ff = Math.max(f + energyDrinkA[i], g);
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g = Math.max(g + energyDrinkB[i], f);
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f = ff;
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}
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return Math.max(f, g);
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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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long long f = energyDrinkA[0], g = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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long long ff = max(f + energyDrinkA[i], g);
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g = max(g + energyDrinkB[i], f);
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f = ff;
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}
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return max(f, g);
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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, g := energyDrinkA[0], energyDrinkB[0]
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for i := 1; i < n; i++ {
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f, g = max(f+energyDrinkA[i], g), max(g+energyDrinkB[i], f)
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}
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return int64(max(f, g))
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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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let [f, g] = [energyDrinkA[0], energyDrinkB[0]];
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for (let i = 1; i < n; ++i) {
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[f, g] = [Math.max(f + energyDrinkA[i], g), Math.max(g + energyDrinkB[i], f)];
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}
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return Math.max(f, g);
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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183269
<!-- problem:end -->
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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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long long f = energyDrinkA[0], g = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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long long ff = max(f + energyDrinkA[i], g);
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g = max(g + energyDrinkB[i], f);
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f = ff;
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}
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return max(f, g);
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}
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};
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func maxEnergyBoost(energyDrinkA []int, energyDrinkB []int) int64 {
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n := len(energyDrinkA)
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f, g := energyDrinkA[0], energyDrinkB[0]
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for i := 1; i < n; i++ {
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f, g = max(f+energyDrinkA[i], g), max(g+energyDrinkB[i], f)
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}
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return int64(max(f, g))
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}
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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 = energyDrinkA[0], g = energyDrinkB[0];
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for (int i = 1; i < n; ++i) {
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long ff = Math.max(f + energyDrinkA[i], g);
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g = Math.max(g + energyDrinkB[i], f);
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f = ff;
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}
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return Math.max(f, g);
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}
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}
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class Solution:
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def maxEnergyBoost(self, energyDrinkA: List[int], energyDrinkB: List[int]) -> int:
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f, g = energyDrinkA[0], energyDrinkB[0]
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for a, b in zip(energyDrinkA[1:], energyDrinkB[1:]):
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f, g = max(f + a, g), max(g + b, f)
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return max(f, g)
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function maxEnergyBoost(energyDrinkA: number[], energyDrinkB: number[]): number {
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const n = energyDrinkA.length;
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let [f, g] = [energyDrinkA[0], energyDrinkB[0]];
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for (let i = 1; i < n; ++i) {
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[f, g] = [Math.max(f + energyDrinkA[i], g), Math.max(g + energyDrinkB[i], f)];
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}
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return Math.max(f, g);
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}

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