给定两个分别为长度m和n的字符串X和Y。问题是找到包含所有元音字符的字符串X和Y的最长公共子序列的长度。
例子：
Input : X = "aieef"
Y = "klaief"
Output : aie
Input : X = "geeksforgeeks"
Y = "feroeeks"
Output : eoee
资料来源： Paytm面试经验(后端开发人员)。
天真的方法：生成两个给定序列的所有子序列，并找到包含所有元音字符的最长匹配子序列。该解决方案在时间复杂度方面是指数级的。
高效方法(动态编程)：这种方法是最长公共子序列的一种变体。 DP-4问题。这篇文章的不同之处在于，共同的子序列字符必须是元音。
C++
// C++ implementation to find the length of longest common
// subsequence which contains all vowel characters
#include
using namespace std;
// function to check whether 'ch'
// is a vowel or not
bool isVowel(char ch)
{
if (ch == 'a' || ch == 'e' || ch == 'i'
|| ch == 'o' || ch == 'u')
return true;
return false;
}
// function to find the length of longest common subsequence
// which contains all vowel characters
int lcs(char* X, char* Y, int m, int n)
{
int L[m + 1][n + 1];
int i, j;
// Following steps build L[m+1][n+1] in bottom up fashion. Note
// that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]
for (i = 0; i <= m; i++) {
for (j = 0; j <= n; j++) {
if (i == 0 || j == 0)
L[i][j] = 0;
else if ((X[i - 1] == Y[j - 1]) && isVowel(X[i - 1]))
L[i][j] = L[i - 1][j - 1] + 1;
else
L[i][j] = max(L[i - 1][j], L[i][j - 1]);
}
}
// L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1]
// which contains all vowel characters
return L[m][n];
}
// Driver program to test above
int main()
{
char X[] = "aieef";
char Y[] = "klaief";
int m = strlen(X);
int n = strlen(Y);
cout << "Length of LCS = "
<< lcs(X, Y, m, n);
return 0;
} Java
// Java implementation to find the
// length of longest common subsequence
// which contains all vowel characters
class GFG
{
// function to check whether 'ch'
// is a vowel or not
static boolean isVowel(char ch)
{
if (ch == 'a' || ch == 'e' ||
ch == 'i' || ch == 'o' ||
ch == 'u')
return true;
return false;
}
// function to find the length of
// longest common subsequence which
// contains all vowel characters
static int lcs(String X, String Y,
int m, int n)
{
int L[][] = new int[m + 1][n + 1];
int i, j;
// Following steps build L[m+1][n+1]
// in bottom up fashion. Note that
// L[i][j] contains length of LCS of
// X[0..i-1] and Y[0..j-1]
for (i = 0; i <= m; i++)
{
for (j = 0; j <= n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if ((X.charAt(i - 1) == Y.charAt(j - 1)) &&
isVowel(X.charAt(i - 1)))
L[i][j] = L[i - 1][j - 1] + 1;
else
L[i][j] = Math.max(L[i - 1][j],
L[i][j - 1]);
}
}
// L[m][n] contains length of LCS
// for X[0..n-1] and Y[0..m-1]
// which contains all vowel characters
return L[m][n];
}
// Driver Code
public static void main(String[] args)
{
String X = "aieef";
String Y = "klaief";
int m = X.length();
int n = Y.length();
System.out.println("Length of LCS = " +
lcs(X, Y, m, n));
}
}
// This code is contributed by BilalPython3
# Python3 implementation to find the
# length of longest common subsequence
# which contains all vowel characters
# function to check whether 'ch'
# is a vowel or not
def isVowel(ch):
if (ch == 'a' or ch == 'e' or
ch == 'i'or ch == 'o' or
ch == 'u'):
return True
return False
# function to find the length of longest
# common subsequence which contains all
# vowel characters
def lcs(X, Y, m, n):
L = [[0 for i in range(n + 1)]
for j in range(m + 1)]
i, j = 0, 0
# Following steps build L[m+1][n+1] in
# bottom up fashion. Note that L[i][j]
# contains length of LCS of X[0..i-1]
# and Y[0..j-1]
for i in range(m + 1):
for j in range(n + 1):
if (i == 0 or j == 0):
L[i][j] = 0
elif ((X[i - 1] == Y[j - 1]) and
isVowel(X[i - 1])):
L[i][j] = L[i - 1][j - 1] + 1
else:
L[i][j] = max(L[i - 1][j],
L[i][j - 1])
# L[m][n] contains length of LCS for
# X[0..n-1] and Y[0..m-1] which
# contains all vowel characters
return L[m][n]
# Driver Code
X = "aieef"
Y = "klaief"
m = len(X)
n = len(Y)
print("Length of LCS =", lcs(X, Y, m, n))
# This code is contributed by Mohit KumarC#
// C# implementation to find the
// length of longest common subsequence
// which contains all vowel characters
using System;
class GFG
{
// function to check whether
// 'ch' is a vowel or not
static int isVowel(char ch)
{
if (ch == 'a' || ch == 'e' ||
ch == 'i' || ch == 'o' ||
ch == 'u')
return 1;
return 0;
}
// find max value
static int max(int a, int b)
{
return (a > b) ? a : b;
}
// function to find the length of
// longest common subsequence which
// contains all vowel characters
static int lcs(String X, String Y,
int m, int n)
{
int [,]L = new int[m + 1, n + 1];
int i, j;
// Following steps build L[m+1,n+1]
// in bottom up fashion. Note that
// L[i,j] contains length of LCS of
// X[0..i-1] and Y[0..j-1]
for (i = 0; i <= m; i++)
{
for (j = 0; j <= n; j++)
{
if (i == 0 || j == 0)
L[i, j] = 0;
else if ((X[i - 1] == Y[j - 1]) &&
isVowel(X[i - 1]) == 1)
L[i, j] = L[i - 1, j - 1] + 1;
else
L[i, j] = max(L[i - 1, j],
L[i, j - 1]);
}
}
// L[m,n] contains length of LCS
// for X[0..n-1] and Y[0..m-1]
// which contains all vowel characters
return L[m, n];
}
// Driver Code
static public void Main(String []args)
{
String X = "aieef";
String Y = "klaief";
int m = X.Length;
int n = Y.Length;
Console.WriteLine("Length of LCS = " +
lcs(X, Y, m, n));
}
}
// This code is contributed by Arnab KunduPHP
输出：
Length of LCS = 3
时间复杂度： O(m * n)。
辅助空间： O(m * n)。