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feat: add solutions to lc problem: No.0948
No.0948.Bag of Tokens
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‎solution/0900-0999/0940.Distinct Subsequences II/README.md

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最后，我们需要对 $dp$ 中的所有元素求和，再对 $10^9+7$ 取余，即为答案。
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时间复杂度 $O(n\times C)$，其中 $n$ 是字符串 $s$ 的长度，$C$ 是字符集的大小，本题中 $C=26$。空间复杂度 $O(C)$。
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时间复杂度 $O(n\times C)$，其中 $n$ 是字符串 $s$ 的长度，而 $C$ 是字符集的大小，本题中 $C=26$。空间复杂度 $O(C)$。
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**方法二：优化的动态规划**
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‎solution/0900-0999/0948.Bag of Tokens/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 排序 + 双指针**
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令牌的使用方法有两种，一种是消耗能量得到分数，一种是消耗分数得到能量。显然，我们应该消耗尽可能少的能量来得到尽可能多的分数。
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因此，我们可以将令牌按照消耗能量的多少进行排序，然后使用双指针，一个指针从左向右遍历，一个指针从右向左遍历，每次遍历都尽可能地消耗能量得到分数，然后更新最大分数。如果当前能量不足以消耗当前令牌，那么我们就尝试使用分数来消耗当前令牌，如果分数不足以消耗当前令牌，那么我们就停止遍历。
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时间复杂度 $O(n\log n)$，空间复杂度 $O(n)$。其中 $n$ 为令牌的数量。
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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 bagOfTokensScore(self, tokens: List[int], power: int) -> int:
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tokens.sort()
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i, j = 0, len(tokens) - 1
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ans = t = 0
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while i <= j:
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if power >= tokens[i]:
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power -= tokens[i]
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i, t = i + 1, t + 1
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ans = max(ans, t)
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elif t:
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power += tokens[j]
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j, t = j - 1, t - 1
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else:
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break
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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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```java
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class Solution {
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public int bagOfTokensScore(int[] tokens, int power) {
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Arrays.sort(tokens);
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int i = 0, j = tokens.length - 1;
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int ans = 0, t = 0;
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while (i <= j) {
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if (power >= tokens[i]) {
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power -= tokens[i++];
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++t;
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ans = Math.max(ans, t);
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} else if (t > 0) {
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power += tokens[j--];
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--t;
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} else {
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break;
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}
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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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int bagOfTokensScore(vector<int>& tokens, int power) {
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sort(tokens.begin(), tokens.end());
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int i = 0, j = tokens.size() - 1;
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int ans = 0, t = 0;
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while (i <= j) {
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if (power >= tokens[i]) {
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power -= tokens[i++];
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ans = max(ans, ++t);
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} else if (t) {
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power += tokens[j--];
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--t;
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} else {
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break;
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}
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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 bagOfTokensScore(tokens []int, power int) int {
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sort.Ints(tokens)
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i, j := 0, len(tokens)-1
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ans, t := 0, 0
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for i <= j {
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if power >= tokens[i] {
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power -= tokens[i]
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i, t = i+1, t+1
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ans = max(ans, t)
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} else if t > 0 {
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power += tokens[j]
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j, t = j-1, t-1
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} else {
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break
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}
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}
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return ans
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}
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func max(a, b int) int {
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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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### **...**

‎solution/0900-0999/0948.Bag of Tokens/README_EN.md

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@@ -62,13 +62,102 @@ There is no need to play the 1<sup>st</sup> token since you cannot play it face
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### **Python3**
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```python
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class Solution:
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def bagOfTokensScore(self, tokens: List[int], power: int) -> int:
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tokens.sort()
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i, j = 0, len(tokens) - 1
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ans = t = 0
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while i <= j:
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if power >= tokens[i]:
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power -= tokens[i]
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i, t = i + 1, t + 1
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ans = max(ans, t)
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elif t:
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power += tokens[j]
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j, t = j - 1, t - 1
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else:
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break
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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 int bagOfTokensScore(int[] tokens, int power) {
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Arrays.sort(tokens);
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int i = 0, j = tokens.length - 1;
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int ans = 0, t = 0;
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while (i <= j) {
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if (power >= tokens[i]) {
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power -= tokens[i++];
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++t;
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ans = Math.max(ans, t);
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} else if (t > 0) {
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power += tokens[j--];
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--t;
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} else {
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break;
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}
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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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int bagOfTokensScore(vector<int>& tokens, int power) {
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sort(tokens.begin(), tokens.end());
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int i = 0, j = tokens.size() - 1;
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int ans = 0, t = 0;
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while (i <= j) {
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if (power >= tokens[i]) {
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power -= tokens[i++];
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ans = max(ans, ++t);
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} else if (t) {
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power += tokens[j--];
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--t;
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} else {
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break;
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}
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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 bagOfTokensScore(tokens []int, power int) int {
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sort.Ints(tokens)
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i, j := 0, len(tokens)-1
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ans, t := 0, 0
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for i <= j {
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if power >= tokens[i] {
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power -= tokens[i]
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i, t = i+1, t+1
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ans = max(ans, t)
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} else if t > 0 {
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power += tokens[j]
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j, t = j-1, t-1
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} else {
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break
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}
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}
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return ans
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}
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func max(a, b int) int {
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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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### **...**

‎solution/0900-0999/0948.Bag of Tokens/Solution.cpp

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class Solution {
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public:
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int bagOfTokensScore(vector<int>& tokens, int power) {
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sort(tokens.begin(), tokens.end());
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int i = 0, j = tokens.size() - 1;
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int ans = 0, t = 0;
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while (i <= j) {
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if (power >= tokens[i]) {
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power -= tokens[i++];
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ans = max(ans, ++t);
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} else if (t) {
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power += tokens[j--];
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--t;
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} else {
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break;
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}
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}
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return ans;
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}
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};

‎solution/0900-0999/0948.Bag of Tokens/Solution.go

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func bagOfTokensScore(tokens []int, power int) int {
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sort.Ints(tokens)
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i, j := 0, len(tokens)-1
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ans, t := 0, 0
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for i <= j {
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if power >= tokens[i] {
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power -= tokens[i]
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i, t = i+1, t+1
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ans = max(ans, t)
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} else if t > 0 {
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power += tokens[j]
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j, t = j-1, t-1
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} else {
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break
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}
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}
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return ans
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}
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func max(a, b int) int {
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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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}

‎solution/0900-0999/0948.Bag of Tokens/Solution.java

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@@ -0,0 +1,20 @@
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class Solution {
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public int bagOfTokensScore(int[] tokens, int power) {
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Arrays.sort(tokens);
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int i = 0, j = tokens.length - 1;
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int ans = 0, t = 0;
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while (i <= j) {
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if (power >= tokens[i]) {
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power -= tokens[i++];
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++t;
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ans = Math.max(ans, t);
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} else if (t > 0) {
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power += tokens[j--];
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--t;
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} else {
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break;
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}
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}
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return ans;
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}
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}

‎solution/0900-0999/0948.Bag of Tokens/Solution.py

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@@ -0,0 +1,16 @@
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class Solution:
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def bagOfTokensScore(self, tokens: List[int], power: int) -> int:
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tokens.sort()
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i, j = 0, len(tokens) - 1
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ans = t = 0
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while i <= j:
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if power >= tokens[i]:
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power -= tokens[i]
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i, t = i + 1, t + 1
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ans = max(ans, t)
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elif t:
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power += tokens[j]
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j, t = j - 1, t - 1
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else:
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break
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return ans

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