DP[i][j] = max(DP[i-1][j], DP[i][j-1])<\/code><\/li>\n<\/ul>\n\n This allows us to calculate the LCS of the two strings.\n<\/p>\n
C++ Implementation<\/h2>\n
\n Now let’s implement the above theory in C++ code.\n<\/p>\n
\n#include <iostream>\n#include <vector>\n#include <string>\n\nusing namespace std;\n\nint calculateLCS(const string &A, const string &B) {\n int m = A.length();\n int n = B.length();\n\n \/\/ Initialize the DP table\n vector<vector<int>> DP(m + 1, vector<int>(n + 1, 0));\n\n \/\/ Fill the DP table\n for (int i = 1; i <= m; i++) {\n for (int j = 1; j <= n; j++) {\n if (A[i - 1] == B[j - 1]) {\n DP[i][j] = DP[i - 1][j - 1] + 1;\n } else {\n DP[i][j] = max(DP[i - 1][j], DP[i][j - 1]);\n }\n }\n }\n\n return DP[m][n];\n}\n\nint main() {\n string A = \"ABCBDAB\";\n string B = \"BDCAB\";\n\n int length = calculateLCS(A, B);\n cout << \"The length of the longest common subsequence is \" << length << \".\" << endl;\n\n return 0;\n}\n<\/code><\/pre>\nCode Explanation<\/h2>\n
\n In the above C++ code, the calculateLCS<\/code> function is used to compute the LCS of the two strings.
\n First, the DP table is initialized, and then comparisons are made for each pair of characters while filling the DP table.
\n Finally, the length of the longest common subsequence is stored in the bottom right cell of the DP table (DP[m][n]).\n<\/p>\nTime Complexity<\/h2>\n
\n The time complexity of this algorithm is O(m * n), and the space complexity is also O(m * n).
\n This is proportional to the lengths of the two strings, so if the input string lengths are long, it may affect performance.\n<\/p>\n
Conclusion<\/h2>\n
\n In this lecture, we covered the problem of finding the longest common subsequence between two strings.
\n We learned how to solve the problem using C++ and dynamic programming, and the longest common subsequence problem is one of the frequently encountered problems in coding tests.
\n If you practice and learn similar algorithms through various problems, you should be able to achieve good results in coding tests.\n<\/p>\n
Thank you!<\/p>\n
<\/body><\/p>\n","protected":false},"excerpt":{"rendered":"
Preparing for coding tests can be somewhat challenging, but systematically mastering algorithm problems can lead to good results. In this course, we will cover the Longest Common Subsequence (LCS) problem. This problem is an important algorithmic question that measures how much two subsequences overlap in strings or sequences. 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