Introduction

In this blog post, we will explore how to solve the LeetCode problem "1467" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given 2n balls of k distinct colors. You will be given an integer array balls of size k where balls[i] is the number of balls of color i. All the balls will be shuffled uniformly at random, then we will distribute the first n balls to the first box and the remaining n balls to the other box (Please read the explanation of the second example carefully). Please note that the two boxes are considered different. For example, if we have two balls of colors a and b, and two boxes [] and (), then the distribution [a] (b) is considered different than the distribution [b] (a) (Please read the explanation of the first example carefully). Return the probability that the two boxes have the same number of distinct balls. Answers within 10-5 of the actual value will be accepted as correct. Example 1: Input: balls = [1,1] Output: 1.00000 Explanation: Only 2 ways to divide the balls equally: - A ball of color 1 to box 1 and a ball of color 2 to box 2 - A ball of color 2 to box 1 and a ball of color 1 to box 2 In both ways, the number of distinct colors in each box is equal. The probability is 2/2 = 1 Example 2: Input: balls = [2,1,1] Output: 0.66667 Explanation: We have the set of balls [1, 1, 2, 3] This set of balls will be shuffled randomly and we may have one of the 12 distinct shuffles with equal probability (i.e. 1/12): [1,1 / 2,3], [1,1 / 3,2], [1,2 / 1,3], [1,2 / 3,1], [1,3 / 1,2], [1,3 / 2,1], [2,1 / 1,3], [2,1 / 3,1], [2,3 / 1,1], [3,1 / 1,2], [3,1 / 2,1], [3,2 / 1,1] After that, we add the first two balls to the first box and the second two balls to the second box. We can see that 8 of these 12 possible random distributions have the same number of distinct colors of balls in each box. Probability is 8/12 = 0.66667 Example 3: Input: balls = [1,2,1,2] Output: 0.60000 Explanation: The set of balls is [1, 2, 2, 3, 4, 4]. It is hard to display all the 180 possible random shuffles of this set but it is easy to check that 108 of them will have the same number of distinct colors in each box. Probability = 108 / 180 = 0.6 Constraints: 1 <= balls.length <= 8 1 <= balls[i] <= 6 sum(balls) is even.

Explanation

Here's the breakdown of the approach, complexity analysis, and the Python code for solving this probability problem:

• High-Level Approach:

  • Calculate the total number of ways to distribute the balls into two boxes of equal size (n). This will be the denominator of our probability.
  • Calculate the number of ways to distribute the balls such that both boxes have the same number of distinct colors. We'll achieve this using a recursive backtracking approach.
  • Divide the number of favorable outcomes (same distinct colors) by the total number of outcomes to get the probability.

• Complexity Analysis:

  • Runtime Complexity: O(k * (sum(balls))^k), where k is the number of colors. This is due to the recursive nature of the backtracking, exploring many possible combinations.
  • Storage Complexity: O(k * sum(balls)), primarily due to the recursion depth and the storage of intermediate counts.

Code

    from collections import Counter
import math

def solve():
    balls = [1, 2, 1, 2]
    print(get_probability(balls))

def get_probability(balls):
    n = sum(balls) // 2
    k = len(balls)
    total_ways = combinations_with_replacement(sum(balls), balls)
    same_distinct_ways = count_same_distinct(balls, n, k)
    return same_distinct_ways / total_ways

def count_same_distinct(balls, n, k):
    def backtrack(index, box1, box2, distinct1, distinct2):
        if index == k:
            if sum(box1) == n and sum(box2) == n and distinct1 == distinct2:
                return 1
            else:
                return 0

        count = 0
        for i in range(balls[index] + 1):
            remaining = balls[index] - i
            new_box1 = box1[:]
            new_box2 = box2[:]
            new_box1.append(i)
            new_box2.append(remaining)

            new_distinct1 = distinct1 + (1 if i > 0 else 0)
            new_distinct2 = distinct2 + (1 if remaining > 0 else 0)
            count += backtrack(index + 1, new_box1, new_box2, new_distinct1, new_distinct2)

        return count

    return backtrack(0, [], [], 0, 0)

def combinations_with_replacement(n, balls):
    total_balls = sum(balls)
    numerator = math.factorial(total_balls)
    denominator = 1
    for ball_count in balls:
        denominator *= math.factorial(ball_count)
    return numerator / denominator

if __name__ == "__main__":
    solve()