Problem Description

You have 2n balls distributed across k different colors. The array balls tells you how many balls of each color you have - specifically, balls[i] represents the number of balls of color i.

The task is to:

1. Shuffle all the balls randomly
2. Put the first n balls into box 1
3. Put the remaining n balls into box 2

You need to calculate the probability that both boxes end up with the same number of distinct colors.

Important clarifications:

• The two boxes are distinguishable (order matters). For example, if you have balls of colors a and b, putting a in box 1 and b in box 2 is different from putting b in box 1 and a in box 2.
• "Same number of distinct colors" means both boxes have an equal count of unique colors. For instance, if box 1 has colors {red, blue} and box 2 has colors {green, yellow}, both have 2 distinct colors.
• The distribution is done after a uniform random shuffle of all balls.
• Your answer should be accurate within 10^-5 of the actual probability.

For example, if you have balls = [1, 1] (one ball of color 0 and one ball of color 1), and you distribute them into two boxes with 1 ball each, the probability that both boxes have the same number of distinct colors (which would be 1 distinct color in each box) can be calculated based on all possible distributions after shuffling.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: This problem involves distributing balls into boxes based on probability calculations, not traversing nodes and edges in a graph structure.

Need to solve for kth smallest/largest?

• No: We're calculating a probability value, not finding the kth smallest or largest element.

Involves Linked Lists?

• No: The problem uses an array to represent ball counts, not linked list operations.

Does the problem have small constraints?

• Yes: The problem has relatively small constraints. We have at most k distinct colors and 2n total balls. Since we need to explore all possible ways to distribute balls into two boxes while maintaining specific conditions, the search space is manageable but requires exploring multiple combinations.

Brute force / Backtracking?

• Yes: Backtracking is the appropriate approach here. We need to:
  • Try all possible ways to distribute balls of each color between the two boxes
  • Track the number of balls in each box (must sum to n each)
  • Track the number of distinct colors in each box
  • Count valid distributions where both boxes have the same number of distinct colors
  • Calculate the probability based on combinatorial counting

Conclusion: The flowchart correctly leads us to use a backtracking approach. The solution explores all possible distributions of balls across colors and boxes, using recursion with memoization to efficiently count valid arrangements. The backtracking nature is evident in how we try different amounts of each color in box 1 (from 0 to balls[i]), recursively solve for remaining colors, and aggregate the results using combinatorial mathematics.

Intuition

To find the probability that both boxes have the same number of distinct colors, we need to think about what we're actually counting.

The key insight is that we can use the formula: Probability = (Favorable Outcomes) / (Total Outcomes)

For total outcomes: After shuffling all 2n balls, we pick n balls for the first box and n for the second. The total number of ways to do this is C(2n, n) - the binomial coefficient for choosing n balls from 2n.

For favorable outcomes: We need to count distributions where both boxes have equal distinct colors. This is where backtracking comes in.

The clever approach is to iterate through each color and decide how to split its balls between the two boxes. For color i with balls[i] balls, we can put anywhere from 0 to balls[i] balls in the first box (the rest go to the second box).

Here's the critical observation:

• If we put all balls of a color in one box, that box gains a distinct color while the other doesn't
• If we put 0 balls in the first box, the second box gets all of them and gains that distinct color
• If we split the balls between boxes, both boxes get that color

We track this using a "difference" counter:

• +1 when only the first box gets a color
• -1 when only the second box gets a color
• 0 when both boxes get the color (or neither, but that's not possible since we have balls of each color)

A valid distribution has diff = 0 (equal distinct colors in both boxes) and exactly n balls in the first box.

For each way to distribute balls of a single color, we multiply by C(balls[i], x) where x is the number we put in the first box. This accounts for which specific balls of that color go to which box.

The recursion explores all combinations, accumulating the count of valid distributions multiplied by their respective combinatorial coefficients. Memoization prevents recalculating the same states.

Learn more about Math, Dynamic Programming, Backtracking and Combinatorics patterns.

