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# @lc app=leetcode id=1143 lang=python3
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# [1143] Longest Common Subsequence
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# https://leetcode.com/problems/longest-common-subsequence/description/
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# algorithms
# Medium (58.51%)
# Likes:    10300
# Dislikes: 122
# Total Accepted:    654.9K
# Total Submissions: 1.1M
# Testcase Example:  '"abcde"\n"ace"'
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# Given two strings text1 and text2, return the length of their longest common
# subsequence. If there is no common subsequence, return 0.
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# 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.
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#
# For example, "ace" is a subsequence of "abcde".
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# A common subsequence of two strings is a subsequence that is common to both
# strings.
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#
# Example 1:
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# Input: text1 = "abcde", text2 = "ace"
# Output: 3
# Explanation: The longest common subsequence is "ace" and its length is 3.
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# Example 2:
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# Input: text1 = "abc", text2 = "abc"
# Output: 3
# Explanation: The longest common subsequence is "abc" and its length is 3.
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# Example 3:
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# Input: text1 = "abc", text2 = "def"
# Output: 0
# Explanation: There is no such common subsequence, so the result is 0.
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# Constraints:
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# 1 <= text1.length, text2.length <= 1000
# text1 and text2 consist of only lowercase English characters.
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#
#


# @lc code=start
class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        dp = [[0 for _ in range(len(text2) + 1)] for _ in range(len(text1) + 1)]
        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[-1][-1]


# @lc code=end
