add Longest Common Subsequence solution

Author: HoonDongKang

longest-common-subsequence/HoonDongKang.ts

@@ -0,0 +1,69 @@
+/**
+ * [Problem]: [1143] Longest Common Subsequence
+ * (https://leetcode.com/problems/longest-common-subsequence/description/)
+ */
+function longestCommonSubsequence(text1: string, text2: string): number {
+    //시간복잡도 O(2^n)
+    //공간복잡도 O(n)
+    // Time Limit Exceeded
+    function dfsFunc(text1: string, text2: string): number {
+        function dfs(i: number, j: number): number {
+            if (i === text1.length || j === text2.length) return 0;
+
+            if (text1[i] === text2[j]) {
+                return 1 + dfs(i + 1, j + 1);
+            } else {
+                return Math.max(dfs(i, j + 1), dfs(i + 1, j));
+            }
+        }
+
+        return dfs(0, 0);
+    }
+
+    //시간복잡도 O(n)
+    //공간복잡도 O(n)
+    function memoizationFunc(text1: string, text2: string): number {
+        const memo = new Map<string, number>();
+
+        function dfs(i: number, j: number) {
+            if (i === text1.length || j === text2.length) return 0;
+
+            const key = `${i}, ${j}`;
+            if (memo.has(key)) return memo.get(key)!;
+
+            let result = 0;
+            if (text1[i] === text2[j]) {
+                result = 1 + dfs(i + 1, j + 1);
+            } else {
+                result = Math.max(dfs(i, j + 1), dfs(i + 1, j));
+            }
+
+            memo.set(key, result);
+            return result;
+        }
+
+        return dfs(0, 0);
+    }
+
+    //시간복잡도 O(n)
+    //공간복잡도 O(n)
+    function dpFunc(tex1: string, text2: string): number {
+        const length1 = text1.length;
+        const length2 = text2.length;
+        const dp: number[][] = Array.from({ length: length1 + 1 }, () =>
+            Array(length2 + 1).fill(0)
+        );
+
+        for (let i = 1; i <= length1; i++) {
+            for (let j = 1; j <= length2; j++) {
+                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[length1][length2];
+    }
+}