Add longest-common-subsequence solution in TS

Author: Jeehay28

longest-common-subsequence/Jeehay28.ts

@@ -0,0 +1,48 @@
+// TC: O(m * n)
+// SC: O(m * n)
+function longestCommonSubsequence(text1: string, text2: string): number {
+  const memo = new Map<string, number>();
+
+  const dfs = (i: number, j: number) => {
+    const key = `${i}-${j}`;
+
+    if (memo.has(key)) return memo.get(key);
+
+    if (i === text1.length || j === text2.length) {
+      // ""
+      memo.set(key, 0);
+    } else if (text1[i] === text2[j]) {
+      memo.set(key, 1 + dfs(i + 1, j + 1)!);
+    } else {
+      memo.set(key, Math.max(dfs(i + 1, j)!, dfs(i, j + 1)!));
+    }
+
+    return memo.get(key);
+  };
+
+  return dfs(0, 0)!;
+}
+
+
+// TC: O(m * n)
+// SC: O(m * n)
+// function longestCommonSubsequence(text1: string, text2: string): number {
+//   const m = text1.length + 1;
+//   const n = text2.length + 1;
+
+//   const dp: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
+
+//   for (let i = 1; i < m; i++) {
+//     for (let j = 1; j < n; j++) {
+//       // Note: text1[i - 1] and text2[j - 1] because dp uses 1-based indexing,
+//       // while the strings use 0-based indexing
+//       if (text1[i - 1] === text2[j - 1]) {
+//         dp[i][j] = 1 + dp[i - 1][j - 1];
+//       } else {
+//         dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
+//       }
+//     }
+//   }
+
+//   return dp[m - 1][n - 1];
+// }