Merge pull request akshitagit#98 from yash0412/LCS-yash

akshitagupta15june · web-flow · commit e768008c6404 · 2020-10-22T10:37:19.000+05:30
Added LCS in dynamic programming
diff --git a/DynamicProgramming/LongestCommonSubsequence.js b/DynamicProgramming/LongestCommonSubsequence.js
@@ -0,0 +1,61 @@
+/*
+
+Longest Common Subsequence is a classical example of how to use dynamic programming with memorization
+to solve a complex problem in a much easier and faster way.
+
+LCS Problem Statement: 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”.
+
+The brute force method requires to find all possible subsequences of the given sequences and finding the longest common subsequence.
+This approach has exponential time complexity.
+
+LCS with Dynamic Programming approach requires only O(mn) time, where m and n are the lengths of the given sequences.
+
+*/
+
+function lcs(p, q, m, n, dp) {
+  var result = 0;
+  if (m === 0 || n === 0) {
+    result = 0;
+  } else if (dp[m - 1][n - 1] !== null) {
+    //If LCS value is already calcualted for this substring
+    result = dp[m - 1][n - 1];
+  } else if (p[m - 1] === q[n - 1]) {
+    //If last character in substrings are the same
+    result = 1 + lcs(p, q, m - 1, n - 1, dp);
+    dp[m - 1][n - 1] = result;
+  } else {
+    //If last character in substrings are different
+
+    //find LCS by removing last character from string 1
+    var temp1 = lcs(p, q, m - 1, n, dp);
+
+    //find LCS by removing last character from string 2
+    var temp2 = lcs(p, q, m, n - 1, dp);
+
+    //take min of both results
+    result = temp1 > temp2 ? temp1 : temp2;
+
+    //set the value in Matrix
+    dp[m - 1][n - 1] = result;
+  }
+  return result;
+}
+
+//input strings of size m and n
+var str1 = "nematode_knowledge";
+var str2 = "empty_bottle";
+
+//Matrix of size mXn to store calculated values
+var dp = [];
+
+//Prefill the matrix with null
+for (var i = 0; i < str1.length; i++) {
+  dp[i] = [];
+  for (var j = 0; j < str2.length; j++) {
+    dp[i].push(null);
+  }
+}
+
+console.log(lcs(str1, str2, str1.length, str2.length, dp));
