## Algorithm

Problem Name: 948. Bag of Tokens

You have an initial power of `power`, an initial score of `0`, and a bag of `tokens` where `tokens[i]` is the value of the `ith` token (0-indexed).

Your goal is to maximize your total score by potentially playing each token in one of two ways:

• If your current power is at least `tokens[i]`, you may play the `ith` token face up, losing `tokens[i]` power and gaining `1` score.
• If your current score is at least `1`, you may play the `ith` token face down, gaining `tokens[i]` power and losing `1` score.

Each token may be played at most once and in any order. You do not have to play all the tokens.

Return the largest possible score you can achieve after playing any number of tokens.

Example 1:

```Input: tokens = , power = 50
Output: 0
Explanation: Playing the only token in the bag is impossible because you either have too little power or too little score.
```

Example 2:

```Input: tokens = [100,200], power = 150
Output: 1
Explanation: Play the 0th token (100) face up, your power becomes 50 and score becomes 1.
There is no need to play the 1st token since you cannot play it face up to add to your score.
```

Example 3:

```Input: tokens = [100,200,300,400], power = 200
Output: 2
Explanation: Play the tokens in this order to get a score of 2:
1. Play the 0th token (100) face up, your power becomes 100 and score becomes 1.
2. Play the 3rd token (400) face down, your power becomes 500 and score becomes 0.
3. Play the 1st token (200) face up, your power becomes 300 and score becomes 1.
4. Play the 2nd token (300) face up, your power becomes 0 and score becomes 2.
```

Constraints:

• `0 <= tokens.length <= 1000`
• `0 <= tokens[i], power < 104`

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int bagOfTokensScore(int[] tokens, int power) {
int score = 0;
Arrays.sort(tokens);
int leftIdx = 0;
int rightIdx = tokens.length - 1;
int currScore = 0;
while (leftIdx <= rightIdx) {
if (power >= tokens[leftIdx]) {
currScore++;
power -= tokens[leftIdx++];
} else {
if (currScore == 0) {
break;
}
currScore--;
power += tokens[rightIdx--];
}
score = Math.max(score, currScore);
}
return score;
}
}
``````
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Input

cmd
tokens = , power = 50

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const bagOfTokensScore = function (tokens, P) {
tokens.sort((a, b) => a - b)
let res = 0,
score = 0,
i = 0,
j = tokens.length - 1
while (i <= j) {
if (P >= tokens[i]) {
P -= tokens[i++]
res = Math.max(res, ++score)
} else if (score > 0) {
score--
P += tokens[j--]
} else break
}
return res
}
``````
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Input

cmd
tokens = , power = 50

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def bagOfTokensScore(self, tokens, P):
"""
:type tokens: List[int]
:type P: int
:rtype: int
"""
tokens.sort()
l, r, score = 0, len(tokens) - 1, 0
while l <= r:
if P >= tokens[l]:
P -= tokens[l]
score += 1
l += 1
elif score and l != r:
P += tokens[r]
score -= 1
r -= 1
else:
break
return score
``````
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Input

cmd
tokens = [100,200], power = 150

Output

cmd
1