1. Introduction<\/h2>\n
\n The key to algorithm problem solving is to develop the ability to understand and solve various problems. Among them, the Longest Common Subsequence (LCS) problem is a fundamental problem that frequently appears in several algorithm tests. In this article, we will take a closer look at the process of solving the LCS problem using Kotlin.\n <\/p>\n<\/section>\n \n Given two strings, the problem is to find the length of the longest subsequence that appears in both strings without deleting any characters. In other words, it is to find the longest subsequence that can be formed while maintaining the order of characters in both strings.\n <\/p>\n \n A common approach to solve the LCS problem is Dynamic Programming. Dynamic Programming is a methodology for solving complex problems by breaking them down into simpler subproblems. Here, we will compare each character of the two strings and calculate the maximum length.\n <\/p>\n \n First, let the lengths of the two strings be \n The table will be filled according to the following rules, comparing characters up to the last character of both strings:\n <\/p>\n \n The last cell \n Now, based on the theoretical explanation, let’s calculate the LCS using Kotlin.\n <\/p>\n \n The above code has a time complexity of O(m * n) and a space complexity of O(m * n). Therefore, as the length of the strings increases, performance may decrease. However, the LCS problem can be improved through various optimization techniques.\n <\/p>\n \n Currently, a 2D array is used, but since only the previous row’s values are needed, it can also be solved with a 1D array. This can reduce the space complexity to O(n).\n <\/p>\n \n The Longest Common Subsequence problem is a very important problem in algorithm study and code testing. It can be efficiently solved using dynamic programming, and we have learned that there are various optimization techniques. I hope this problem provides an opportunity to develop algorithmic thinking and practice Kotlin programming.\n <\/p>\n<\/section>\n2. Problem Description<\/h2>\n
Example<\/h3>\n
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3. Approach to Problem Solving<\/h2>\n
3.1. Defining the Dynamic Programming Table<\/h3>\n
m<\/code> and n<\/code>, and define dp[i][j]<\/code> as the length of the LCS of the first i<\/code> characters of string 1 and the first j<\/code> characters of string 2. The size of this table is (m+1) x (n+1)<\/code>. The first row and column of the table are initialized to 0.\n <\/p>\n3.2. Setting the Recurrence Relation<\/h3>\n
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dp[i][j] = dp[i-1][j-1] + 1<\/code><\/li>\ndp[i][j] = max(dp[i-1][j], dp[i][j-1])<\/code><\/li>\n<\/ul>\n3.3. Extracting the Result<\/h3>\n
dp[m][n]<\/code> of the completed dp<\/code> table contains the length of the LCS for the two strings.\n <\/p>\n<\/section>\n4. Kotlin Implementation<\/h2>\n
\nfun longestCommonSubsequence(s1: String, s2: String): Int {\n val m = s1.length\n val n = s2.length\n val dp = Array(m + 1) { IntArray(n + 1) }\n\n for (i in 1..m) {\n for (j in 1..n) {\n if (s1[i - 1] == s2[j - 1]) {\n dp[i][j] = dp[i - 1][j - 1] + 1\n } else {\n dp[i][j] = maxOf(dp[i - 1][j], dp[i][j - 1])\n }\n }\n }\n\n return dp[m][n]\n}\n\nfun main() {\n val s1 = \"ABCBDAB\"\n val s2 = \"BDCAB\"\n println(\"Length of Longest Common Subsequence: ${longestCommonSubsequence(s1, s2)}\")\n}\n <\/code><\/pre>\n<\/section>\n5. Analysis<\/h2>\n
5.1. Improving Space Efficiency<\/h3>\n
\nfun longestCommonSubsequenceOptimized(s1: String, s2: String): Int {\n val m = s1.length\n val n = s2.length\n val dp = IntArray(n + 1)\n\n for (i in 1..m) {\n val prev = IntArray(n + 1)\n for (j in 1..n) {\n if (s1[i - 1] == s2[j - 1]) {\n dp[j] = prev[j - 1] + 1\n } else {\n dp[j] = maxOf(dp[j - 1], prev[j])\n }\n prev[j] = dp[j]\n }\n }\n return dp[n]\n}\n <\/code><\/pre>\n<\/section>\n6. Conclusion<\/h2>\n