Problem Description

You have a binary array possible of length n that represents levels in a game. Alice and Bob will play through all these levels, with Alice playing first from level 0, then Bob playing the remaining levels.

The scoring system works as follows:

• If possible[i] == 1, the level can be cleared and the player gains 1 point
• If possible[i] == 0, the level is impossible to clear and the player loses 1 point

Both players play optimally to maximize their own points. Alice goes first and plays some number of levels in order starting from level 0, then Bob plays all the remaining levels.

The goal is to find the minimum number of levels Alice needs to play to ensure she gets more points than Bob. Each player must play at least 1 level.

Return the minimum number of levels Alice should play to have more points than Bob. If it's impossible for Alice to get more points than Bob regardless of how many levels she plays, return -1.

For example, if possible = [1, 0, 1, 0]:

• If Alice plays 1 level: Alice gets 1 point (level 0), Bob gets -1 + 1 + (-1) = -1 points (levels 1, 2, 3). Alice wins with 1 > -1.
• So the answer would be 1.

The key insight is that since both players play optimally and the levels are fixed, we just need to find the split point where Alice's cumulative score exceeds Bob's cumulative score, using the minimum number of levels for Alice.

Intuition

The key observation is that once we decide how many levels Alice plays, the scores for both players are completely determined since the array is fixed and both play optimally.

Let's think about this problem step by step:

1. First, we need to understand what "playing optimally" means here. Since the levels are fixed and must be played in order, there's no choice in strategy - each player simply plays their assigned levels and gets the corresponding points. The "optimal" play is already built into the problem structure.

2. The total score available in the game is fixed. If we convert each level to its point value (1 becomes +1, 0 becomes -1), the sum of all these values represents the total points distributed between Alice and Bob.

3. For Alice to win, she needs to get more than half of the total points. Mathematically, if Alice's score is t and the total is s, then Bob's score is s - t. Alice wins when t > s - t, which simplifies to t > s/2 or 2t > s.

4. Since Alice wants to play the minimum number of levels, we should check each possible split point from left to right. We start with Alice playing just 1 level, then 2 levels, and so on, until we find the first point where Alice's cumulative score exceeds Bob's.

5. We can't let Alice play all n levels because Bob must play at least 1 level. So we only check up to n-1 levels for Alice.

6. As we iterate through each possible number of levels for Alice, we maintain a running sum of her score. Once this running sum exceeds the remaining score (Bob's score), we've found our answer.

This greedy approach works because once Alice's score exceeds Bob's at some split point, playing more levels would only increase the number of levels Alice plays without providing any benefit (we want the minimum).

Learn more about Prefix Sum patterns.

Solution Approach

The solution implements the enumeration approach described in the reference:

1. Calculate the total score s: We first convert the binary array to scores and sum them up. Each 1 contributes +1 point and each 0 contributes -1 point. This can be written as:

   s = sum(-1 if x == 0 else 1 for x in possible)

   This gives us the total points available in the game.

2. Initialize Alice's running score t: We start with t = 0 to track Alice's cumulative score as we iterate through the levels.

3. Enumerate possible split points: We iterate through the array from index 0 to n-2 (since Bob must play at least 1 level). For each level i:

   • Add the score of level i to Alice's total: t += -1 if x == 0 else 1
   • Check if Alice wins: If t > s - t (Alice's score > Bob's score), return the current number of levels i

   The enumeration uses enumerate(possible[:-1], 1) which:

   • possible[:-1] ensures we don't include the last level (Bob needs at least 1)
   • Starting index 1 gives us the count of levels directly (1-indexed)

4. Return the result:

   • If we find a split point where Alice wins, we immediately return that index i (the minimum number of levels)
   • If we've checked all valid split points and Alice never wins, return -1

The algorithm has a time complexity of O(n) where we traverse the array at most twice (once for calculating total sum, once for finding the split point), and space complexity of O(1) as we only use a few variables to track the scores.

