LCS formed by consecutive segments of at least length K

Given two strings s1, s2 and K, find the length of the longest subsequence formed by consecutive segments of at least length K.

Examples:

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. 



Prerequisite : Longest Common Subsequence

Create a LCS[][] array where LCSi, j denotes the length of the longest common subsequence formed by characters of s1 till i and s2 till j having consecutive segments of at least length K. Create a cnt[][] array to count the length of the common segment. cnti, j= cnti-1, j-1+1 when s1[i-1]==s2[j-1]. If characters are not equal then segments are not equal hence mark cnti, j as 0.
When cnti, j>=k, then update the lcs value by adding the value of cnti-a, j-a where a is the length of the segments a<=cnti, j. The answer for the longest subsequence with consecutive segments of at least length k will be stored in lcs[n][m] where n and m are the length of string1 and string2.

C++

// CPP program to find the Length of Longest 
// subsequence formed by consecutive segments
// of at least length K
#include <bits/stdc++.h>
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 fucntion
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 fucntion
    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.

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 fucntion
    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.


Output:

8




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