LeetCode 3096 - Minimum Levels to Gain More Points

This problem is asking us to determine the minimum number of levels Alice should play in order to score more points than Bob, given that both play optimally and that some levels may be impossible to clear. The input is a binary array possible of length n.

leetcodemediumarrayprefix-sum

LeetCode Problem 3096

Difficulty: 🟡 Medium
Topics: Array, Prefix Sum

Solution

Problem Understanding

This problem is asking us to determine the minimum number of levels Alice should play in order to score more points than Bob, given that both play optimally and that some levels may be impossible to clear. The input is a binary array possible of length n. A 1 indicates a level can always be cleared, and a 0 indicates a level that cannot be cleared by either player. Each level gives +1 point if cleared and -1 point if failed.

Alice plays the first k levels, and Bob plays the remaining n - k levels. Both must play at least one level. The goal is to find the minimum k such that Alice's total score exceeds Bob's total score. If no such k exists, return -1.

Constraints indicate the input can be as large as 10^5, so solutions with quadratic time complexity are too slow. Important edge cases include arrays where all levels are impossible ([0,0,...]), all levels are possible, or arrays with alternating possible and impossible levels.

Approaches

The brute-force approach would try every possible split of levels from 1 to n-1, calculating Alice's score for the first i levels and Bob's score for the remaining levels. While this is correct, it has O(n^2) complexity because computing Bob's score each time requires summing a subarray, which is too slow for n = 10^5.

The optimal approach uses a prefix sum technique. Convert the array so that each level contributes +1 if possible and -1 if impossible. Compute a prefix sum array for Alice, then compute the total sum for the array. For each candidate split k, Alice’s score is the prefix sum of the first k levels, and Bob's score is total - Alice_score. We find the minimum k such that Alice_score > Bob_score. Using prefix sums reduces the repeated summation and gives an O(n) solution.

Approach Time Complexity Space Complexity Notes
Brute Force O(n^2) O(1) Try every split and sum scores, too slow for large n
Optimal O(n) O(n) Use prefix sums and total sum to compute Alice/Bob scores efficiently

Algorithm Walkthrough

  1. Convert the possible array to a score array, where each 1 becomes +1 and each 0 becomes -1. This simplifies score calculations.
  2. Compute the prefix sum array of score. The i-th prefix sum represents Alice’s score if she plays levels 0 to i.
  3. Compute the total sum of score, which represents the combined score if Alice plays all levels.
  4. Iterate over valid splits k from 1 to n-1 (both players must play at least one level). For each k, calculate Alice's score as prefix_sum[k-1] and Bob’s score as total - prefix_sum[k-1].
  5. If Alice’s score exceeds Bob’s score, return k immediately as the minimum required levels.
  6. If no valid split is found after iterating through all possibilities, return -1.

Why it works: By transforming levels into scores and using prefix sums, we can calculate any split efficiently in O(1) time. Iterating through the prefix sums ensures we find the minimum k where Alice beats Bob, which guarantees optimality.

Python Solution

from typing import List

class Solution:
    def minimumLevels(self, possible: List[int]) -> int:
        n = len(possible)
        # Convert possible array to scores: 1 -> +1, 0 -> -1
        score = [1 if x == 1 else -1 for x in possible]
        
        # Compute prefix sums
        prefix_sum = [0] * n
        prefix_sum[0] = score[0]
        for i in range(1, n):
            prefix_sum[i] = prefix_sum[i-1] + score[i]
        
        total = prefix_sum[-1]
        
        # Check all valid splits where both play at least one level
        for k in range(1, n):
            alice_score = prefix_sum[k-1]
            bob_score = total - alice_score
            if alice_score > bob_score:
                return k
        
        return -1

This implementation first converts levels to numeric scores and computes prefix sums for efficient split calculation. Then it iterates through possible splits, computing Alice and Bob’s scores and returning the minimum k when Alice wins.

Go Solution

func minimumLevels(possible []int) int {
    n := len(possible)
    score := make([]int, n)
    
    for i := 0; i < n; i++ {
        if possible[i] == 1 {
            score[i] = 1
        } else {
            score[i] = -1
        }
    }
    
    prefixSum := make([]int, n)
    prefixSum[0] = score[0]
    for i := 1; i < n; i++ {
        prefixSum[i] = prefixSum[i-1] + score[i]
    }
    
    total := prefixSum[n-1]
    
    for k := 1; k < n; k++ {
        alice := prefixSum[k-1]
        bob := total - alice
        if alice > bob {
            return k
        }
    }
    
    return -1
}

In Go, slices are used for prefix sums and scores. The logic mirrors Python, with explicit integer handling and slice initialization.

Worked Examples

Example 1: possible = [1,0,1,0]

Convert to scores: [1, -1, 1, -1]

Prefix sums: [1, 0, 1, 0]

Total sum: 0

Iterate splits:

  • k=1 → Alice=1, Bob=0-1= -1 → Alice > Bob → return 1

Example 2: possible = [1,1,1,1,1]

Convert to scores: [1,1,1,1,1]

Prefix sums: [1,2,3,4,5]

Total sum: 5

Iterate splits:

  • k=1 → Alice=1, Bob=5-1=4 → no
  • k=2 → Alice=2, Bob=3 → no
  • k=3 → Alice=3, Bob=2 → yes → return 3

Example 3: possible = [0,0]

Convert to scores: [-1,-1]

Prefix sums: [-1,-2]

Total sum: -2

Iterate splits:

  • k=1 → Alice=-1, Bob=-1 → no

Return -1

Complexity Analysis

Measure Complexity Explanation
Time O(n) Converting to scores, computing prefix sums, and iterating splits all take O(n)
Space O(n) Prefix sum array stores n elements, score array stores n elements

The approach is efficient and suitable for the maximum input size of 10^5.

Test Cases

# Provided examples
assert Solution().minimumLevels([1,0,1,0]) == 1  # Alice can win by playing 1 level
assert Solution().minimumLevels([1,1,1,1,1]) == 3  # Alice needs 3 levels
assert Solution().minimumLevels([0,0]) == -1  # Impossible to win

# Edge cases
assert Solution().minimumLevels([1,0]) == 1  # Minimum input size, Alice wins by first level
assert Solution().minimumLevels([0,1]) == -1  # Alice cannot win
assert Solution().minimumLevels([1,1,0,1,0,1]) == 3  # Mixed pattern
assert Solution().minimumLevels([1]*100000) == 50001  # Large n, all possible levels
assert Solution().minimumLevels([0]*100000) == -1  # Large n, all impossible
Test Why
[1,0,1,0] Validates Alice can win quickly with mixed levels
[1,1,1,1,1] All levels possible, requires precise split
[0,0] All levels impossible, ensures function returns -1
[1,0] Minimum n edge case
[0,1] Alice cannot win with first level impossible
[1]*100000 Stress test for performance
[0]*100000 Stress test for impossible scenario

Edge Cases

All impossible levels: If all elements are 0, both Alice and Bob score equally negative. The function must correctly return -1.

All possible levels: If all elements are 1, the minimum split may be just over half the levels. The prefix sum ensures we correctly compute the split without recomputation.

Alternating possible/impossible levels: Arrays like [1,0,1,0,1] test