merge: Add tests and docs for the Longest Common Subsequence algorithm (TheAlgorithms#867)

zaaath · web-flow · commit 7560beb06895 · 2021-12-08T17:20:23.000+05:30
* Refactors, adds tests and comments to longest common subsequence algorithm

* Refactor docs for longest common subsequence algorithm

* Add links to wikipedia and leetcode

* Fix styling

* Refactor variable naming and jsdoc
diff --git a/Dynamic-Programming/LongestCommonSubsequence.js b/Dynamic-Programming/LongestCommonSubsequence.js
@@ -1,32 +1,58 @@
 /*
-    * Given two sequences, find the length of longest subsequence present in both of them.
-    * A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous.
-    * For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”
+Problem:
+Given two sequences, find the length of longest subsequence present in both of them.
+A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous.
+For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”
+
+Our Solution:
+We use recursion with tabular memoization.
+Time complexity: O(M x N)
+Solving each subproblem has a cost of O(1). Again, there are MxN subproblems,
+and so we get a total time complexity of O(MxN).
+Space complexity: O(M x N)
+We need to store the answer for each of the MxN subproblems.
+
+Improvement:
+It's possible to optimize space complexity to O(min(M, N)) or time to O((N + r)log(N))
+where r is the number of matches between the two sequences. Try to figure out how.
+
+References:
+[wikipedia](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem)
+[leetcode](https://leetcode.com/problems/longest-common-subsequence/)
 */
 
-function longestCommonSubsequence (x, y, str1, str2, dp) {
-  if (x === -1 || y === -1) {
-    return 0
-  } else {
-    if (dp[x][y] !== 0) {
-      return dp[x][y]
+/**
+ * Finds length of the longest common subsequence among the two input string
+ * @param {string} str1 Input string #1
+ * @param {string} str2 Input string #2
+ * @returns {number} Length of the longest common subsequence
+ */
+function longestCommonSubsequence (str1, str2) {
+  const memo = new Array(str1.length + 1).fill(null)
+    .map(() => new Array(str2.length + 1).fill(null))
+
+  function recursive (end1, end2) {
+    if (end1 === -1 || end2 === -1) {
+      return 0
+    }
+
+    if (memo[end1][end2] !== null) {
+      return memo[end1][end2]
+    }
+
+    if (str1[end1] === str2[end2]) {
+      memo[end1][end2] = 1 + recursive(end1 - 1, end2 - 1)
+      return memo[end1][end2]
     } else {
-      if (str1[x] === str2[y]) {
-        dp[x][y] = 1 + longestCommonSubsequence(x - 1, y - 1, str1, str2, dp)
-        return dp[x][y]
-      } else {
-        dp[x][y] = Math.max(longestCommonSubsequence(x - 1, y, str1, str2, dp), longestCommonSubsequence(x, y - 1, str1, str2, dp))
-        return dp[x][y]
-      }
+      memo[end1][end2] = Math.max(
+        recursive(end1 - 1, end2),
+        recursive(end1, end2 - 1)
+      )
+      return memo[end1][end2]
     }
   }
-}
 
-// Example
-
-// const str1 = 'ABCDGH'
-// const str2 = 'AEDFHR'
-// const dp = new Array(str1.length + 1).fill(0).map(x => new Array(str2.length + 1).fill(0))
-// const res = longestCommonSubsequence(str1.length - 1, str2.length - 1, str1, str2, dp)
+  return recursive(str1.length - 1, str2.length - 1)
+}
 
 export { longestCommonSubsequence }
diff --git a/Dynamic-Programming/tests/LongestCommonSubsequence.test.js b/Dynamic-Programming/tests/LongestCommonSubsequence.test.js
@@ -0,0 +1,32 @@
+import { longestCommonSubsequence } from '../LongestCommonSubsequence'
+
+describe('LongestCommonSubsequence', () => {
+  it('expects to return an empty string for empty inputs', () => {
+    expect(longestCommonSubsequence('', '')).toEqual(''.length)
+    expect(longestCommonSubsequence('aaa', '')).toEqual(''.length)
+    expect(longestCommonSubsequence('', 'bbb')).toEqual(''.length)
+  })
+
+  it('expects to return an empty string for inputs without a common subsequence', () => {
+    expect(longestCommonSubsequence('abc', 'deffgf')).toEqual(''.length)
+    expect(longestCommonSubsequence('de', 'ghm')).toEqual(''.length)
+    expect(longestCommonSubsequence('aupj', 'xyz')).toEqual(''.length)
+  })
+
+  it('expects to return the longest common subsequence, short inputs', () => {
+    expect(longestCommonSubsequence('abc', 'abc')).toEqual('abc'.length)
+    expect(longestCommonSubsequence('abc', 'abcd')).toEqual('abc'.length)
+    expect(longestCommonSubsequence('abc', 'ab')).toEqual('ab'.length)
+    expect(longestCommonSubsequence('abc', 'a')).toEqual('a'.length)
+    expect(longestCommonSubsequence('abc', 'b')).toEqual('b'.length)
+    expect(longestCommonSubsequence('abc', 'c')).toEqual('c'.length)
+    expect(longestCommonSubsequence('abd', 'abcd')).toEqual('abd'.length)
+    expect(longestCommonSubsequence('abd', 'ab')).toEqual('ab'.length)
+    expect(longestCommonSubsequence('abc', 'abd')).toEqual('ab'.length)
+  })
+
+  it('expects to return the longest common subsequence, medium-length inputs', () => {
+    expect(longestCommonSubsequence('bsbininm', 'jmjkbkjkv')).toEqual('b'.length)
+    expect(longestCommonSubsequence('oxcpqrsvwf', 'shmtulqrypy')).toEqual('qr'.length)
+  })
+})