Solution Approach

The solution uses dynamic programming with memoization through a recursive backtracking approach. Let's break down the implementation:

Core Function Structure: The main recursive function dfs(i, j, diff) takes three parameters:

• i: Current color index we're processing (0 to k-1)
• j: Remaining balls needed for the first box
• diff: Difference in distinct colors between boxes

Base Cases:

1. When i >= k (processed all colors):
   • Return 1 if j == 0 (first box has exactly n balls) AND diff == 0 (equal distinct colors)
   • Return 0 otherwise
2. When j < 0: Return 0 (invalid state - too many balls in first box)

Recursive Logic: For each color i, we try all possible distributions:

for x in range(balls[i] + 1):

Where x is the number of balls of color i going to the first box.

Tracking Distinct Colors: The variable y tracks how the distribution affects distinct colors:

• y = 1 if x == balls[i] (all balls go to first box, only first box gets this color)
• y = -1 if x == 0 (all balls go to second box, only second box gets this color)
• y = 0 otherwise (both boxes get this color)

Combinatorial Counting: For each distribution choice, we multiply by comb(balls[i], x) which represents the number of ways to choose which specific x balls of color i go to the first box.

State Accumulation:

ans += dfs(i + 1, j - x, diff + y) * comb(balls[i], x)

• Move to next color (i + 1)
• Reduce remaining balls for first box by x (j - x)
• Update the difference counter (diff + y)
• Multiply by combinatorial coefficient

Final Probability Calculation:

return dfs(0, n, 0) / comb(n << 1, n)

• Start with color 0, need n balls in first box, difference of 0
• Divide favorable outcomes by total outcomes C(2n, n)

Optimization: The @cache decorator memoizes results, preventing redundant calculations for the same (i, j, diff) states.

Variable Initialization:

• n = sum(balls) >> 1: Total balls divided by 2 (using bit shift for efficiency)
• k = len(balls): Number of distinct colors

This approach efficiently explores all valid distributions while avoiding redundant calculations, counting exactly those arrangements where both boxes have the same number of distinct colors.

Example Walkthrough

Let's walk through a small example with balls = [1, 2, 1] representing:

• 1 ball of color 0 (red)
• 2 balls of color 1 (blue)
• 1 ball of color 2 (green)

Total: 4 balls, so each box gets 2 balls.

Step 1: Calculate total possible outcomes Total ways to choose 2 balls from 4 = C(4,2) = 6

Step 2: Use DFS to find favorable outcomes

Starting with dfs(0, 2, 0) - process color 0, need 2 balls for box 1, difference = 0.

For color 0 (1 red ball), we can put:

• 0 balls in box 1: dfs(1, 2, -1) * C(1,0) = dfs(1, 2, -1) * 1
  • Box 2 gets red, diff becomes -1
• 1 ball in box 1: dfs(1, 1, 1) * C(1,1) = dfs(1, 1, 1) * 1
  • Box 1 gets red, diff becomes +1

Following path dfs(1, 2, -1) - color 1 (2 blue balls):

• 0 balls in box 1: Would make diff = -2 (box 2 has 2 colors, box 1 has 0)
• 1 ball in box 1: dfs(2, 1, -1) * C(2,1) = dfs(2, 1, -1) * 2
  • Both boxes get blue, diff stays -1
• 2 balls in box 1: dfs(2, 0, 0) * C(2,2) = dfs(2, 0, 0) * 1
  • Box 1 gets blue, diff becomes 0

Continuing dfs(2, 0, 0) - color 2 (1 green ball):

• 0 balls in box 1: dfs(3, 0, -1) * 1 → Returns 0 (diff ≠ 0)
• 1 ball in box 1: Would need -1 balls (invalid)

So this path contributes: 1 * 1 * 1 = 1

Following similar logic for all paths:

Valid distributions where diff = 0 and j = 0:

1. Red→Box1, Blue(1)→Box1, Blue(1)→Box2, Green→Box2

   • Box 1: {red, blue}, Box 2: {blue, green} ✓ (2 colors each)
   • Contribution: 1 * 2 = 2

2. Red→Box2, Blue(1)→Box1, Blue(1)→Box2, Green→Box1

   • Box 1: {blue, green}, Box 2: {red, blue} ✓ (2 colors each)
   • Contribution: 1 * 2 = 2

Total favorable outcomes = 2 + 2 = 4

Step 3: Calculate probability Probability = 4/6 = 2/3 ≈ 0.66667

The algorithm systematically explores all possible distributions, using the difference counter to track when both boxes have equal distinct colors, and multiplying by combinatorial coefficients to account for which specific balls go where.

