Problem Description

You are given a binary array possible of length n.

Alice and Bob are playing a game that consists of n levels. Some of the levels are impossible to clear, while others can always be cleared. Specifically, if possible[i] == 0, then the i-th level is impossible for both players to clear. A player gains 1 point for clearing a level and loses 1 point if they fail to clear it.

At the start of the game, Alice will play some levels in the given order starting from the 0-th level, after which Bob will play the remaining levels.

Alice wants to know the minimum number of levels she should play to gain more points than Bob if both players play optimally to maximize their points.

Return the minimum number of levels Alice should play to gain more points. If this is not possible, return -1.

Note that each player must play at least 1 level.

Intuition

To determine the minimum number of levels Alice should play to ultimately score more points than Bob, we need to conceptualize the point accumulation process over the levels. Here's how we arrive at the solution:

1. Understand Total Points: First, calculate the total possible points (s) both players together can achieve by summing up the point values for each level based on the array possible. If possible[i] is 1, then it adds 1 point, and if it is 0, it deducts 1 point since the level is impossible to clear.

2. Iterate through Levels for Alice: Start determining how many initial levels Alice should play by iterating through the possible array. At each step, compute the cumulative points (t) Alice can gain up to that level.

3. Compare Points Optimally: Compare Alice's cumulative points (t) with Bob's potential points in the remaining levels, which would be s - t. As Alice moves through the levels, if at any point for some level i Alice's score (t) surpasses the remaining potential score for Bob (s - t), then the minimum number of levels Alice should play to guarantee more points than Bob is found, which will be i.

4. Check Feasibility: If iterating through all levels until one before the last (n-1), Alice still cannot have a significant advantage over Bob, return -1 to indicate it's impossible for Alice to score more points than Bob.

This strategic approach ensures that both Alice and Bob always act optimally to maximize their collective scores. By incrementing the number of levels Alice plays, you can systematically determine the minimum number she needs to gain an advantage if possible.

Learn more about Prefix Sum patterns.

Solution Approach

To solve the problem efficiently, an enumeration strategy is employed, which involves iterating over the levels to determine how many levels Alice needs to play before she has more points than Bob. Here's a detailed breakdown of the solution approach:

1. Calculate Total Points (s):

   • Create a variable s to represent the total sum of potential points that can be earned by both players in the game. Iterate through the possible array and calculate the total score using a generator expression or a loop: s = sum(-1 if x == 0 else 1 for x in possible).
   • This computation provides the net score assuming all levels have been processed by both players. A 1 is added to the total for each level that can be cleared and a -1 for each impossible level.

2. Iterate and Track Points for Alice (t):

   • Initialize a variable t to keep track of the cumulative points Alice has as she starts playing from level 0.
   • Use a loop to iterate over each level from 0 to n-2 (excluding the last level since Bob needs to play at least one level). For each level indexed by i:
     • Update Alice’s cumulative score t: If possible[i] == 0, subtract 1, otherwise, add 1, similar to the calculation used for s.
     • Compare Alice’s points with what Bob could potentially earn in the remaining game: Check if t > s - t, which translates to Alice having more points than Bob after level i.

3. Determine Minimum Levels Alice Needs to Play:

   • If for some level i, the condition t > s - t is met, return i, signifying that the minimum number of levels Alice needs to play to maintain a lead over Bob is i + 1, since indices are 0-based.

4. Return -1 when Necessary:

   • If the loop completes without finding a suitable i where Alice's points exceed Bob's possible points, return -1. This implies it is impossible for Alice to secure a lead with any combination of levels.

The implemented code as follows:

class Solution:
    def minimumLevels(self, possible: List[int]) -> int:
        s = sum(-1 if x == 0 else 1 for x in possible)
        t = 0
        for i, x in enumerate(possible[:-1], 1):
            t += -1 if x == 0 else 1
            if t > s - t:
                return i
        return -1

This efficient approach ensures a linear time complexity, O(n), where n is the number of levels, making it suitable for potentially large values of n.

Example Walkthrough

Consider a small example where possible = [1, 0, 1, 1]. Let's walk through the solution approach using this example:

1. Calculate Total Points (s):
   Iterate through the possible array and compute the total points s.

   • For possible[0] = 1, add 1.
   • For possible[1] = 0, subtract 1.
   • For possible[2] = 1, add 1.
   • For possible[3] = 1, add 1.

   Thus, s = 1 - 1 + 1 + 1 = 2.

2. Iterate and Track Points for Alice (t):
   Initialize t to 0 and iterate over the array until the second last element to ensure Bob plays at least one level.

   • Level 0 (possible[0] = 1):
     Update Alice's score t = 0 + 1 = 1.
     Compare Alice's score with Bob's potential score:

     • t = 1
     • Bob's potential score = s - t = 2 - 1 = 1.
       Since t is not greater than Bob's potential score, Alice doesn't yet have more points than Bob.

   • Level 1 (possible[1] = 0):
     Update Alice's score t = 1 - 1 = 0.
     Compare Alice's score with Bob's potential score:

     • t = 0
     • Bob's potential score = s - t = 2 - 0 = 2.
       Alice's score is not more than Bob's.

   • Level 2 (possible[2] = 1):
     Update Alice's score t = 0 + 1 = 1.
     Compare Alice's score with Bob's potential score:

     • t = 1
     • Bob's potential score = s - t = 2 - 1 = 1.
       Now t > s - t, indicating Alice can have more points than Bob.

3. Determine Minimum Levels Alice Needs to Play:
   Since Alice has more points than Bob at level 2 when t > s - t, the minimum number of levels Alice should play is i + 1 = 3.

The solution yields a result of 3, meaning Alice should play the first three levels to secure more points than Bob.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the input list possible. This complexity stems from the fact that the algorithm iterates through the list once using a for loop. Each iteration performs constant-time operations, leading to a linear time complexity relative to the size of the input.

The space complexity of the code is O(1). This is because the algorithm only uses a fixed amount of additional space beyond the input list, namely the integer variables s, t, and i. The storage requirement does not depend on the size of the input list, resulting in constant space complexity.

Learn more about how to find time and space complexity quickly.