CPMagical CP Library

Problem solution · Java

Count the Number of Infection Sequences

Count the Number of Infection Sequences: 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
58 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 Count the Number of Infection Sequences, 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

  • 58 lines of Java from the credited upstream file 2954.java.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks 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 codeCount the Number of Infection Sequences · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int numberOfSequence(int n, int[] sick) {    final long[][] factAndInvFact = getFactAndInvFact(n - sick.length);    final long[] fact = factAndInvFact[0];    final long[] invFact = factAndInvFact[1];    long ans = fact[n - sick.length]; // the number of infected children    int prevSick = -1;     for (int i = 0; i < sick.length; ++i) {      // The segment [prevSick + 1, sick - 1] are the current non-infected      // children.      final int nonInfected = sick[i] - prevSick - 1;      prevSick = sick[i];      if (nonInfected == 0)        continue;      ans *= invFact[nonInfected];      ans %= MOD;      if (i > 0) {        // There're two choices per second since the children at the two        // endpoints can both be the infect candidates. So, there are        // 2^{nonInfected - 1} ways to infect all children in the current        // segment.        ans *= modPow(2, nonInfected - 1);        ans %= MOD;      }    }     final int nonInfected = n - sick[sick.length - 1] - 1;    ans *= invFact[nonInfected];    return (int) (ans % MOD);  }   private static final int MOD = 1_000_000_007;   private long[][] getFactAndInvFact(int n) {    long[] fact = new long[n + 1];    long[] invFact = new long[n + 1];    long[] inv = new long[n + 1];    fact[0] = invFact[0] = 1;    inv[0] = inv[1] = 1;    for (int i = 1; i <= n; ++i) {      if (i >= 2)        inv[i] = MOD - MOD / i * inv[MOD % i] % MOD;      fact[i] = fact[i - 1] * i % MOD;      invFact[i] = invFact[i - 1] * inv[i] % MOD;    }    return new long[][] {fact, invFact};  }   private long modPow(long x, long n) {    if (n == 0)      return 1;    if (n % 2 == 1)      return x * modPow(x % MOD, (n - 1)) % MOD;    return modPow(x * x % MOD, (n / 2)) % MOD;  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