Dadas dos strings s1, s2 y K, encuentre la longitud de la subsecuencia más larga formada por segmentos consecutivos de al menos longitud K.
Ejemplos:
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.
Requisito previo : Subsecuencia común más larga
Cree una array LCS[][] donde LCS i, j denota la longitud de la subsecuencia común más larga formada por caracteres de s1 hasta i y s2 hasta j que tengan segmentos consecutivos de al menos una longitud K. Cree un cnt[ ][] array para contar la longitud del segmento común. cnt i, j = cnt i-1, j-1 +1 cuando s1[i-1]==s2[j-1]. Si los caracteres no son iguales, los segmentos no son iguales, por lo tanto, marque cnt i, j como 0.
Cuando cnt i, j >=k , actualice el valor de lcs agregando el valor de lcs i-a, ja donde a es la longitud de los segmentos a<=cnt i, j. La respuesta para la subsecuencia más larga con segmentos consecutivos de al menos una longitud k se almacenará en lcs[n][m] donde n y m son la longitud de string1 y 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 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.
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 segments
function 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 function
var k = 4;
var s1 = "aggasdfa";
var s2 = "aggajasdfa";
document.write( longestSubsequenceCommonSegment(k, s1, s2));
</script>
Producción:
8
Complejidad de tiempo : O (N * M), ya que estamos usando bucles anidados para atravesar N * M veces.
Espacio auxiliar : O(N*M), ya que estamos usando espacio adicional para lcs y cnt .
Donde N y M son la longitud de las cuerdas.