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📜  由长度至少为K的连续段形成的LCS

📅  最后修改于: 2021-05-07 05:06:13             🧑  作者: Mango

给定两个字符串s1,s2和K,找到至少长度为K的连续段形成的最长子序列的长度。

例子:

Input : s1 = aggayxysdfa
        s2 = aggajxaaasdfa 
         k = 4
Output : 8 
Explanation: aggasdfa is the longest
subsequence that can be formed by taking
consecutive segments, minimum of length 4.
Here segments are "agga" and "sdfa" which 
are of length 4 which is included in making 
the longest subsequence. 

Input : s1 = aggasdfa 
        s2 = aggajasdfaxy
         k = 5
Output : 5 

Input: s1 = "aabcaaaa" 
       s2 = "baaabcd"  
        k = 3
Output: 4 
Explanation: "aabc" is the longest subsequence that 
is formed by taking segment of minimum length 3. 
The segment is of length 4. 

先决条件:最长的公共子序列

创建一个LCS [] []数组,其中LCS i,j表示由s1到i和s2到j的字符形成的最长公共子序列的长度,这些字符具有至少为K的连续段。创建一个cnt [] []数组,以计算公共段的长度。当s1 [i-1] == s2 [j-1]时 cnt i,j = cnt i-1,j-1 +1 。如果字符不相等,则段不相等,因此将cnt i,j标记为0。
cnt i,j > = k时,则通过添加cnt ia,ja的值来更新lcs值其中a是段a <= cnt i,j的长度。具有至少长度为k的连续段的最长子序列的答案将存储在lcs [n] [m]中,其中n和m为string1和string2的长度。

C++
// CPP program to find the Length of Longest 
// subsequence formed by consecutive segments
// of at least length K
#include 
using namespace std;
  
// Returns the length of the longest common subsequence
// with a minimum of length of K consecutive segments
int longestSubsequenceCommonSegment(int k, string s1, 
                                           string s2)
{
    // length of strings
    int n = s1.length();
    int m = s2.length();
  
    // declare the lcs and cnt array
    int lcs[n + 1][m + 1];
    int cnt[n + 1][m + 1];
  
    // initialize the lcs and cnt array to 0
    memset(lcs, 0, sizeof(lcs));
    memset(cnt, 0, sizeof(cnt));
  
    // iterate from i=1 to n and j=1 to j=m
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
  
            // stores the maximum of lcs[i-1][j] and lcs[i][j-1]
            lcs[i][j] = max(lcs[i - 1][j], lcs[i][j - 1]);
  
            // when both the characters are equal
            // of s1 and s2
            if (s1[i - 1] == s2[j - 1])
                cnt[i][j] = cnt[i - 1][j - 1] + 1; 
  
            // when length of common segment is
            // more than k, then update lcs answer 
            // by adding that segment to the answer
            if (cnt[i][j] >= k) {
  
                // formulate for all length of segments
                // to get the longest subsequence with 
                // consecutive Common Segment of length 
                // of min k length
                for (int a = k; a <= cnt[i][j]; a++) 
  
                    // update lcs value by adding segment length
                    lcs[i][j] = max(lcs[i][j], 
                                    lcs[i - a][j - a] + a);
                  
            }
        }
    }
  
    return lcs[n][m];
}
  
// driver code to check the above function
int main()
{
    int k = 4;
    string s1 = "aggasdfa";
    string s2 = "aggajasdfa";
    cout << longestSubsequenceCommonSegment(k, s1, s2);
    return 0;
}


Java
// Java program to find the Length of Longest 
// subsequence formed by consecutive segments
// of at least length K
  
class GFG {
  
    // Returns the length of the longest common subsequence
    // with a minimum of length of K consecutive segments
    static int longestSubsequenceCommonSegment(int k, String s1, 
                                               String s2)
    {
        // length of strings
        int n = s1.length();
        int m = s2.length();
      
        // declare the lcs and cnt array
        int lcs[][] = new int[n + 1][m + 1];
        int cnt[][] = new int[n + 1][m + 1];
      
      
        // iterate from i=1 to n and j=1 to j=m
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
      
                // stores the maximum of lcs[i-1][j] and lcs[i][j-1]
                lcs[i][j] = Math.max(lcs[i - 1][j], lcs[i][j - 1]);
      
                // when both the characters are equal
                // of s1 and s2
                if (s1.charAt(i - 1) == s2.charAt(j - 1))
                    cnt[i][j] = cnt[i - 1][j - 1] + 1; 
      
