We should use power on the smallest tokens (face up) to gain score, and use score on the largest tokens (face down) to gain the most power. This greedy approach works perfectly with Two Pointers on a sorted array.

1. Sort tokens.
2. l = 0, r = len(tokens) - 1, score = 0, max_score = 0.
3. While l <= r:
   • If power >= tokens[l]:
     • power -= tokens[l], score += 1, l += 1, max_score = max(max_score, score).
   • Else if score > 0:
     • power += tokens[r], score -= 1, r -= 1.
   • Else:
     • Break.
4. Return max_score.

• Time Complexity: O(N log N).
• Space Complexity: O(1).

def bag_of_tokens_score(tokens, power):
    tokens.sort()
    l, r = 0, len(tokens) - 1
    score = 0
    res = 0
    
    while l <= r:
        if power >= tokens[l]:
            power -= tokens[l]
            score += 1
            l += 1
            res = max(res, score)
        elif score > 0:
            power += tokens[r]
            score -= 1
            r -= 1
        else:
            break
            
    return res