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Problem solution · Java

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

Probability of a Two Boxes Having The Same Number of Distinct Balls: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
35 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Probability of a Two Boxes Having The Same Number of Distinct Balls, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 35 lines of Java from the credited upstream file 1467.java.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeProbability of a Two Boxes Having The Same Number of Distinct Balls · JavaJava
Use this to learn the idea, then write your own version.
enum BoxCase { EQUAL_BALLS, EQUAL_DISTANT_BALLS } class Solution {  public double getProbability(int[] balls) {    final int n = Arrays.stream(balls).sum() / 2;    return cases(balls, 0, 0, 0, 0, 0, n, BoxCase.EQUAL_DISTANT_BALLS) /        cases(balls, 0, 0, 0, 0, 0, n, BoxCase.EQUAL_BALLS);  }   private int[] fact = {1, 1, 2, 6, 24, 120, 720};   // Assume we have two boxes A and B.  double cases(int[] balls, int i, int ballsCountA, int ballsCountB, int colorsCountA,               int colorsCountB, int n, BoxCase boxCase) {    if (ballsCountA > n || ballsCountB > n)      return 0;    if (i == balls.length)      return boxCase == BoxCase.EQUAL_BALLS ? 1 : (colorsCountA == colorsCountB ? 1 : 0);     double ans = 0;     // balls taken from A for `balls[i]`    for (int ballsTakenA = 0; ballsTakenA <= balls[i]; ++ballsTakenA) {      final int ballsTakenB = balls[i] - ballsTakenA;      final int newcolorsCountA = colorsCountA + (ballsTakenA > 0 ? 1 : 0);      final int newcolorsCountB = colorsCountB + (ballsTakenB > 0 ? 1 : 0);      ans += cases(balls, i + 1, ballsCountA + ballsTakenA, ballsCountB + ballsTakenB,                   newcolorsCountA, newcolorsCountB, n, boxCase) /             (fact[ballsTakenA] * fact[ballsTakenB]);    }     return ans;  }}

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