TopcoderArchive
TCO11 Apr 2010
TCO10 Round 5

OverlappingConcatenation

Problem statement, definition, constraints, and public examples.

Problem Statement

An overlapping concatenation of an ordered pair of Strings (A, B) is a String S such that A is a prefix of S and B is a suffix of S. A shortest overlapping concatenation (SOC) of (A, B) is an overlapping concatenation whose length is as small as possible. For example, if A is "cabab" and B is "ababc", the shortest overlapping concatenation is "cababc" (all quotes for clarity). You are given two Strings A and B, in which some of the characters are known and some are unknown. Each character in A and B will be a lowercase letter ('a' - 'z') if it is known or '*' if it is unknown.

Assuming that each unknown character will be selected uniformly at random from the characters contained in alphabet, return the expected length of the shortest overlapping concatenation of A and B.

Definition

Class:
OverlappingConcatenation
Method:
shortestExpected
Parameters:
String, String, String
Returns:
double
Method signature:
double shortestExpected(String A, String B, String alphabet)
(be sure your method is public)

Constraints

  • alphabet will contain between 1 and 26 lowercase letters ('a' - 'z'), inclusive.
  • The characters in alphabet will be distinct.
  • A and B will contain between 1 and 32 characters, inclusive.
  • A and B will contain the same number of characters.
  • Each character in A and B will either be contained in alphabet or it will be '*'.

Examples

  1. "aa*aa"
    "a*aaa"
    "ab"
    Returns: 7.0
    The 4 equal-probability possibilities for the shortest overlapping concatenation (SOC) are as follows: A | B | SOC aaaaa | aaaaa | aaaaa aabaa | aaaaa | aabaaaaa aaaaa | abaaa | aaaaabaaa aabaa | abaaa | aabaaa The result is therefore (5 + 8 + 9 + 6) / 4 = 7.
  2. "**"
    "**"
    "ab"
    Returns: 3.125
    There are 16 possibilities for A and B. Of these, 4 give a SOC of length 2; 6 give a SOC of length 3; and 6 give a SOC of length 4. The answer is therefore (4*2 + 6*3 + 6*4) / 16 = 3.125.
  3. "**a*cd"
    "cde***"
    "abcdefghijklmnopqrstuvwxyz"
    Returns: 10.0
    It makes no difference what the unknown characters are; the SOC is always the same length.
  4. "h*phz*"
    "hzph*p"
    "zph"
    Returns: 10.555555555555557
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