DP29 最长相同子串 Longest Common Substring @geeksforgeeks

Given two strings ‘X’ and ‘Y’, find the length of the longest common substring. For example, if the given strings are “GeeksforGeeks” and “GeeksQuiz”, the output should be 5 as longest common substring is “Geeks”

Let m and n be the lengths of first and second strings respectively.

A simple solution is to one by one consider all substrings of first string and for every substring check if it is a substring in second string. Keep track of the maximum length substring. There will be O(m^2) substrings and we can find whether a string is subsring on another string in O(n) time (See this). So overall time complexity of this method would be O(n * m²)

Dynamic Programming can be used to find the longest common substring in O(m*n) time. The idea is to find length of the longest common suffix for all substrings of both strings and store these lengths in a table.

The longest common suffix has following optimal substructure property
   LCSuff(X, Y, m, n) = LCSuff(X, Y, m-1, n-1) + 1 if X[m-1] = Y[n-1]
                        0  Otherwise (if X[m-1] != Y[n-1])

The maximum length Longest Common Suffix is the longest common substring.
   LCSubStr(X, Y, m, n)  = Max(LCSuff(X, Y, i, j)) where 1 <= i <= m
                                                     and 1 <= j <= n

The longest substring can also be solved in O(n+m) time using Suffix Tree. We will be covering Suffix Tree based solution in a separate post.

思路：

1 暴力枚举 O(m^2 * n)：在长度为m的string里枚举所有子串需要O(m^2)，对于每一个子串要在另一个长度为n的string里找相同，用KMP需要O(n)

2 DP 把求相同子串的问题转为求相同后缀的问题 O(m*n)

3 后缀树: O(m+n)

参考 https://en.wikipedia.org/wiki/Longest_common_substring_problem

DP:

package DP;


public class LongestCommonSubstring {


public static void main(String[] args) {
String X = "OldSite:GeeksforGeeks.org";
String Y = "NewSite:GeeksQuiz.com";
int m = X.length();
int n = Y.length();
System.out.println(LCSubstr(X, Y, m, n));
}

// Time Complexity: O(m*n)  Time Complexity: O(m*n)
public static int LCSubstr(String X, String Y, int m, int n){
// Create a table to store lengths of longest common suffixes of
   // substrings.   Note that LCSuff[i][j] contains length of longest
   // common suffix of X[0..i-1] and Y[0..j-1]. The first row and
   // first column entries have no logical meaning, they are used only
   // for simplicity of program
int[][] LCSuff = new int[m+1][n+1];
int longest = 0; // To store length of the longest common substring

/* Following steps build LCSuff[m+1][n+1] in bottom up fashion. */
for(int i=0; i<=m; i++){
for(int j=0; j<=n; j++){
if(i==0 || j==0){
LCSuff[i][j] = 0;
}else if(X.charAt(i-1) == Y.charAt(j-1)){
LCSuff[i][j] = LCSuff[i-1][j-1] + 1;
longest = Math.max(longest, LCSuff[i][j]);
}else{
LCSuff[i][j] = 0;
}
}
}

return longest;
}


}

http://www.geeksforgeeks.org/longest-common-substring/