1467. Probability of a Two Boxes Having The Same Number of Distinct Balls


Problem Description

In this problem, we are given 2n balls of k different colors. We're also provided with an array called balls, which has k elements where each element, balls[i], represents the count of balls of color i.

Our task is to shuffle all the balls uniformly at random, then distribute the first n balls into one box and the remaining n balls into a second box. We must calculate the probability that both boxes end up with the same number of distinct colors.

To provide further clarity, the boxes are distinct such that arrangement matters – putting a red ball in the first box and a blue ball in the second box is different from putting the blue ball in the first box and the red ball in the second box. Our goal is to find the likelihood of both boxes having an equal variety of colors after the distribution is completed.

It's a challenging combinatorial probability question, as we need to consider all possible distributions of the balls into the boxes and then determine the proportion where the number of unique colors per box is identical.

Intuition

The intuition behind the solution is linked to depth-first search (DFS) and combinatorics concepts. Essentially, we use DFS to traverse through all possible combinations of distributions of balls into the two boxes, and we calculate the probability of those distributions where the number of distinct balls in each box matches.

We solve the problem recursively by considering one color at a time with the following parameters:

  1. i - the current color index.
  2. j - the number of balls to be placed in the first box. Initially, this is n, half of all balls.
  3. diff - the difference in the number of distinct colors between the two boxes.

We employ the memoization technique (using @cache) to prevent recalculating results for the same state, which significantly reduces the complexity.

At each recursive step (dfs function), we explore all possible distributions of the current color's balls into the two boxes. For example, if we have 3 balls of the current color, we can place 0, 1, 2, or 3 balls in the first box (and the rest will go to the second box automatically). We adjust the value of diff accordingly; it increases by 1 if all balls of the current color go into one box, decreases by 1 if none go into the first box, and remains unchanged otherwise.

For each such configuration, we calculate the probabilities using the comb function, which computes binomial coefficients, and sum these probabilities up to get the total probability for the current state.

The base case for the recursion is when all colors have been considered (i >= k). If at this point, there are no more balls to place (j == 0) and the difference in distinct colors is zero (diff == 0), it means we have found a valid distribution and return 1; otherwise, we return 0 as it's an invalid distribution.

Finally, the probability of having the same number of distinct balls in each box is the result of our dfs function divided by the total number of ways to distribute 2n balls into two groups of n (which is the binomial coefficient comb(2n, n)).

In short, the solution combines combinatorial math with recursive DFS and memoization to efficiently count the number of valid distributions where each box has the same number of unique colored balls.

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

Solution Approach

The problem is approached with a combination of recursive depth-first search (DFS), dynamic programming through memoization, and combinatorial mathematics. Here's a step-by-step breakdown of the getProbability method:

  1. Memoization Decorator (@cache): This is a Python feature used to store the results of costly function calls and return the cached result when the same inputs occur again. By decorating the dfs method with @cache, we ensure we only compute each state once, even though it might be reached by different paths in the recursion tree. This is key to improving performance and is part of a dynamic programming strategy.

  2. Recursive DFS (dfs): The dfs function is the core of this recursive approach. It takes three parameters - i, which is the current index of the colors being considered; j, which indicates the number of balls to be placed in the first box; and diff, a value representing the difference in the number of distinct colors between the two boxes.

  3. Base Case: The recursion stops when all color indices have been considered (i >= k). At this point, if j (the number of balls left to place in the first box) is zero and diff is zero, it means both boxes have an equal number of balls and an equal number of distinct colors, respectively. Therefore, it is a valid distribution, and a probability of 1 is returned. Otherwise, a probability of 0 is returned since it doesn't meet the conditions.

  4. Ball Distribution Loop: For each color, we iterate through all possible distributions of the current color's balls (ranging from 0 to the total number of balls of that color). The loop variable x represents the number of balls of the current color placed in the first box.

  5. Difference Update: The diff is updated based on whether all balls of the current color go into one box (diff + 1), or none do (diff - 1), or the balls are split between the boxes.

  6. Combination Calculation (comb): We calculate the number of combinations for placing x balls of the current color in the first box using the comb function from Python's math library, which computes binomial coefficients. This represents the number of ways to choose x from balls[i].

  7. Recursive Call: We make the recursive call with updated parameters and multiply the result by the calculated combinations to get the probability for that particular distribution.

  8. Probability Calculation: After traversing all colors and possible distributions of balls, the accumulated result is the number of valid distributions where the number of distinct balls is the same in each box. To convert this into a probability, we divide it by the total number of ways to distribute all balls into two groups of n (computed by comb(2n, n)).

The returned value of dfs(0, n, 0) / comb(n << 1, n) is the final probability that we need to output.

