What is the time complexity of Probability of a Two Boxes Having The Same Number of Distinct Balls?

The optimal solution for Probability of a Two Boxes Having The Same Number of Distinct Balls runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only need to iterate through the colors and calculate combinations without generating all distributions.

What is the brute force solution for Probability of a Two Boxes Having The Same Number of Distinct Balls?

The brute force approach for Probability of a Two Boxes Having The Same Number of Distinct Balls is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible distributions of balls into two boxes and count how many of those distributions have the same number of distinct colors in both boxes. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Probability of a Two Boxes Having The Same Number of Distinct Balls?

Always clarify the problem requirements, especially regarding distinct counts and box identity. Think about combinatorial mathematics and how it can simplify counting problems. Practice similar problems to get comfortable with combinatorial reasoning.

What are common mistakes when solving Probability of a Two Boxes Having The Same Number of Distinct Balls?

Failing to account for the distinct nature of boxes when counting distributions. Overcomplicating the problem by trying to generate all combinations instead of using combinatorial logic.

#1467

Probability of a Two Boxes Having The Same Number of Distinct Balls

Hard

ArrayMathDynamic ProgrammingBacktrackingCombinatoricsProbability and StatisticsCombinatorial CountingDynamic Programming

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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)

Instead of generating all combinations, we can use combinatorial mathematics to calculate the number of valid distributions directly based on the counts of balls of each color. This approach is efficient and avoids redundant calculations.

⚙️

Algorithm

4 steps

1Step 1: Calculate the total number of ways to choose n balls from 2n balls using combinatorial formulas.
2Step 2: For each color, determine how many can go into each box while maintaining distinct counts.
3Step 3: Use dynamic programming to keep track of valid distributions and their probabilities.
4Step 4: Calculate the final probability of having the same number of distinct colors in both boxes.

solution.py18 lines

1# Full working Python code
2from math import factorial
3
4def probabilitySameDistinct(balls):
5    n = sum(balls) // 2
6    total_ways = factorial(sum(balls)) // (factorial(n) * factorial(sum(balls) - n))
7    same_distinct_count = 0
8
9    def countWays(index, box1_count, box2_count):
10        if index == len(balls):
11            return box1_count == box2_count
12        ways = 0
13        for i in range(min(balls[index], n - box1_count) + 1):
14            ways += countWays(index + 1, box1_count + i, n - box1_count - i)
15        return ways
16
17    same_distinct_count = countWays(0, 0, 0)
18    return same_distinct_count / total_ways

ℹ

Complexity note: The time complexity is O(n) because we only need to iterate through the colors and calculate combinations without generating all distributions.

1Understanding combinatorial distributions is key to solving this problem efficiently.
2The distinct count of colors can be tracked using combinations rather than brute-force enumeration.

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