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feat: add solutions to lc problem: No.2383
No.2383.Minimum Hours of Training to Win a Competition
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‎solution/2300-2399/2383.Minimum Hours of Training to Win a Competition/README.md

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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 是对手的数量。
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**方法二：贪心**
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我们可以先在初始时，把精力直接补充到足够击败这 $n$ 个对手，因此初始训练小时数为 $ans = \max(0, (\sum_{i=0}^{n-1} energy[i]) - initialEnergy + 1)$。
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接下来我们只需考虑经验值的问题。遍历 $n$ 个对手，若当前经验不足以超过对手，则将经验补到刚好能超过该对手，击败对手后，把对手的经验值加到自己身上。
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遍历结束，返回训练的小时数。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 是对手的数量。
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<!-- tabs:start -->
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### **Python3**
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return ans
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```
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```python
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class Solution:
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def minNumberOfHours(self, initialEnergy: int, initialExperience: int, energy: List[int], experience: List[int]) -> int:
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ans = max(0, sum(energy) - initialEnergy + 1)
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for x in experience:
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if initialExperience <= x:
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ans += x - initialExperience + 1
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initialExperience = x + 1
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initialExperience += x
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return ans
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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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}
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```
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```java
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class Solution {
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public int minNumberOfHours(int initialEnergy, int initialExperience, int[] energy, int[] experience) {
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int s = 0;
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for (int x : energy) {
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s += x;
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}
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int ans = Math.max(0, s - initialEnergy + 1);
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for (int x : experience) {
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if (initialExperience <= x) {
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ans += x - initialExperience + 1;
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initialExperience = x + 1;
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}
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initialExperience += x;
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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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};
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```
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```cpp
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class Solution {
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public:
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int minNumberOfHours(int initialEnergy, int initialExperience, vector<int>& energy, vector<int>& experience) {
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int s = accumulate(energy.begin(), energy.end(), 0);
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int ans = max(0, s - initialEnergy + 1);
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for (int x : experience) {
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if (initialExperience <= x) {
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ans += x - initialExperience + 1;
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initialExperience = x + 1;
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}
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initialExperience += x;
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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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}
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```
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```go
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func minNumberOfHours(initialEnergy int, initialExperience int, energy []int, experience []int) (ans int) {
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s := 0
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for _, x := range energy {
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s += x
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}
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if y := s - initialEnergy + 1; y > 0 {
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ans = y
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}
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for _, x := range experience {
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if initialExperience <= x {
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ans += x - initialExperience + 1
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initialExperience = x + 1
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}
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initialExperience += x
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}
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return
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}
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```
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### **TypeScript**
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```ts
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}
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```
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```ts
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function minNumberOfHours(
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initialEnergy: number,
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initialExperience: number,
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energy: number[],
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experience: number[],
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): number {
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const s = energy.reduce((a, b) => a + b, 0);
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let ans = Math.max(0, s - initialEnergy + 1);
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for (const x of experience) {
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if (initialExperience <= x) {
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ans += x - initialExperience + 1;
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initialExperience = x + 1;
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}
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initialExperience += x;
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}
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return ans;
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}
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```
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### **C**
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```c

‎solution/2300-2399/2383.Minimum Hours of Training to Win a Competition/README_EN.md

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return ans
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```
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```python
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class Solution:
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def minNumberOfHours(self, initialEnergy: int, initialExperience: int, energy: List[int], experience: List[int]) -> int:
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ans = max(0, sum(energy) - initialEnergy + 1)
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for x in experience:
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if initialExperience <= x:
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ans += x - initialExperience + 1
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initialExperience = x + 1
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initialExperience += x
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return ans
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```
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### **Java**
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```java
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}
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```
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```java
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class Solution {
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public int minNumberOfHours(int initialEnergy, int initialExperience, int[] energy, int[] experience) {
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int s = 0;
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for (int x : energy) {
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s += x;
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}
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int ans = Math.max(0, s - initialEnergy + 1);
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for (int x : experience) {
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if (initialExperience <= x) {
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ans += x - initialExperience + 1;
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initialExperience = x + 1;
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}
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initialExperience += x;
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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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};
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```
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```cpp
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class Solution {
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public:
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int minNumberOfHours(int initialEnergy, int initialExperience, vector<int>& energy, vector<int>& experience) {
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int s = accumulate(energy.begin(), energy.end(), 0);
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int ans = max(0, s - initialEnergy + 1);
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for (int x : experience) {
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if (initialExperience <= x) {
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ans += x - initialExperience + 1;
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initialExperience = x + 1;
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}
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initialExperience += x;
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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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}
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```
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```go
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func minNumberOfHours(initialEnergy int, initialExperience int, energy []int, experience []int) (ans int) {
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s := 0
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for _, x := range energy {
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s += x
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}
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if y := s - initialEnergy + 1; y > 0 {
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ans = y
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}
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for _, x := range experience {
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if initialExperience <= x {
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ans += x - initialExperience + 1
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initialExperience = x + 1
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}
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initialExperience += x
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}
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return
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}
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```
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### **TypeScript**
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```ts
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}
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```
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```ts
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function minNumberOfHours(
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initialEnergy: number,
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initialExperience: number,
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energy: number[],
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experience: number[],
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): number {
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const s = energy.reduce((a, b) => a + b, 0);
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let ans = Math.max(0, s - initialEnergy + 1);
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for (const x of experience) {
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if (initialExperience <= x) {
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ans += x - initialExperience + 1;
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initialExperience = x + 1;
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}
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initialExperience += x;
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}
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return ans;
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}
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```
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### **C**
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```c

‎solution/2300-2399/2384.Largest Palindromic Number/README.md

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我们去除回文串的前导零，若回文串为空，则返回“0”。否则返回该回文串。
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时间复杂度 $O(n)$。
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时间复杂度 $O(n)$，其中 $n$ 为 $num$ 的长度。
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<!-- tabs:start -->
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