The data structure used in the algorithm is primarily recursion with memoization, while the combinatorics part of it taps into mathematical formulas for computing binomial coefficients. The @cache decorator seamlessly accomplishes memoization without additional code for managing a memoization table.

By leveraging the principles of divide-and-conquer along with dynamic programming, the implementation successfully handles the complexity of the problem by breaking it down into smaller, manageable subproblems.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

What is the running time of the following code?

1int sqrt(int n) {
2  for (int guess = 1; guess * guess <= n; guess++) {
3    if (guess * guess == n) {
4      return guess;
5    }
6  }
7  return -1;
8}

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have 2n = 4 balls and k = 2 different colors. Our balls array, representing the count of balls for each color, is [2, 2], which means we have 2 balls of color 1 and 2 balls of color 2.

We want to distribute these 4 balls into two boxes, each holding 2 balls (n = 2) and find the probability that both boxes end up with the same number of distinct colors, which in this case must be 1 color each, for them to have the same amount.

Let's follow the steps of the solution approach:

  1. Recursive DFS (dfs): We start with the first color (i=0), needing to place n = 2 balls in the first box and a diff of 0 since initially, no box has any balls.

  2. Ball Distribution Loop: We explore all distributions of color 1's balls:

    • Place 0 balls in the first box: dfs(1, 2, 1) because no balls are placed in the first box.
    • Place 1 ball in the first box: dfs(1, 1, 0) because we placed one and have one left for the other box.
    • Place 2 balls in the first box: dfs(1, 0, 1) because all balls of this color are in the first box.
  3. Recursive Calls and Combination Calculation:

    • For dfs(1, 2, 1): We move to the next color and again distribute balls with updated parameters.
    • For dfs(1, 1, 0): We do the same; this might lead to a scenario where each box gets one ball of color 1 and one ball of color 2, which matches our condition for equal distinct colors.
    • For dfs(1, 0, 1): Similar to above, but since all balls of the first color are in the first box, we expect to place all balls of the second color in the second box to keep diff at zero.
  4. Base Case: We reach the base case when i = k (2) - when we've considered all colors:

    • If j is not zero or diff is not zero, this configuration is invalid (probability 0).
    • If j is zero and diff is zero, this configuration is valid (probability 1).
  5. Accumulating Probabilities: Each valid state contributes to the probability. We add the probabilities of all valid distributions.

  6. Total Probability: Finally, we calculate the total number of ways to distribute the 2n balls into two boxes (comb(4, 2) in this example), which is 6. We divide the accumulated valid distributions by this number to get the probability.

For this particular example, there will be 2 valid distributions where each box ends up with one ball of color 1 and one ball of color 2:

  • One distribution comes from the dfs(1, 1, 0) branch.
  • The other distribution is from the opposite scenario, which is symmetric given the uniform random distribution of balls.

So the probability of both boxes having the same number of distinct colors is 2 / comb(4, 2) = 2 / 6 = 1 / 3 or approximately 0.3333.

This small example captures the essence of the DFS and combinatorics involved in the approach for calculating probabilities of color distribution.

