Thursday Code Puzzler: Longest Common Subsequence
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Longest Common Subsequence
The LCS of two groups is the longest group that are common between both and in the same order in each group. Let's look at an example:
Group A: dzone puzzles
Group B: dropzone
The LCS here is dzone
Another example:
Group A: 1234
Group B: 12222354
If you're still confused, there's a good explanation over at Wikipedia.
What you need to do in this challenge is write a method that identifies the LCS between two String parameters.
Catch up on all our previous puzzlers here.
Comments
Vijay Nathani replied on Thu, 2012/08/30 - 4:05am
Groovy solution
def maxSubSequence(a,b) { def convert = { it.toList().subsequences().collect {it.join('')} } def sequencesA = convert(a) sequencesA.retainAll(convert(b)) sequencesA.max {it.size()} } println maxSubSequence('dzone puzzles','dropzone') //prints dopzeprintln maxSubSequence('1234','12222354') //prints 1234
James Nute replied on Thu, 2012/08/30 - 3:27pm
#include <iostream> #include <string> using namespace std; std::string LCS(std::string groupA, std::string groupB){ std::string result; int lengthOfA = groupA.length()-1; int lengthOfB = groupB.length()-1; char endOfA = groupA[lengthOfA]; char endOfB = groupB[lengthOfB]; if(lengthOfA < 0 || lengthOfB < 0){ return result; } if(endOfA==endOfB){ result.insert(0,1,endOfA); result.insert(0,LCS(groupA.substr(0,lengthOfA),groupB.substr(0,lengthOfB))); } else{ std::string lcs1(LCS(groupA.substr(0,lengthOfA+1),groupB.substr(0,lengthOfB))); std::string lcs2(LCS(groupA.substr(0,lengthOfA),groupB.substr(0,lengthOfB+1))); if(lcs1.length() > lcs2.length()){ result.insert(0,lcs1); } else{ result.insert(0,lcs2); } } return result; } int main() { std::string groupA("dzone puzzles"); std::string groupB("dropzone"); cout << "LCS is " << LCS(groupA,groupB) <<endl; return 0; }sun east replied on Thu, 2012/08/30 - 7:29pm
A short but not efficient solution written by Scala:
def lcs(a: String, b: String): String = { if(a.size > 0 && b.size > 0) { val pre = if(a(0) == b(0)) ""+a(0) else "" Seq(lcs(a,b.tail),lcs(a.tail,b), pre + lcs(a.tail,b.tail)).maxBy(_.size) } else "" }Vijay Nathani replied on Thu, 2012/08/30 - 11:22pm
in response to:
James Nute
Awesome. Your recursive algorithm in groovy
def longerString(s1,s2) { s1.size() > s2.size()? s1 : s2 } def lcs(a,b) { if (!a || !b) return "" (a[0] == b[0])? (a[0] + lcs(a.substring(1),b.substring(1))): longerString(lcs(a,b.substring(1)), lcs(a.substring(1),b)) } println lcs('dzone puzzles','dropzone') //prints dopzeprintln lcs('1234','122222354') //prints 1234