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 lcsi-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.
Implementation:
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 segmentsint 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 functionint 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 Kclass 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 = 4s1 = "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 Kusing 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. |
Javascript
<script>// JavaScript 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 segmentsfunction longestSubsequenceCommonSegment(k, s1, s2){ // length of strings var n = s1.length; var m = s2.length; // declare the lcs and cnt array var lcs = Array.from(Array(n+1), ()=>Array(m+1).fill(0)); var cnt = Array.from(Array(n+1), ()=>Array(m+1).fill(0)); // iterate from i=1 to n and j=1 to j=m for (var i = 1; i <= n; i++) { for (var 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 (var 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 functionvar k = 4;var s1 = "aggasdfa";var s2 = "aggajasdfa";document.write( longestSubsequenceCommonSegment(k, s1, s2));</script> |
Output
8
Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.
Auxiliary Space: O(N*M), as we are using extra space for lcs and cnt.
Where N and M are the length of the strings.
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