added 0008 Longest Common Subsequence to ( L-A )

Author: harisdev-netizen

L-A/0008 Longest Common Subsequence ( L-A )/README.md

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+# 0005 Longest Common Subsequence ( L-A )
+
+## Problem
+
+Given two strings, find the length of their longest common subsequence (LCS). A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.
+
+## Example 1
+
+```
+Input:
+str1 = "ABCDGH"
+str2 = "AEDFHR"
+
+Output:
+The longest common subsequence is "ADH" with a length of 3.
+
+Input:
+str1 = "AGGTAB"
+str2 = "GXTXAYB"
+
+Output:
+The longest common subsequence is "GTAB" with a length of 4.
+```
+
+## Solution Pseudocode
+
+```javascript
+function longestCommonSubsequence(str1, str2) {
+  const m = str1.length;
+  const n = str2.length;
+
+  // Initialize a 2D array with 0
+  const lcs = Array(m + 1)
+    .fill()
+    .map(() => Array(n + 1).fill(0));
+
+  // Fill the 2D array with LCS lengths
+  for (let i = 1; i <= m; i++) {
+    for (let j = 1; j <= n; j++) {
+      if (str1[i - 1] === str2[j - 1]) {
+        lcs[i][j] = lcs[i - 1][j - 1] + 1;
+      } else {
+        lcs[i][j] = Math.max(lcs[i - 1][j], lcs[i][j - 1]);
+      }
+    }
+  }
+
+  // Return the length of LCS
+  return lcs[m][n];
+}
+
+// Example usage
+const str1 = "ABCDGH";
+const str2 = "AEDFHR";
+const result = longestCommonSubsequence(str1, str2);
+console.log(result); // Output: 3
+
+// Another example
+const str3 = "AGGTAB";
+const str4 = "GXTXAYB";
+const result2 = longestCommonSubsequence(str3, str4);
+console.log(result2); // Output: 4
+```
+
+## How it works
+
+- The code returns an object with two properties: `length`, which is the length of the longest common subsequence, and `sequence`, which is the actual subsequence itself.
+- It also includes a section of code to trace back the 2D array and find the actual LCS string.
+- The time complexity of this algorithm is O(mn), where m and n are the lengths of the input strings.
+
+## References
+
+- [Google](https://www.google.com/search?client=opera&q=Longest+Common+Subsequence&sourceid=opera&ie=UTF-8&oe=UTF-8)
+
+## Problem Added By
+
+- [Haris](https://github.com/harisdev-netizen)
+
+## Contributing
+
+Pull requests are welcome. For major changes, please open an issue first to discuss what you would like to change.
+
+Please make sure to update tests as appropriate.

L-A/0008 Longest Common Subsequence ( L-A )/longestCommonSubsequence.js

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+function longestCommonSubsequence(str1, str2) {
+  const m = str1.length;
+  const n = str2.length;
+
+  // Initialize a 2D array with 0
+  const lcs = Array(m + 1)
+    .fill()
+    .map(() => Array(n + 1).fill(0));
+
+  // Fill the 2D array with LCS lengths
+  for (let i = 1; i <= m; i++) {
+    for (let j = 1; j <= n; j++) {
+      if (str1[i - 1] === str2[j - 1]) {
+        lcs[i][j] = lcs[i - 1][j - 1] + 1;
+      } else {
+        lcs[i][j] = Math.max(lcs[i - 1][j], lcs[i][j - 1]);
+      }
+    }
+  }
+
+  // Find the LCS string by tracing back the 2D array
+  let i = m;
+  let j = n;
+  let lcsStr = "";
+
+  while (i > 0 && j > 0) {
+    if (str1[i - 1] === str2[j - 1]) {
+      lcsStr = str1[i - 1] + lcsStr;
+      i--;
+      j--;
+    } else if (lcs[i - 1][j] > lcs[i][j - 1]) {
+      i--;
+    } else {
+      j--;
+    }
+  }
+
+  return {
+    length: lcs[m][n],
+    sequence: lcsStr,
+  };
+}
+
+// Example usage
+const str1 = "ABCDGH";
+const str2 = "AEDFHR";
+const result = longestCommonSubsequence(str1, str2);
+console.log(result); // Output: { length: 3, sequence: 'ADH' }
+
+// Another example
+const str3 = "AGGTAB";
+const str4 = "GXTXAYB";
+const result2 = longestCommonSubsequence(str3, str4);
+console.log(result2); // Output: { length: 4, sequence: 'GTAB' }