Longest Common Subsequence (2 Strings) - Dynamic Programming & Competing Subproblems
Why Stack Overflow recommends it
Stack Overflow users link it as the explanation of the LCS dynamic programming table when answering or asking about LCS implementations in JavaScript, Swift and Java.
Recommended by 2 people across 2 different months, 2019 to 2021.
What it is
A 25-minute walkthrough by Back To Back SWE of the Longest Common Subsequence problem for two strings. It builds up the dynamic programming table, explains the competing subproblems when characters differ, and states the O(nm) time and space complexity.
What people said
Recommends ielyamani · 2019-02 · Stack Overflow · on “Find most common substring from an Array of Strings”
The longest common subsequence between two strings can be defined recursively as follows : func lcs(_ str1: String, _ str2: String, _ i: String.Index, _ j: String.Index) -> Int { if (i == str1.startIndex || j == str2.startIndex) { return 0 } let beforeI = str1.index(before: i) let beforeJ = str2.index(before: j) if str1[beforeI] == str2[beforeJ] { return 1 + lcs(str1, str2, beforeI, beforeJ) } else { return max(lcs(str1, str2, i, beforeJ), lcs(str1, str2, beforeI, j)) } } You can find a complete explanation of how this dynamic programming algorithm works here [this video] . So, given an array of strings and a search text : let array = ["ram", "gopal", "varma", "govind", "ravan", "alan", "logan"]
Recommends Uygar · 2020-09 · Stack Overflow · on “JavaScript Longest common subsequence”
I recommend you to watch this video first, the solution is very well explained in it. [this video] The code is returning the max length of the string, you can modify it in order to serve your purpose. function longestCommonSubsequence(s1, s2) { // string to array const arr1 = [...s1] const arr2 = [...s2] // define n x m sized array filled with 0's let matrix = [...Array(arr1.length+1)].map(e => Array(arr2.length+1).fill(0)) // fill the matrix for(let i = 1; i <= arr1.length; i++) { for(let j = 1; j <= arr2.length; j++) { if(arr1[i-1] == arr2[j-1]) { matrix[i][j] = matrix[i-1][j] + 1} else matrix[i][j] = Math.max(matrix[i-1][j], matrix[i][j-1]) } } // return the max which is at the right bottom co
Cites Edon · 2019-09 · Stack Overflow · on “javascript least common subsequence, implementing the cache”
I am trying to implement the javascript version of the ' Longest Common Subsequence [LINK] ' problem. I am following the java implementation here [this video] : // working solution class BottomUp { public int longestCommonsubsequenceLength(String s1, String s2) { int cache[][] = new int[s2.length() + 1][s1.length() + 1]; for (int s2Row = 0; s2Row <= s2.length(); s2Row++) { for (int s1Col = 0; s1Col <= s1.length(); s1Col++) { if (s2Row == 0 || s1Col == 0) { cache[s2Row][s1Col] = 0; } else if (s2.charAt(s2Row - 1) == s1.charAt(s1Col - 1)) { cache[s2Row][s1Col] = cache[s2Row - 1][s1Col - 1] + 1; } else { cache[s2Row][s1Col] = Math.max(cache[s2Row - 1][s1Col], cache[s2Row][s1Col - 1]); } } } return c
Cites Kevin Lu · 2021-10 · Stack Overflow · on “Longest Common Subsequence with Restricted Number of Substring Parts”
Recently I am learning Dynamic Programming (DP) and this video [this video] shows building up the DP table to get the Longest Common Subsequence (lcs). What would this table be if there is another parameter m that restricts the number of substring parts? For example, suppose we have a="xxyzxyz" and b="xyz" and we are only allowed to break the substrings to only 2 parts, so m=2. And the result generates the maximum number of combination that the substrings can place in the string. In this case, we have the following, [x]xyx[yz] x[x]yx[yz] xxy[x][yz] x[xy]x[z] Note that the letters inside [] are considered 1 part, so we have 2 []'s in accordance with m=2. So in this case, the result is 4. I am wond