Solution Implementation

1from functools import lru_cache  # Import lru_cache for memoization
2from math import comb  # Import comb to calculate binomial coefficients
3
4class Solution:
5    def getProbability(self, balls: List[int]) -> float:
6        # Define a recursive function with memoization through lru_cache
7        @lru_cache(None)
8        def dfs(index: int, balls_in_first: int, color_difference: int) -> float:
9            # Base case: when all balls have been distributed
10            if index >= number_of_colors:
11                return 1 if balls_in_first == 0 and color_difference == 0 else 0
12            # If the number of balls in the first urn is negative, the combination is impossible
13            if balls_in_first < 0:
14                return 0
15          
16            # Initialize the answer for the current recursive step
17            answer = 0
18            # Iterate over all possible distributions of balls for the current color
19            for balls_to_first_urn in range(balls[index] + 1):
20                # Check if this distribution would change the color difference
21                color_diff_change = 1 if balls_to_first_urn == balls[index] else (-1 if balls_to_first_urn == 0 else 0)
22                # Recursive call for the next color, adjusting the accumulated balls and color difference
23                answer += dfs(index + 1, balls_in_first - balls_to_first_urn, color_difference + color_diff_change) * comb(balls[index], balls_to_first_urn)
24            return answer
25
26        # Calculate the total number of balls and the number of colors
27        total_balls = sum(balls)
28        half_total_balls = total_balls >> 1  # Shifting right to divide by 2
29        number_of_colors = len(balls)
30
31        # Calculate the valid arrangements over all possible arrangements (combinatorial total)
32        valid_arrangements = dfs(0, half_total_balls, 0)
33        total_arrangements = comb(total_balls, half_total_balls)
34      
35        # Return the probability as the ratio of valid to total arrangements
36        return valid_arrangements / total_arrangements
37
1import java.util.HashMap;
2import java.util.List;
3import java.util.Map;
4
5public class Solution {
6    private int halfTotalBalls;
7    private long[][] binomialCoefficients;
8    private int[] balls;
9    private Map<List<Integer>, Long> cache = new HashMap<>();
10
11    // Calculates the probability of splitting the balls into two fair halves
12    public double getProbability(int[] balls) {
13        int maxBalls = 0;
14        for (int ballCount : balls) {
15            halfTotalBalls += ballCount; // Total number of balls
16            maxBalls = Math.max(maxBalls, ballCount); // Max number of balls in any single color
17        }
18        halfTotalBalls >>= 1; // Half of the total balls
19        this.balls = balls; // Assign the balls array
20        int matrixSize = Math.max(maxBalls, halfTotalBalls << 1);
21        binomialCoefficients = new long[matrixSize + 1][matrixSize + 1];
22      
23        // Calculate binomial coefficients using dynamic programming
24        for (int i = 0; i <= matrixSize; ++i) {
25            binomialCoefficients[i][0] = 1;
26            for (int j = 1; j <= i; ++j) {
27                binomialCoefficients[i][j] = binomialCoefficients[i - 1][j - 1] + binomialCoefficients[i - 1][j];
28            }
29        }
30      
31        // Calculate the probability using depth-first search
32        long validCombinationsCount = dfs(0, halfTotalBalls, 0);
33        long totalCombinationsCount = binomialCoefficients[halfTotalBalls << 1][halfTotalBalls];
34        return (double) validCombinationsCount / totalCombinationsCount;
35    }
36
37    // Helper for depth-first search to calculate valid combinations
38    private long dfs(int index, int remainingBalls, int colorDifference) {
39        // Base case: if all balls are used and color difference is 0, count as valid
40        if (index >= balls.length) {
41            return (remainingBalls == 0 && colorDifference == 0) ? 1 : 0;
42        }
43        if (remainingBalls < 0) {
44            return 0;
45        }
46      
47        // Create a key for caching
48        List<Integer> key = List.of(index, remainingBalls, colorDifference);
49        if (cache.containsKey(key)) {
50            return cache.get(key);
51        }
52      
53        long validCombinations = 0;
54        for (int x = 0; x <= balls[index]; ++x) {
55            int colorDiffAdjustment = (x == balls[index]) ? 1 : (x == 0 ? -1 : 0);
56            validCombinations += dfs(index + 1, remainingBalls - x, colorDifference + colorDiffAdjustment)
57                                * binomialCoefficients[balls[index]][x];
58        }
59      
60        cache.put(key, validCombinations);
61        return validCombinations;
62    }
63}
64
1#include <vector>
2#include <numeric>
3#include <algorithm>
4#include <cstring>
5#include <functional>
6
7class Solution {
8public:
9    double getProbability(std::vector<int>& balls) {
10        // Calculate the total number of half-balls given that balls are divided into two halves.
11        int halfBalls = std::accumulate(balls.begin(), balls.end(), 0) / 2;
12        // Find the maximum number of balls of any color.
13        int maxBalls = *std::max_element(balls.begin(), balls.end());
14        // m is the maximum value for computing combinations, accounting for both cases.
15        int maxCombinations = std::max(maxBalls, halfBalls * 2);
16        // Initialize the combination table using Pascal's triangle.
17        long long combination[maxCombinations + 1][maxCombinations + 1];
18        std::memset(combination, 0, sizeof(combination));
19        for (int i = 0; i <= maxCombinations; ++i) {
20            combination[i][0] = 1;
21            for (int j = 1; j <= i; ++j) {
22                combination[i][j] = combination[i - 1][j - 1] + combination[i - 1][j];
23            }
24        }
25        // The number of different colors.
26        int colorCount = balls.size();
27        // Memoization array to store intermediate results.
28        long long memo[colorCount][halfBalls + 1][colorCount * 2 + 1];
29        std::memset(memo, -1, sizeof(memo));
30      
31        // Depth-first search function to try all possible distributions of balls.
32        std::function<long long(int, int, int)> dfs = [&](int index, int remaining, int colorDiff) -> long long {
33            // If all balls have been distributed, check if each side has the same number of balls
34            // and the same number of colors.
35            if (index >= colorCount) {
36                return (remaining == 0 && colorDiff == colorCount) ? 1 : 0;
37            }
38            // If the remaining number of balls is negative, then it's an invalid distribution.
39            if (remaining < 0) {
40                return 0;
41            }
42            // If this state has been computed before, return the stored result.
43            if (memo[index][remaining][colorDiff] != -1) {
44                return memo[index][remaining][colorDiff];
45            }
46            long long ans = 0;
47            // Try all possible distributions of the current color.
48            for (int x = 0; x <= balls[index]; ++x) {
49                // If all balls of current color go to one side, increment the colorDiff; otherwise, decrement or leave unchanged.
50                int colorBias = (x == balls[index]) ? 1 : (x == 0 ? -1 : 0);
51                // Recursively explore the next color while adjusting the remaining balls and color difference.
52                ans += dfs(index + 1, remaining - x, colorDiff + colorBias) * combination[balls[index]][x];
53            }
54            // Store the computed result.
55            return memo[index][remaining][colorDiff] = ans;
56        };
57      
58        // Calculate the probability by dividing the number of valid combinations by the total combinations for distributing balls.
59        return static_cast<double>(dfs(0, halfBalls, colorCount)) / combination[halfBalls * 2][halfBalls];
60    }
61};
62
1function getProbability(balls: number[]): number {
2    // Calculate the total number of balls and half of it.
3    const halfTotalBalls = balls.reduce((a, b) => a + b, 0) >> 1;
4    // Determine the maximum number of balls in any single color (bucket).
5    const maxBallsSingleColor = Math.max(...balls);
6    // Compute maximum value using both max in any single color and total balls.
7    const maxValue = Math.max(maxBallsSingleColor, halfTotalBalls << 1);
8    // Precompute all combinations using Pascal's Triangle.
9    const combination: number[][] = Array(maxValue + 1)
10        .fill(0)
11        .map(() => Array(maxValue + 1).fill(0));
12    for (let i = 0; i <= maxValue; ++i) {
13        combination[i][0] = 1;
14        for (let j = 1; j <= i; ++j) {
15            combination[i][j] = combination[i - 1][j - 1] + combination[i - 1][j];
16        }
17    }
18    // Number of colors.
19    const colorCount = balls.length;
20    // Initialize memoization array for Dynamic Programming.
21    const dp: number[][][] = Array(colorCount)
22        .fill(0)
23        .map(() =>
24            Array(halfTotalBalls + 1)
25                .fill(0)
26                .map(() => Array((colorCount << 1) | 1).fill(-1)),
27        );
28      
29    // Depth-first search function for recursive solution.
30    const dfs = (index: number, remaining: number, colorDifference: number): number => {
31        // Base case: all colors processed.
32        if (index >= colorCount) {
33            return remaining === 0 && colorDifference === colorCount ? 1 : 0;
34        }
35        // Not enough balls to distribute.
36        if (remaining < 0) {
37            return 0;
38        }
39        // Return memoized results to avoid recomputing.
40        if (dp[index][remaining][colorDifference] !== -1) {
41            return dp[index][remaining][colorDifference];
42        }
43        let ans = 0;
44        // Iterate over number of balls to put in the first half.
45        for (let x = 0; x <= balls[index]; ++x) {
46            const colorDiff = x === balls[index] ? 1 : x === 0 ? -1 : 0;
47             ans += dfs(index + 1, remaining - x, colorDifference + colorDiff) * combination[balls[index]][x];
48        }
49        // Memoize and return the computed result.
50        return (dp[index][remaining][colorDifference] = ans);
51    };
52  
53    // Calculate the probability using the dfs function and combinations.
54    return dfs(0, halfTotalBalls, colorCount) / combination[halfTotalBalls << 1][halfTotalBalls];
55}
56

