lcs

Consider 2 sequences X[1..m] and Y[1..n]. The memoization algorithm would compute the LCS in time O(m*n). Is there any better algorithm to find out LCS wrt time? I guess memoization done diagonally can give us O(min(m,n)) time complexity. Gene Myers in 1986 came up with a very nice algorithm for this, described here: An O(ND) Difference Algorithm and Its Variations . This algorithm takes time proportional to the edit distance between sequences, so it is much faster when the difference is small. It works by looping over all possible edit distances, starting from 0, until it finds a distance for

In order to find the minimal number of insertions required to convert a given string(s) to palindrome I find the longest common subsequence of the string(lcs_string) and its reverse. Therefore the number of insertions to be made is length(s) - length(lcs_string) What method should be employed to find the equivalent palindrome string on knowing the number of insertions to be made? For example : 1) azbzczdzez Number of insertions required : 5 Palindrome string : azbzcezdzeczbza Although multiple palindrome strings may exist for the same string but I want to find only one palindrome? Let S[i, j]

Regarding how they work, I was wondering low-level working stuff: What will trigger a merge conflict? Is the context also used by the tools in order to apply the patch? How do they deal with changes that do not actually modify source code behavior? For example, swapping function definition places. Regarding safety, truth be told, the huge Linux kernel repository is a testament for their safety. But I wondering about the following points: Are there any caveats/limitations regarding the tools that the user should be aware of? Have the algorithms been proven to not generate wrong results? If not,

题意： 给出两个字符串，求出最长的公共子序列大小。 思路： 算是最经典的LCS问题吧。 设 \(X=(x_1,x_2,.....x_n) 和 Y=(y_1,y_2,.....y_m)\) 是两个序列，将 X 和 Y 的最长公共子序列记为 \(lcs(X,Y)\) ,找出 \(lcs(X,Y)\) 就是一个最优问题 然后我们需要将其分解成子问题并求出子问题的最优解： (寻找子问题来推出当前问题，正是寻找状态转移方程最重要的一步) 1）如果 \(x_n=y_m\) ，即X的最后一个元素与Y的最后一个元素相同，这说明该元素一定位于公共子序列中。因此，现在只需要找： \(lcs(X_{n-1}，Y_{m-1})\) \(lcs(X_{n-1}，Y_{m-1})\) 就是原问题的一个子问题 2）如果 \(x_n != y_m\) 就往回递归寻找两个子问题： \(lcs(X_{n-1}，Y{m}) 和 lcs(X_{n}，Y_{m-1})\) \(lcs(X_{n-1}，Y_m)\) 表示：最长公共序列可以在 \((x_1,x_2,....x_{n-1}) 和 (y_1,y_2,...y_n)\) 中找 \(lcs(X_n，Y_{m-1})\) 表示：最长公共序列可以在 \((x_1,x_2,....x_n) 和 (y_1,y_2,...y_{n-1})\) 中找 最后就可以得到下面的 递推

You are given a sequence of numbers and you need to find a longest increasing subsequence from the given input(not necessary continuous). I found the link to this( Longest increasing subsequence on Wikipedia ) but need more explanation. If anyone could help me understand the O(n log n) implementation, that will be really helpful. If you could explain the algo with an example, that will be really appreciated. I saw the other posts as well and what I did not understand is: L = 0 for i = 1, 2, ... n: binary search for the largest positive j ≤ L such that X[M[j]] < X[i] (or set j = 0 if no such

The longest common subsequence problem is a classic computer science problem, algorithms to solve it are the root of version control systems and wiki engines. Two basic algorithms are the Hunt–McIlroy algorithm which was used to create the original version of diff , and the Myers diff algorithm which is used by the GNU diff utility currently. Both seem to work more or less by finding the shortest path through a graph that represents the edit space between the two strings or text files. The edit space is the number of inserts or deletes necessary to transform one sequence into the other. So

最长公共子序列 给定两个字符串S1和S2，求两个字符串的最长公共子序列的长度。 输入样例 ABCD AEBD 输出样例 3 解释 S1和S2的最长公共子序列为ABD，长度为3 思路 动态规划 L C S ( m , n ) LCS(m ,n) L C S ( m , n ) 表示 S 1 [ 0... m ] S1[0...m] S 1 [ 0 . . . m ] 和 S 2 [ 0... n ] S2[0...n] S 2 [ 0 . . . n ] 的最长公共子序列的长度 S 1 [ m ] = = S 2 [ n ] : L C S ( m , n ) = 1 + L C S ( m − 1 , n − 1 ) S1[m] == S2[n]: LCS(m, n) = 1 + LCS(m - 1, n - 1) S 1 [ m ] = = S 2 [ n ] : L C S ( m , n ) = 1 + L C S ( m − 1 , n − 1 ) S 1 [ m ] ! = S 2 [ n ] : L C S ( m , n ) = m a x ( L C S ( m − 1 , n ) , L C S ( m , n − 1 ) ) S1[m] != S2[n]: LCS(m, n) = max(LCS(m - 1, n), LCS(m, n - 1)) S 1 [ m