包含元音的最长公共子序列的长度

给定两个长度分别为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 面试体验(后端开发人员)。
朴素方法：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呈指数级增长。

高效方法(动态规划)：这种方法是最长公共子序列的变体 | DP-4 问题。

这篇文章的区别只是公共子序列字符必须都是元音。

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 Bilal

Python3

# 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 Kumar

C#

// 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 Kundu

Javascript

 
// Javascript implementation to find the
// length of longest common subsequence
// which contains all vowel characters
 
// Function to check whether 'ch'
// is a vowel or not
function isVowel(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
function lcs(X, Y, m, n)
{
    let L = new Array(m + 1);
    let 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++)
    {
        L[i] = new Array(n + 1);
        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] = 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
let X = "aieef";
let Y = "klaief";
let m = X.length;
let n = Y.length;
 
document.write("Length of LCS = " + lcs(X, Y, m, n));
 
// This code is contributed by avanitrachhadiya2155
 

输出：

Length of LCS = 3

时间复杂度： O(m*n)。
辅助空间： O(m*n)。

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