                // when length of common segment is
                // more than k, then update lcs answer 
                // by adding that segment to the answer
                if (cnt[i][j] >= k) 
                {
      
                    // formulate for all length of segments
                    // to get the longest subsequence with 
                    // consecutive Common Segment of length 
                    // of min k length
                    for (int a = k; a <= cnt[i][j]; a++) 
      
                        // update lcs value by adding 
                        // segment length
                        lcs[i][j] = Math.max(lcs[i][j], 
                                        lcs[i - a][j - a] + a);
                      
                }
            }
        }
      
        return lcs[n][m];
    }
      
    // driver code to check the above function
    public static void main(String[] args)
    {
        int k = 4;
        String s1 = "aggasdfa";
        String s2 = "aggajasdfa";
        System.out.println(longestSubsequenceCommonSegment(k, s1, s2));
    }
}
  
// This code is contributed by prerna saini.


Python3
# Python3 program to find the Length of Longest 
# subsequence formed by consecutive segments 
# of at least length K 
  
# Returns the length of the longest common subsequence 
# with a minimum of length of K consecutive segments 
def longestSubsequenceCommonSegment(k, s1, s2) :
      
    # length of strings 
    n = len(s1) 
    m = len(s2) 
  
    # declare the lcs and cnt array 
    lcs = [[0 for x in range(m + 1)] for y in range(n + 1)] 
    cnt = [[0 for x in range(m + 1)] for y in range(n + 1)]
  
  
    # iterate from i=1 to n and j=1 to j=m 
    for i in range(1, n + 1) :
        for j in range(1, m + 1) :
            # stores the maximum of lcs[i-1][j] and lcs[i][j-1] 
            lcs[i][j] = max(lcs[i - 1][j], lcs[i][j - 1])
  
            # when both the characters are equal 
            # of s1 and s2 
            if (s1[i - 1] == s2[j - 1]):
                cnt[i][j] = cnt[i - 1][j - 1] + 1; 
  
            # when length of common segment is 
            # more than k, then update lcs answer 
            # by adding that segment to the answer 
            if (cnt[i][j] >= k) :
                  
                # formulate for all length of segments 
                # to get the longest subsequence with 
                # consecutive Common Segment of length 
                # of min k length 
                for a in range(k, cnt[i][j] + 1) :
                      
                    # update lcs value by adding 
                    # segment length 
                    lcs[i][j] = max(lcs[i][j],lcs[i - a][j - a] + a)
                      
    return lcs[n][m] 
  
  
# Driver code  
k = 4
s1 = "aggasdfa"
s2 = "aggajasdfa"
print(longestSubsequenceCommonSegment(k, s1, s2)) 
  
# This code is contributed by Nikita Tiwari.


C#
// C# program to find the Length of Longest 
// subsequence formed by consecutive segments
// of at least length K
using System;
  
class GFG {
  
    // Returns the length of the longest common subsequence
    // with a minimum of length of K consecutive segments
    static int longestSubsequenceCommonSegment(int k, string s1, 
                                            string s2)
    {
        // length of strings
        int n = s1.Length;
        int m = s2.Length;
      
        // declare the lcs and cnt array
        int [,]lcs = new int[n + 1,m + 1];
        int [,]cnt = new int[n + 1,m + 1];
      
      
        // iterate from i=1 to n and j=1 to j=m
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
      
                // stores the maximum of lcs[i-1][j] and lcs[i][j-1]
                lcs[i,j] = Math.Max(lcs[i - 1,j], lcs[i,j - 1]);
      
                // when both the characters are equal
                // of s1 and s2
                if (s1[i - 1] == s2[j - 1])
                    cnt[i,j] = cnt[i - 1,j - 1] + 1; 
      
                // when length of common segment is
                // more than k, then update lcs answer 
                // by adding that segment to the answer
                if (cnt[i,j] >= k) 
                {
      
                    // formulate for all length of segments
                    // to get the longest subsequence with 
                    // consecutive Common Segment of length 
                    // of min k length
                    for (int a = k; a <= cnt[i,j]; a++) 
      
                        // update lcs value by adding 
                        // segment length
                        lcs[i,j] = Math.Max(lcs[i,j], 
                                        lcs[i - a,j - a] + a);
                      
                }
            }
        }
      
        return lcs[n,m];
    }
      
    // driver code to check the above function
    public static void Main()
    {
        int k = 4;
        string s1 = "aggasdfa";
        string s2 = "aggajasdfa";
    Console.WriteLine(longestSubsequenceCommonSegment(k, s1, s2));
    }
}
  
// This code is contributed by vt_m.


输出:

8