Minimum Levels to Gain More Points

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Arrays

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 i^th 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 0^th 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 <= 10⁵
possible[i] is either 0 or 1.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the values in the `points` array and the target value `k`?
Can the `points` array be empty, and if so, what should the function return?
Are all the values in the `points` array non-negative?
If the sum of all points is less than `k`, what value should be returned?
Is the input `k` guaranteed to be a non-negative integer?

Brute Force Solution

Approach

The brute force approach figures out the fewest levels needed to get more points by trying absolutely every combination of levels we could possibly choose. We test each combination to see if it gets us enough points, and then pick the combination that uses the fewest levels.

Here's how the algorithm would work step-by-step:

Start by considering no levels at all. Check if we already have enough points without needing to gain any.
Next, consider gaining just one level. Try each level individually to see if any of them give us enough points.
Then, consider gaining two levels. Try every possible pair of levels to see if any combination works.
Continue this process, trying three levels, four levels, and so on, testing all possible combinations at each step.
For each combination of levels, calculate the total points we would have after gaining those levels.
If a combination gives us enough points, remember how many levels were used in that combination.
Once we've tried all combinations, find the combination that used the fewest levels while still giving us enough points. That's the answer.

Code Implementation

def minimum_levels_to_gain_more_points(levels, points_needed):
    number_of_levels = len(levels)
    minimum_levels_required = float('inf')

    for i in range(2 ** number_of_levels):
        # Create a bitmask to represent level upgrades
        level_upgrades = []
        levels_upgraded_count = 0

        for j in range(number_of_levels):
            if (i >> j) & 1:
                level_upgrades.append(1)
                levels_upgraded_count += 1
            else:
                level_upgrades.append(0)

        total_points = 0
        for k in range(number_of_levels):
            total_points += levels[k] + level_upgrades[k]

        # If enough points are gained, update the minimum
        if total_points >= points_needed:

            minimum_levels_required = min(minimum_levels_required, levels_upgraded_count)

    # If no solution found return -1
    if minimum_levels_required == float('inf'):
        return -1
    else:
        return minimum_levels_required

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach iterates through all possible combinations of levels. For n levels, there are 2^n possible subsets (each level can either be chosen or not). For each subset, we calculate the total points which takes O(n) time in the worst case (if we pick all levels), although we can consider it constant as compared to 2^n. Therefore, the dominant factor is the generation and testing of 2^n combinations, making the overall time complexity O(2^n).

Space Complexity

O(1) – The provided brute force solution primarily iterates and compares values without persistently storing a large amount of data. While generating combinations, each combination is evaluated and discarded. No auxiliary data structures scale with the input size. The space complexity is therefore dominated by a few integer variables (for loops, storing current points, storing best count), which is constant regardless of the number of levels.

Optimal Solution

Approach

To find the fewest levels needed to gain more points, we should prioritize the levels that give us the most points with the least cost. It's like picking the best deals first to maximize our benefit quickly. We then sort the levels based on some calculated priority.

Here's how the algorithm would work step-by-step:

First, figure out how efficient each level is by dividing the points it gives by the cost to complete it. This tells you how much 'bang for your buck' you get from each level.
Next, sort all the levels from most efficient to least efficient. This puts the best 'deals' at the front.
Start completing the levels in this sorted order, from most efficient to least efficient. Keep track of how many total points you've earned.
Stop as soon as your total points reach the required goal. The number of levels you completed is the minimum number needed.

Code Implementation

def min_levels_to_gain_points(levels, required_points):
    number_of_levels = len(levels)
    point_per_level = []
    for level_index in range(number_of_levels):
        point_per_level.append((levels[level_index][0] / levels[level_index][1], level_index))

    # Sort by points per level to process best values first
    point_per_level.sort(reverse=True)

    current_points = 0
    levels_picked_count = 0
    levels_picked = []

    level_index = 0
    while current_points < required_points and level_index < number_of_levels:
        best_level_index = point_per_level[level_index][1]

        # Check if picking this level exceeds target
        if current_points + levels[best_level_index][0] <= required_points:
            current_points += levels[best_level_index][0]
            levels_picked_count += 1
            levels_picked.append(best_level_index)

        level_index += 1

    # If target cannot be reached return -1
    if current_points < required_points:
        return -1

    return levels_picked_count

Big(O) Analysis

Time Complexity

O(n log n) – Calculating the efficiency (points/cost) for each of the n levels takes O(n) time. Sorting the levels based on their efficiency is the dominant operation, which has a time complexity of O(n log n) using an efficient sorting algorithm like merge sort or quicksort. Iterating through the sorted levels to accumulate points and count the minimum levels needed takes O(n) time in the worst case. Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).

Space Complexity

O(N) – The algorithm sorts the levels based on their efficiency. This sorting operation generally requires auxiliary space. The dominant space usage comes from creating a sorted copy or using an in-place sorting algorithm that might require O(N) auxiliary space in the worst case (e.g., merge sort) for temporary storage during the sorting process. Therefore, the auxiliary space complexity is proportional to the number of levels, N, where N represents the number of levels. The space complexity is O(N).

Edge Cases

Case	How to Handle
Empty points array	Return 0 since no levels need to be played to reach k points when there are no levels.
k is zero or negative	Return 0 since no levels need to be played to reach a non-positive k value.
Points array with one element and element value is greater than or equal to k	Return 1, as a single level is sufficient to reach the required points.
Points array with all zero values and k is greater than 0	Return the length of the points array since all levels must be played but still may not reach the goal.
All points are very small, and a large number of levels are needed to reach k	The algorithm should iterate through the entire array summing the points until k is reached or the end of the array is reached.
Points array contains negative values, although this shouldn't happen per problem description	The algorithm should still function, although accumulating negative points might make reaching k impossible.
The sum of all points is less than k	Return the total number of levels in the input array since all must be played but reaching the target is impossible.
Integer overflow when summing points values	Use a data type with a larger range to store the sum (e.g., long) to prevent overflow.

Empty points array

How to Handle:

Return 0 since no levels need to be played to reach k points when there are no levels.

k is zero or negative

How to Handle:

Return 0 since no levels need to be played to reach a non-positive k value.

Points array with one element and element value is greater than or equal to k

How to Handle:

Return 1, as a single level is sufficient to reach the required points.

Points array with all zero values and k is greater than 0

How to Handle:

Return the length of the points array since all levels must be played but still may not reach the goal.

All points are very small, and a large number of levels are needed to reach k

How to Handle:

The algorithm should iterate through the entire array summing the points until k is reached or the end of the array is reached.

Points array contains negative values, although this shouldn't happen per problem description

How to Handle:

The algorithm should still function, although accumulating negative points might make reaching k impossible.

The sum of all points is less than k

How to Handle:

Return the total number of levels in the input array since all must be played but reaching the target is impossible.

Integer overflow when summing points values

How to Handle:

Use a data type with a larger range to store the sum (e.g., long) to prevent overflow.