Create 1143_Longest_Common_Subsequence.md

Author: koji

1143_Longest_Common_Subsequence.md

@@ -0,0 +1,59 @@
+## 1143. Longest Common Subsequence
+
+```
+Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.
+
+A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.
+
+For example, "ace" is a subsequence of "abcde".
+A common subsequence of two strings is a subsequence that is common to both strings.
+
+ 
+
+Example 1:
+
+Input: text1 = "abcde", text2 = "ace" 
+Output: 3 
+Explanation: The longest common subsequence is "ace" and its length is 3.
+Example 2:
+
+Input: text1 = "abc", text2 = "abc"
+Output: 3
+Explanation: The longest common subsequence is "abc" and its length is 3.
+Example 3:
+
+Input: text1 = "abc", text2 = "def"
+Output: 0
+Explanation: There is no such common subsequence, so the result is 0.
+ 
+
+Constraints:
+
+1 <= text1.length, text2.length <= 1000
+text1 and text2 consist of only lowercase English characters.
+```
+
+```python
+def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+ size1 = len(text1)
+ size2 = len(text2)
+ 
+ dp = [[0] * (size2+1) for _ in range(size1+1)]
+ 
+ # print(dp)
+ 
+ for i in range(1, size1+1):
+ for j in range(1, size2+1):
+ 
+ if text1[i-1] == text2[j-1]:
+ dp[i][j] = dp[i-1][j-1]+1
+ else:
+ dp[i][j] = max(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])
+ 
+ return dp[-1][-1]
+```
+
+```
+Runtime: 476 ms, faster than 38.20% of Python3 online submissions for Longest Common Subsequence.
+Memory Usage: 21.9 MB, less than 78.67% of Python3 online submissions for Longest Common Subsequence.
+```