Solution Implementation

Time and Space Complexity

Time Complexity: O(k * n * k * max(balls[i]))

The analysis breaks down as follows:

• The DFS function has three parameters: i (ranges from 0 to k), j (ranges from -max(balls) to n), and diff (ranges from -k to k)
• Due to memoization with @cache, each unique state (i, j, diff) is computed only once
• The number of unique states is O(k * n * k) where:
  • i has k possible values
  • j has at most n+1 relevant values (though it can go negative, early termination occurs)
  • diff has at most 2k+1 possible values (from -k to k)
• For each state, we iterate through all possible distributions of balls[i], which takes O(max(balls[i])) time
• The combination calculation comb(balls[i], x) can be considered O(1) with preprocessing or O(balls[i]) without
• Overall: O(k * n * k * max(balls[i]))

Space Complexity: O(k * n * k)

The space analysis:

• The recursion depth is at most O(k) since we process k different ball colors
• The memoization cache stores all unique states (i, j, diff), which is O(k * n * k)
• The cache dominates the space complexity
• Total space: O(k * n * k)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Color Difference Tracking

Pitfall: A common mistake is incorrectly calculating how the distribution of balls affects the difference in distinct colors between boxes. Developers often misunderstand when to increment, decrement, or keep the difference unchanged.

Wrong Implementation:

# Incorrect logic that doesn't properly track color distribution
if balls_in_first_box > 0:
    color_diff_change = 1
elif balls_in_first_box < balls[color_index]:
    color_diff_change = -1
else:
    color_diff_change = 0

Correct Implementation:

# Proper tracking: only change diff when ONE box gets ALL balls of a color
if balls_in_first_box == balls[color_index]:
    color_diff_change = 1  # Only first box has this color
elif balls_in_first_box == 0:
    color_diff_change = -1  # Only second box has this color
else:
    color_diff_change = 0  # Both boxes have this color

2. Floating Point Precision Issues

Pitfall: Using regular division throughout the recursion can accumulate floating-point errors, especially with large factorials.

Wrong Approach:

def dfs(i, j, diff):
    # Dividing at each step causes precision loss
    return (dfs(i + 1, j - x, diff + y) * comb(balls[i], x)) / some_value

Better Approach:

# Keep calculations as integers until the final division
def dfs(i, j, diff):
    total_ways += dfs(i + 1, j - x, diff + y) * comb(balls[i], x)
    return total_ways

# Only divide once at the end
return dfs(0, n, 0) / comb(2 * n, n)

3. Missing or Incorrect Base Case Handling

Pitfall: Forgetting to check if remaining_balls < 0 or incorrectly handling the terminal condition can lead to wrong counts or infinite recursion.

Wrong Implementation:

def dfs(color_index, remaining_balls, color_difference):
    if color_index >= num_colors:
        return 1 if color_difference == 0 else 0  # Forgot to check remaining_balls

Correct Implementation:

def dfs(color_index, remaining_balls, color_difference):
    if color_index >= num_colors:
        # Must check BOTH conditions
        return 1 if remaining_balls == 0 and color_difference == 0 else 0
  
    if remaining_balls < 0:  # Important pruning condition
        return 0

4. Inefficient State Space Exploration

Pitfall: Not pruning impossible states early, leading to unnecessary recursive calls.

Inefficient:

def dfs(i, j, diff):
    if i >= k:
        return 1 if j == 0 and diff == 0 else 0
  
    # Explores all states even when j is already negative
    for x in range(balls[i] + 1):
        # ...

Optimized:

def dfs(i, j, diff):
    if i >= k:
        return 1 if j == 0 and diff == 0 else 0
  
    if j < 0:  # Early termination
        return 0
  
    # Additional optimization: check if remaining balls are sufficient
    remaining_total = sum(balls[i:])
    if j > remaining_total:  # Impossible to fill first box
        return 0

5. Incorrect Combinatorial Calculation

Pitfall: Using the wrong formula for total possible distributions or forgetting that the boxes are distinguishable.

Wrong:

# Treating boxes as indistinguishable
total_distributions = comb(2*n, n) // 2  # Wrong!

Correct:

# Boxes ARE distinguishable
total_distributions = comb(2*n, n)  # C(2n, n) ways to choose n balls for first box