Скачать или смотреть ✅Day 25| LeetCode 1143.Longest Common Subsequence |

🌟 Welcome to the Leetcode Daily Challenge Solution! 🚀

In today's coding adventure, we delve into the realm of dynamic programming to unravel the mysteries of finding the longest common subsequence between two strings. Our journey involves crafting a dynamic programming table and navigating through the matrix to uncover the solution.

🎯 Problem Overview:
We are presented with two strings, text1 and text2, and our quest is to determine the length of their longest common subsequence. A subsequence is a sequence of characters that appears in the same order within both strings, but not necessarily consecutively.

🚀 Algorithm Breakdown:

Initialize a 2D array dp to store the lengths of the longest common subsequences. The dimensions of dp are (n + 1) x (m + 1), where n and m are the lengths of text1 and text2 respectively.

Set the base cases: If either of the strings is empty, the length of the common subsequence is 0. Therefore, set dp[i][0] = 0 and dp[0][j] = 0 for all i and j.

Traverse through the strings with nested loops, and at each character comparison, update the dp table:

If text1[i - 1] == text2[j - 1], the current characters match. Increment the length of the common subsequence by 1: dp[i][j] = 1 + dp[i-1][j-1].
If the characters don't match, take the maximum of the lengths obtained by excluding one character from either of the strings: dp[i][j] = max(dp[i-1][j], dp[i][j-1]).
The final result is stored in dp[n][m], representing the length of the longest common subsequence.

📊 Complexity Analysis:

Time Complexity: O(n * m) - Traversing the entire 2D array.
Space Complexity: O(n * m) - Additional space used for the dynamic programming table.

Feel free to explore the solution, modify it, and deepen your understanding of dynamic programming and string manipulation! 🔗✨

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