Problem Statement
A string S is a subsequence of a string T if we can obtain S from T by erasing some (possibly all or none) of its characters. For example, "000" is a subsequence of "01010".
The longest common subsequence (LCS) of two strings A and B is a string C that is a subsequence of each of them and has the largest length among all strings with this property. Let f(A,B) be the length of the LCS of strings A and B. For example, we have f("101", "111000") = 2, f("101", "110011") = 3, and f("00", "1111") = 0.
You are given three small positive integers ab, bc, and ca. Please find three strings A, B, C such that:
You may assume that a solution always exist. If there are multiple solutions you may return any of them.
The longest common subsequence (LCS) of two strings A and B is a string C that is a subsequence of each of them and has the largest length among all strings with this property. Let f(A,B) be the length of the LCS of strings A and B. For example, we have f("101", "111000") = 2, f("101", "110011") = 3, and f("00", "1111") = 0.
You are given three small positive integers ab, bc, and ca. Please find three strings A, B, C such that:
- Each of the strings contains only the characters '0' and '1'.
- The length of each string is between 1 and 1,000, inclusive.
- f(A, B) = ab
- f(B, C) = bc
- f(C, A) = ca
You may assume that a solution always exist. If there are multiple solutions you may return any of them.
Definition
- Class:
- ConstructLCSEasy
- Method:
- construct
- Parameters:
- int, int, int
- Returns:
- String
- Method signature:
- String construct(int ab, int bc, int ca)
- (be sure your method is public)
Constraints
- ab will be between 1 and 50, inclusive.
- bc will be between 1 and 50, inclusive.
- ca will be between 1 and 50, inclusive.
- ab <= bc <= ca.
Examples
2
3
4
Returns: "1111 101 1010101"
The returned string corresponds to the following solution: A = "1111" B = "101" C = "1010101" We can easily verify that the only LCS of A and B is "11", the only LCS of B and C is "101", and the only LCS of C and A is "1111".4
4
7
Returns: "10101010 1011 1010101"
Another solution is: a = "0000111", b = "0000", c = "0000111".6
7
8
Returns: "10101010 1111010 110101010"
7
8
8
Returns: "10101010 010101101 110101001011"
15
17
19
Returns: "000100101101111011000 11110111010011101010 100100001010101001010101000011111"
50
50
50
Returns: "11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111"