Introduction

In this blog post, we will explore how to solve the LeetCode problem "948" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You start with an initial power of power, an initial score of 0, and a bag of tokens given as an integer array tokens, where each tokens[i] denotes the value of tokeni. Your goal is to maximize the total score by strategically playing these tokens. In one move, you can play an unplayed token in one of the two ways (but not both for the same token): Face-up: If your current power is at least tokens[i], you may play tokeni, losing tokens[i] power and gaining 1 score. Face-down: If your current score is at least 1, you may play tokeni, gaining tokens[i] power and losing 1 score. Return the maximum possible score you can achieve after playing any number of tokens. Example 1: Input: tokens = [100], power = 50 Output: 0 Explanation: Since your score is 0 initially, you cannot play the token face-down. You also cannot play it face-up since your power (50) is less than tokens[0] (100). Example 2: Input: tokens = [200,100], power = 150 Output: 1 Explanation: Play token1 (100) face-up, reducing your power to 50 and increasing your score to 1. There is no need to play token0, since you cannot play it face-up to add to your score. The maximum score achievable is 1. 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: Play token0 (100) face-up, reducing power to 100 and increasing score to 1. Play token3 (400) face-down, increasing power to 500 and reducing score to 0. Play token1 (200) face-up, reducing power to 300 and increasing score to 1. Play token2 (300) face-up, reducing power to 0 and increasing score to 2. The maximum score achievable is 2. Constraints: 0 <= tokens.length <= 1000 0 <= tokens[i], power < 104

Explanation

• Greedy Approach: Sort the tokens. Play tokens face-up (increasing score) when power allows, prioritizing smaller tokens. When power is insufficient but score is greater than 0, play larger tokens face-down (increasing power, decreasing score).
  • Two Pointers: Maintain two pointers, one at the beginning (smallest tokens) and one at the end (largest tokens) of the sorted array. Use these pointers to decide which token to play next to maximize the score.
  • Termination Condition: Stop when no more tokens can be played, either because power is too low to play any remaining tokens face-up or because score is 0 and no tokens can be played face-down.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(1) (in-place sorting if possible, otherwise O(n) for a sorted copy).

Code

    def max_tokens(tokens, power):
    """
    Calculates the maximum possible score achievable by strategically playing tokens.

    Args:
        tokens: A list of integers representing the values of the tokens.
        power: An integer representing the initial power.

    Returns:
        An integer representing the maximum possible score.
    """

    tokens.sort()
    left = 0
    right = len(tokens) - 1
    score = 0
    max_score = 0

    while left <= right:
        if power >= tokens[left]:
            power -= tokens[left]
            score += 1
            left += 1
            max_score = max(max_score, score)
        elif score > 0:
            power += tokens[right]
            score -= 1
            right -= 1
        else:
            break

    return max_score