Minimum Levels to Gain More Points
MedPrompt
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 in the game are impossible to clear while others can always be cleared. In particular, if possible[i] == 0, then the ith level is impossible to clear for both the players. A player gains 1 point on clearing a level and loses 1 point if the player fails to clear it.
At the start of the game, Alice will play some levels in the given order starting from the 0th level, after which Bob will play for the rest of the 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.
Example 1:
Input: possible = [1,0,1,0]
Output: 1
Explanation:
Let's look at all the levels that Alice can play up to:
- If Alice plays only level 0 and Bob plays the rest of the levels, Alice has 1 point, while Bob has -1 + 1 - 1 = -1 point.
- If Alice plays till level 1 and Bob plays the rest of the levels, Alice has 1 - 1 = 0 points, while Bob has 1 - 1 = 0 points.
- If Alice plays till level 2 and Bob plays the rest of the levels, Alice has 1 - 1 + 1 = 1 point, while Bob has -1 point.
Alice must play a minimum of 1 level to gain more points.
Example 2:
Input: possible = [1,1,1,1,1]
Output: 3
Explanation:
Let's look at all the levels that Alice can play up to:
- If Alice plays only level 0 and Bob plays the rest of the levels, Alice has 1 point, while Bob has 4 points.
- If Alice plays till level 1 and Bob plays the rest of the levels, Alice has 2 points, while Bob has 3 points.
- If Alice plays till level 2 and Bob plays the rest of the levels, Alice has 3 points, while Bob has 2 points.
- If Alice plays till level 3 and Bob plays the rest of the levels, Alice has 4 points, while Bob has 1 point.
Alice must play a minimum of 3 levels to gain more points.
Example 3:
Input: possible = [0,0]
Output: -1
Explanation:
The only possible way is for both players to play 1 level each. Alice plays level 0 and loses 1 point. Bob plays level 1 and loses 1 point. As both players have equal points, Alice can't gain more points than Bob.
Constraints:
2 <= n == possible.length <= 105possible[i]is either0or1.
Approaches
2 approaches with complexity analysis and trade-offs.
This approach directly simulates the game for every possible split point. For each number of levels k that Alice could play (from 1 to n-1), we calculate her score and Bob's score from scratch by iterating through their respective level ranges and then compare them.
Algorithm
- Iterate through all possible numbers of levels
kthat Alice can play, from1ton-1. - For each
k, initializealiceScoreandbobScoreto0. - Calculate
aliceScoreby iterating from level0tok-1. Add1for a possible level (1) and subtract1for an impossible level (0). - Calculate
bobScoreby iterating from levelkton-1using the same scoring logic. - If
aliceScoreis strictly greater thanbobScore, it means we have found a valid split. Since we are iteratingkin increasing order, this is the minimum number of levels. Returnk. - If the loop completes without finding such a
k, return-1.
Walkthrough
The brute-force method involves checking every possible scenario. Alice must play at least one level, and Bob must also play at least one. This means Alice can play k levels, where k ranges from 1 to n-1.
For each potential value of k, we perform two separate calculations:
- Alice's Score: We loop from level
0tok-1. For each leveli, ifpossible[i]is1, we add 1 toaliceScore; otherwise, we subtract 1. - Bob's Score: We loop from level
kton-1. Similarly, for each levelj, ifpossible[j]is1, we add 1 tobobScore; otherwise, we subtract 1.
After calculating both scores, we check if aliceScore > bobScore. Since our outer loop for k starts from 1 and goes up, the very first time this condition is met, we have found the minimum k. We can immediately return this value. If the loop finishes and we haven't found any such k, it's impossible for Alice to win, so we return -1.
class Solution { public int minimumLevels(int[] possible) { int n = possible.length; // k is the number of levels Alice plays. // Alice must play at least 1 level, and Bob must play at least 1. // So, k can range from 1 to n-1. for (int k = 1; k < n; k++) { int aliceScore = 0; // Calculate Alice's score for levels 0 to k-1 for (int i = 0; i < k; i++) { if (possible[i] == 1) { aliceScore++; } else { aliceScore--; } } int bobScore = 0; // Calculate Bob's score for levels k to n-1 for (int i = k; i < n; i++) { if (possible[i] == 1) { bobScore++; } else { bobScore--; } } if (aliceScore > bobScore) { return k; } } return -1; }}Complexity
Time
O(n^2) - The outer loop runs `n-1` times. For each iteration, we perform two inner loops that, combined, iterate through all `n` elements to calculate the scores. This results in a quadratic time complexity.
Space
O(1) - We only use a constant amount of extra space for variables like `aliceScore`, `bobScore`, and loop counters.
Trade-offs
Pros
Simple to understand and implement as it directly translates the problem description into code.
It is guaranteed to be correct, as it exhaustively checks all possibilities.
Cons
Highly inefficient due to repeated calculations for scores in each iteration.
The time complexity of O(n^2) will result in a 'Time Limit Exceeded' error for large inputs as specified in the constraints.
Solutions
Solution
class Solution {public int minimumLevels(int[] possible) { int s = 0; for (int x : possible) { s += x == 0 ? -1 : 1; } int t = 0; for (int i = 1; i < possible.length; ++i) { t += possible[i - 1] == 0 ? -1 : 1; if (t > s - t) { return i; } } return -1; }}Video walkthrough
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