Given two strings text1 and text2, return the length of their longest common subsequence.
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. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.
If there is no common subsequence, return 0.
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 <= 1000
1 <= text2.length <= 1000
The input strings consist of lowercase English characters only.
class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        dp=[[-1 for i in range(len(text2)+1)]for j in range(len(text1)+1)]
        for i in range(len(text1)+1):
            for j in range(len(text2)+1):
                if i==0 or j==0:
                    dp[i][j]=0
        for i in range(1,len(text1)+1):
            for j in range(1,len(text2)+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])
        return dp[len(text1)][len(text2)]
class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        dp=[[-1 for i in range(len(text2)+1)]for j in range(len(text1)+1)]
        for i in range(len(text1)+1):
            for j in range(len(text2)+1):
                if i==0 or j==0:
                    dp[i][j]=0
        for i in range(1,len(text1)+1):
            for j in range(1,len(text2)+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])
        return dp[len(text1)][len(text2)]