The key efficiency comes from the early termination - as soon as we find the first point where Alice wins, we return immediately since we want the minimum number of levels.

Example Walkthrough

Let's walk through the solution with possible = [1, 0, 1, 0].

Step 1: Calculate total score

• Convert to points: [1, 0, 1, 0] → [+1, -1, +1, -1]
• Total score s = 1 + (-1) + 1 + (-1) = 0

Step 2: Initialize Alice's score

• Start with t = 0 (Alice hasn't played any levels yet)

Step 3: Try each split point

Split 1: Alice plays 1 level

• Alice plays level 0: t = 0 + 1 = 1
• Bob gets remaining: s - t = 0 - 1 = -1
• Check: Is t > s - t? Is 1 > -1? YES!
• Alice wins with score 1 vs Bob's score -1
• Return 1 (minimum levels needed)

Since we found a winning split on the first try, we return immediately without checking further splits.

Let's verify this is correct:

• Alice: Level 0 → score = +1
• Bob: Levels 1, 2, 3 → scores = -1, +1, -1 → total = -1
• Alice wins 1 to -1 ✓

For comparison, if we had continued:

• Split 2: Alice plays levels 0-1 → Alice gets 0 points, Bob gets 0 points (tie)
• Split 3: Alice plays levels 0-2 → Alice gets 1 point, Bob gets -1 point (Alice still wins but needs more levels)

The algorithm correctly identifies that Alice only needs to play 1 level to guarantee victory.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array possible. The algorithm first computes the sum of transformed values in a single pass through the array, which takes O(n) time. Then it iterates through the array once more (excluding the last element) to find the minimum number of levels, which also takes O(n) time. Therefore, the overall time complexity is O(n) + O(n) = O(n).

The space complexity is O(1). The algorithm only uses a constant amount of extra space for variables s, t, i, and x, regardless of the input size. No additional data structures that grow with the input size are created.

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

Common Pitfalls

1. Forgetting that Bob must play at least one level

One of the most common mistakes is iterating through the entire array, including the last element. This would allow Alice to play all levels, leaving Bob with zero levels, which violates the constraint that each player must play at least one level.

Incorrect approach:

# WRONG: This allows Alice to play all n levels
for index, value in enumerate(possible, start=1):
    current_score += -1 if value == 0 else 1
    if current_score > total_score - current_score:
        return index

Correct approach:

# CORRECT: Stop before the last element to ensure Bob plays at least 1 level
for index, value in enumerate(possible[:-1], start=1):
    current_score += -1 if value == 0 else 1
    if current_score > total_score - current_score:
        return index

2. Using incorrect comparison (>= instead of >)

The problem asks for Alice to have more points than Bob, not equal or more. Using >= would return cases where both players tie, which doesn't satisfy the winning condition.

Incorrect approach:

# WRONG: This accepts ties as wins
if current_score >= total_score - current_score:
    return index

Correct approach:

# CORRECT: Alice must strictly have more points than Bob
if current_score > total_score - current_score:
    return index

3. Off-by-one error in counting levels

Another pitfall is incorrectly counting the number of levels Alice plays. Since we're using 0-indexed arrays but need to return the count of levels (1-indexed), careful attention is needed.

Incorrect approach:

# WRONG: Returns 0-indexed position instead of count
for index in range(len(possible) - 1):
    current_score += -1 if possible[index] == 0 else 1
    if current_score > total_score - current_score:
        return index  # This returns 0 for first level instead of 1

Correct approach:

# CORRECT: Use enumerate with start=1 or add 1 to index
for index, value in enumerate(possible[:-1], start=1):
    current_score += -1 if value == 0 else 1
    if current_score > total_score - current_score:
        return index  # Returns correct count

4. Not handling the edge case where Alice can never win

Forgetting to return -1 when it's impossible for Alice to win would cause the function to return None implicitly, which is incorrect.

Solution: Always ensure there's a return statement after the loop to handle the case where no valid split point exists:

# After the loop, if no winning split was found
return -1