Minimum Levels to Gain More Points - Practice Coding Problems

Minimum Levels to Gain More Points - Problem

Imagine a gaming tournament where Alice and Bob are competing in a strategy game with n levels. The twist? Some levels are impossible to clear!

You're given a binary array possible where:

possible[i] = 1: Level i can be cleared (gain +1 point)
possible[i] = 0: Level i is impossible (lose -1 point)

The game rules are:

Alice plays first, starting from level 0, for some consecutive levels
Bob plays the remaining levels
Both players play optimally to maximize their own score
Each player must play at least 1 level

Goal: Find the minimum number of levels Alice should play to score more points than Bob. Return -1 if impossible.

This is a classic prefix sum optimization problem disguised as a game theory challenge!

Input & Output

example_1.py — Basic Case

$ Input: possible = [1,0,1,0]

› Output: 3

💡 Note: Alice plays first 3 levels: scores 1 + (-1) + 1 = 1. Bob plays last level: scores -1. Alice (1) > Bob (-1), so Alice needs minimum 3 levels.

example_2.py — Impossible Case

$ Input: possible = [0,0]

› Output: -1

💡 Note: No matter how they split, both players will have negative scores, but Alice can never score MORE than Bob. Alice plays 1 level (-1), Bob plays 1 level (-1). Alice (-1) = Bob (-1), not greater.

example_3.py — Alice Needs All But One

$ Input: possible = [1,1,1,1,1,0]

› Output: 5

💡 Note: Alice needs to play 5 levels to score 5 points, leaving Bob with the impossible level (-1 point). Alice (5) > Bob (-1).

Constraints

2 ≤ possible.length ≤ 10⁵
possible[i] is either 0 or 1
Each player must play at least 1 level
Alice plays consecutive levels starting from index 0

Visualization

Tap to expand

Understanding the Visualization

Convert Game Points

Transform clearable levels (1) to +1 points and impossible levels (0) to -1 points

Calculate Total Score Pool

Find the sum of all level points - this represents the total points available

Track Alice's Running Score

As Alice plays each level, maintain her cumulative score using prefix sum

Calculate Bob's Remaining Score

Bob gets all remaining points: Total - Alice's current score

Find Optimal Split Point

Return the minimum levels where Alice's score exceeds Bob's score

Key Takeaway

🎯 Key Insight: By using prefix sums, we can efficiently calculate Alice's score and derive Bob's score in O(n) time, eliminating the need for nested loops and redundant calculations.

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

This problem is solved optimally using prefix sums in O(n) time. Convert the binary array to +1/-1 points, calculate the total sum once, then iterate through split points using a running prefix sum for Alice's score and derive Bob's score as total - alice_score. Return the first position where Alice scores more than Bob.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Split Points)	O(n²)	O(1)	Try every possible number of levels Alice can play and calculate scores
Prefix Sum (Optimal)	O(n)	O(1)	Use prefix sums to calculate scores in O(n) time

Brute Force (Try All Split Points) — Algorithm Steps

Loop through each possible split point i (Alice plays levels 0 to i-1)
Calculate Alice's score by summing her levels
Calculate Bob's optimal score by summing his levels
Check if Alice's score > Bob's score
Return the minimum valid split point, or -1 if none exists

Visualization

Tap to expand

Step-by-Step Walkthrough

Try split at position 1

Alice plays level 0, Bob plays levels 1-n

Try split at position 2

Alice plays levels 0-1, Bob plays levels 2-n

Continue until valid split found

Keep trying until Alice's score > Bob's score

Code -

solution.c — C

int minimumLevels(int* possible, int possibleSize) {
    int n = possibleSize;
    if (n < 2) return -1;
    
    // Try each possible split point
    for (int aliceLevels = 1; aliceLevels < n; aliceLevels++) {
        // Calculate Alice's score
        int aliceScore = 0;
        for (int i = 0; i < aliceLevels; i++) {
            aliceScore += (possible[i] == 1) ? 1 : -1;
        }
        
        // Calculate Bob's score
        int bobScore = 0;
        for (int i = aliceLevels; i < n; i++) {
            bobScore += (possible[i] == 1) ? 1 : -1;
        }
        
        if (aliceScore > bobScore) {
            return aliceLevels;
        }
    }
    
    return -1;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n-1 possible splits, we calculate sums of up to n elements

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track scores and minimum result

✓ Linear Space

19.6K Views

Medium Frequency

~18 min Avg. Time

834 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Split Points) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler