Length of longest common subsequence containing vowels

Given two strings X and Y of length m and n respectively. The problem is to find the length of the longest common subsequence of strings X and Y which contains all vowel characters.

Examples:

Input : X = "aieef" 
        Y = "klaief"
Output : aie

Input : X = "geeksforgeeks" 
        Y = "feroeeks"
Output : eoee

Source: Paytm Interview Experience ( Backend Developer ).

Naive Approach: Generate all subsequences of both given sequences and find the longest matching subsequence which contains all vowel characters. This solution is exponential in term of time complexity.

Efficient Approach (Dynamic Programming): This approach is a variation to Longest Common Subsequence | DP-4 problem.

The difference in this post is just that the common subsequence characters must all be vowels. 

Implementation:

C++

// C++ implementation to find the length of longest common
// subsequence which contains all vowel characters
#include <bits/stdc++.h>

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

PHP

<?php
// PHP 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)
{
    $L = array_fill(0, $m + 1, array_fill(0, $n + 1, NULL));

    // 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 Code
$X = "aieef";
$Y = "klaief";

$m = strlen($X);
$n = strlen($Y);

echo "Length of LCS = " . lcs($X, $Y, $m, $n);

// This code is contributed by ita_c
?>

Javascript

<script>

// 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

</script>

Output

Length of LCS = 3

Complexity Analysis:

• Time Complexity: O(m*n). 
• Auxiliary Space: O(m*n).

Efficient approach : Space optimization

In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.

Implementation: 

C++

// C++ implementation to find the length of longest common
// subsequence which contains all vowel characters
#include <bits/stdc++.h>
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)
{

    // initialize vector dp to store
    // computations of subproblems
    vector<int> dp(n + 1, 0);

    // iterating over subproblems to get the
    // current value from previous computations
    for (int i = 1; i <= m; i++) {
        // to store just previous value
        int prev = 0;
        for (int j = 1; j <= n; j++) {

            // current value
            int curr = dp[j];
            if ((X[i - 1] == Y[j - 1]) && isVowel(X[i - 1]))
                dp[j] = prev + 1;
            else
                dp[j] = max(dp[j], dp[j - 1]);

            // assigning values to iterate further
            prev = curr;
        }
    }

    // return answer
    return dp[n];
}

// Driver Code
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;
}

Output

Length of LCS = 3

Time Complexity: O(m*n). 
Auxiliary Space: O(n).