update: longest-common-subsequence

Author: maxyzli

dynamic-programming/longest-common-subsequence/README.md

@@ -4,106 +4,39 @@
 
 DFS
 
-```java
-/**
- * Question   : 1143. Longest Common Subsequence
- * Complexity : Time: O(2^n) ; Space: O(n)
- * Topics     : DP, Backtracking
- */
-class Solution {
-    int longest = 0;
-
-    public int longestCommonSubsequence(String text1, String text2) {
-        longest = 0;
-        longestCommonSubsequence(text1, 0, text2, 0, 0);
-        return longest;
-    }
-
-    private void longestCommonSubsequence(String text1, int p1, String text2, int p2, int count) {
-        if (p1 == text1.length() || p2 == text2.length()) {
-            return;
-        }
-        if (text1.charAt(p1) == text2.charAt(p2)) {
-            longest = Math.max(longest, count + 1);
-            longestCommonSubsequence(text1, p1 + 1, text2, p2 + 1, count + 1);
-        } else {
-            longestCommonSubsequence(text1, p1 + 1, text2, p2, count);
-            longestCommonSubsequence(text1, p1, text2, p2 + 1, count);
-        }
-    }
-}
-```
-
-## Solution 2
-
-DFS
-
 ```java
 /**
  * Question   : 1143. Longest Common Subsequence
  * Complexity : Time: O(2^n) ; Space: O(n)
  * Topics     : DP
  */
-public class Solution {
-    public int longestCommonSubsequence(String text1, String text2) {
-        return longestCommonSubsequence(text1, 0, text2, 0);
-    }
-
-    private int longestCommonSubsequence(String text1, int i, String text2, int j) {
-        if (i >= text1.length() || j >= text2.length()) {
-            return 0;
-        }
-
-        if (text1.charAt(i) == text2.charAt(j)) {
-            return 1 + longestCommonSubsequence(text1, i + 1, text2, j + 1);
-        }
-        return Math.max(longestCommonSubsequence(text1, i + 1, text2, j), longestCommonSubsequence(text1, i, text2, j + 1));
-    }
-}
-```
-
-## Solution 3
-
-Top-down DP
-
-```java
-/**
- * Question   : 1143. Longest Common Subsequence
- * Complexity : Time: O(n^2) ; Space: O(n^2)
- * Topics     : DP
- */
-public class Solution {
+class Solution {
     public int longestCommonSubsequence(String text1, String text2) {
-        Integer[][] dprize = new Integer[text1.length() + 1][text2.length() + 1];
-        return longestCommonSubsequence(text1, 0, text2, 0, dprize);
+        int n1 = text1.length();
+        int n2 = text2.length();
+        return longestCommonSubsequenceUtil(text1, n1, text2, n2);
     }
 
-    private int longestCommonSubsequence(String text1, int i, String text2, int j, Integer[][] dprize) {
-        if (i >= text1.length() || j >= text2.length()) {
+    private int longestCommonSubsequenceUtil(String text1, int n1, String text2, int n2) {
+        if (n1 == 0 || n2 == 0) {
             return 0;
         }
 
-        if (dprize[i][j] != null) {
-            return dprize[i][j];
-        }
-
-        if (text1.charAt(i) == text2.charAt(j)) {
-            dprize[i][j] = 1 + longestCommonSubsequence(text1, i + 1, text2, j + 1, dprize);
+        if (text1.charAt(n1 - 1) == text2.charAt(n2 - 1)) {
+            return 1 + longestCommonSubsequenceUtil(text1, n1 - 1, text2, n2 - 1);
         } else {
-            dprize[i][j] = Math.max(
-                    longestCommonSubsequence(text1, i + 1, text2, j, dprize),
-                    longestCommonSubsequence(text1, i, text2, j + 1, dprize)
+            return Math.max(
+                    longestCommonSubsequenceUtil(text1, n1 - 1, text2, n2),
+                    longestCommonSubsequenceUtil(text1, n1, text2, n2 - 1)
             );
         }
-
-        return dprize[i][j];
     }
 }
 ```
 
-## Solution 4
+## Solution 2
 
-Bottom-up DP
+DP
 
 ```java
 /**
@@ -113,14 +46,14 @@ Bottom-up DP
  */
 class Solution {
     public int longestCommonSubsequence(String text1, String text2) {
-        if (text1.length() == 0 || text2.length() == 0) {
-            return 0;
-        }
+        int n1 = text1.length();
+        int n2 = text2.length();
 
-        int[][] dp = new int[text1.length() + 1][text2.length() + 1];
+        // dp[i][j]: Longest common subsequence given i and j indexes.
+        int[][] dp = new int[n1 + 1][n2 + 1];
 
-        for (int i = 1; i <= text1.length(); i++) {
-            for (int j = 1; j <= text2.length(); j++) {
+        for (int i = 1; i <= n1; i++) {
+            for (int j = 1; j <= n2; j++) {
                 if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
                     dp[i][j] = 1 + dp[i - 1][j - 1];
                 } else {
@@ -129,7 +62,7 @@ class Solution {
             }
         }
 
-        return dp[text1.length()][text2.length()];
+        return dp[n1][n2];
     }
 }
 ```