LeetCode 948 - Bag of Tokens

https://leetcode.com/problems/bag-of-tokens/

You have an initial power P, an initial score of 0 points, and a bag of tokens.

Each token can be used at most once, has a value token[i], and has potentially two ways to use it.

• If we have at least token[i] power, we may play the token face up, losing token[i] power, and gaining 1point.
• If we have at least 1 point, we may play the token face down, gaining token[i] power, and losing 1 point.

Return the largest number of points we can have after playing any number of tokens.

Example 1:

Input: tokens = [100], P = 50
Output: 0

Example 2:

Input: tokens = [100,200], P = 150
Output: 1

Example 3:

Input: tokens = [100,200,300,400], P = 200
Output: 2

Note:

1. tokens.length <= 1000
2. 0 <= tokens[i] < 10000
3. 0 <= P < 10000

Approach 1: Greedy

If we play a token face up, we might as well play the one with the smallest value. Similarly, if we play a token face down, we might as well play the one with the largest value.

Algorithm

We don't need to play anything until absolutely necessary. Let's play tokens face up until we can't, then play a token face down and continue.

We must be careful, as it is easy for our implementation to be incorrect if we do not handle corner cases correctly. We should always play tokens face up until exhaustion, then play one token face down and continue.

Our loop must be constructed with the right termination condition: we can either play a token face up or face down.

Our final answer could be any of the intermediate answers we got after playing tokens face up (but before playing them face down.)

• Time Complexity: $O(N \log N)$, where $N$ is the length of tokens.
• Space Complexity: $O(N)$. 

  public int bagOfTokensScore(int[] tokens, int P) {

    Arrays.sort(tokens);

    int lo = 0, hi = tokens.length - 1;

    int points = 0, ans = 0;

    while (lo <= hi && (P >= tokens[lo] || points > 0)) {

      while (lo <= hi && P >= tokens[lo]) {

        P -= tokens[lo++];

        points++;

      }

      ans = Math.max(ans, points);

      if (lo <= hi && points > 0) {

        P += tokens[hi--];

        points--;

      }

    }

    return ans;

  }

  public int bagOfTokensScore(int[] tokens, int point) {

    Arrays.sort(tokens);

    ArrayDeque<Integer> queue = new ArrayDeque<>(Arrays.stream(tokens).boxed().collect(Collectors.toList()));

    int score = 0;

    while (queue.size() > 1) {

      // without changed, forever loop

      boolean changed = false;

      // bugs: missing !queue.isEmpty(), >= not >

      while (!queue.isEmpty() && point >= queue.peekFirst()) {

        score++;

        point -= queue.removeFirst();

        changed = true;

      }

      if (score > 0 && queue.size() > 1) {

        score--;

        point += queue.removeLast();

        changed = true;

      }

      if (!changed)

        break;

    }

    if (!queue.isEmpty() && point > queue.peekFirst()) {

      score++;

    }

    return score;

  }

https://leetcode.com/problems/bag-of-tokens/discuss/197696/C%2B%2BJavaPython-Greedy-%2B-Two-Pointers

Sort tokens.
Buy at the cheapest and sell at the most expensive.

    public int bagOfTokensScore(int[] tokens, int P) {
        Arrays.sort(tokens);
        int res = 0, points = 0, i = 0, j = tokens.length - 1;
        while (i <= j) {
            if (P >= tokens[i]) {
                P -= tokens[i++];
                res = Math.max(res, ++points);
            } else if (points > 0) {
                points--;
                P += tokens[j--];
            } else {
                break;
            }
        }
        return res;
    }

https://leetcode.com/problems/bag-of-tokens/discuss/197669/Python-two-pointers-O(N*logN)-time-O(1)-space

• Solution is straightforward.
• Always aim for highest point
• Sort the tokens so we can buy from lowest & sell from highest which means =>
  • If we have enough power, no worries. Just lose token[l(eft)], and increase score by 1.
  • If we have at least 1 score and we are not in the last processed token, gain token[r(ight)] and decrease score by 1.
  • Otherwise, we are finished.

class Solution:
    def bagOfTokensScore(self, tokens, P):
        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