How do you solve Minimum Levels to Gain More Points on LeetCode?

Minimum Levels to Gain More Points (LeetCode #3096) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Transforming the problem can simplify calculations.

What is the brute force solution for Minimum Levels to Gain More Points?

The brute force approach for Minimum Levels to Gain More Points is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will simulate every possible prefix of levels that Alice can play and calculate her score against Bob's score for each prefix. This will help us determine the minimum number of levels Alice needs to play to have a score greater than Bob's. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Levels to Gain More Points?

Minimum Levels to Gain More Points uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: .

What are the interview tips for Minimum Levels to Gain More Points?

Clearly explain your thought process while writing code. Consider edge cases, such as all levels being impossible. Practice transforming problems into simpler forms to identify patterns.

What are common mistakes when solving Minimum Levels to Gain More Points?

Not accounting for the minimum levels each player must play. Overlooking the transformation of the array for easier calculations.

#3096

Minimum Levels to Gain More Points

Medium

ArrayPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can optimize the solution by transforming the `possible` array into a new array where 0s become -1s. This allows us to use a prefix sum to quickly calculate Alice's and Bob's scores and find the minimum prefix length where Alice's score exceeds Bob's.

⚙️

Algorithm

5 steps

1Step 1: Transform the `possible` array by replacing all 0s with -1s.
2Step 2: Calculate the total score of the transformed array to determine Bob's score if Alice plays all levels.
3Step 3: Iterate through the transformed array, maintaining a running sum for Alice's score.
4Step 4: For each level, check if Alice's score exceeds half of the total score (Bob's score).
5Step 5: Return the index + 1 of the first level where Alice's score exceeds Bob's score.

solution.py9 lines

1def min_levels(possible):
2    transformed = [1 if x == 1 else -1 for x in possible]
3    total_score = sum(transformed)
4    alice_score = 0
5    for i in range(len(transformed)):
6        alice_score += transformed[i]
7        if alice_score > total_score - alice_score:
8            return i + 1
9    return -1

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times, and the space complexity is O(n) due to the transformed array.

1Transforming the problem can simplify calculations.
2Using prefix sums allows for efficient score comparisons.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.