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@nahurst
Created February 21, 2011 21:27
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Longest Common Subsequence
/**
* Find a longest common subsenquence.
*
* Note: this is not necessarily the only possible longest common subsequence though!
*/
function lcs(listX, listY) {
return lcsBackTrack(
lcsLengths(listX, listY),
listX, listY,
listX.length, listY.length);
}
/**
* Iteratively memoize a matrix of longest common subsequence lengths.
*/
function lcsLengths(listX, listY) {
var lenX = listX.length;
var lenY = listY.length;
// Initialize a lenX+1 x lenY+1 matrix
var memo = [lenX+1];
for (var i = 0; i < lenX+1; i++) {
memo[i] = [lenY+1];
for (var j = 0; j < lenY+1; j++) {
memo[i][j] = 0;
}
}
// Memoize the lcs length at each position in the matrix
for (var i = 1; i < lenX+1; i++) {
for (var j = 1; j < lenY+1; j++) {
if (listX[i-1] == listY[j-1]) {
memo[i][j] = memo[i-1][j-1] + 1;
}
else {
memo[i][j] = Math.max(
memo[i][j-1],
memo[i-1][j]);
}
}
}
return memo;
}
/**
* Recursively read back a memoized matrix of longest common subsequence lengths
* to find a longest common subsequence.
*/
function lcsBackTrack(memo, listX, listY, posX, posY) {
// base case
if (posX == 0 || posY == 0) {
return "";
}
// matcth => go up and left
else if (listX [posX-1] == listY[posY-1]) {
return lcsBackTrack(memo, listX, listY, posX-1, posY-1) + listX[posX-1];
}
else {
// go up
if (memo[posX][posY-1] > memo[posX-1][posY]) {
return lcsBackTrack(memo, listX, listY, posX, posY-1);
}
// go left
else {
return lcsBackTrack(memo, listX, listY, posX-1, posY);
}
}
}
/*
var lcsArray = lcs(["ab","cd","ef"], ["aa", "ab", "ef", "bb"]);
var lcsString = lcs("antler", "lantern");
*/
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