Time and Space Complexity

The time complexity of this recursive solution is based on several factors: the number of recursive calls dfs can potentially make, the limitation of the parameters j (the number of balls in one half), and diff (the difference in the number of unique balls between the two halves). Each state is uniquely defined by the tuple (i, j, diff), where i ranges from 0 to k, j ranges from 0 to n, and diff can theoretically range from -k to k. However, since the difference in unique balls is evaluated by adding 1, subtracting 1, or leaving it unchanged at each step - the actual range of diff is limited to the number of balls in balls[i].

The actual time complexity is O(k * n * min(n, k) * maxBallCount), where maxBallCount is the maximum value in balls, representing the maximum branching factor at each recursive call, due to the number of ways x can be selected from balls[i].

The @cache decorator applies memoization, which means each of the O(k * n * min(n, k)) states is computed only once. Thus, the space complexity is the same as the time complexity without considering the maxBallCount factor, which is O(k * n * min(n, k)). However, we should note that the space required for storing the intermediate combinations (comb(balls[i], x)) must also be considered. Since the combination values are calculated once and stored, this does not significantly change the space complexity.

Another point to consider is the computation of combinations using comb. Each call to comb takes O(min(x, balls[i] - x)) time, but since this is memoized as well, this computation contributes a constant factor in terms of maxBallCount, under the assumption that comb uses memoization, which is common in combination computations.

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


Fast Track Your Learning with Our Quick Skills Quiz:

What's the output of running the following function using the following tree as input?

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12
1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18
1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

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