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 =aggayxysdfas2 =aggajxaaasdfak = 4 Output : 8 Explanation:aggasdfais 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 = aggasdfas2 = 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 LCS_{i, 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. **cnt _{i, j}= cnt_{i-1, j-1}+1** when s1[i-1]==s2[j-1]. If characters are not equal then segments are not equal hence mark cnt

_{i, j}as 0.

When

**cnt**, then update the lcs value by adding the value of cnt

_{i, j}>=k_{i-a, j-a}where a is the length of the segments

**a<=cnt**. The answer for the longest subsequence with consecutive segments of at least length k will be stored in

_{i, j}**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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