Find length of longest subsequence of one string which is substring of another string

Given two strings X and Y. The task is to find the length of the longest subsequence of string X which is a substring in sequence Y.

Examples: 

Input : X = "ABCD",  Y = "BACDBDCD"
Output : 3
"ACD" is longest subsequence of X which
is substring of Y.

Input : X = "A",  Y = "A"
Output : 1

PREREQUISITES: Longest common subsequence problem will help you understand this problem in a snap 🙂

Method 1 (Brute Force): 

Use brute force to find all the subsequences of X and for each subsequence check whether it is a substring of Y or not. If it is a substring of Y, maintain a maximum length variable and compare its length. 

Time Complexity: O(2^n *(n*m)) 

Method 2: (Recursion):

Let n be the length of X and m be the length of Y. We will make a recursive function as follows with 4 arguments and the return type is int as we will get the length of a maximum possible subsequence of X which is the Substring of Y. We will take judgment for further process based on the last char in strings by using n-1 and m-1.

int maxSubsequenceSubstring(string &X,string &Y,int n,int m)
{
    ....
    return ans;
}

For recursion we need 2 things, we will make 1st the base case and 2nd is calls on smaller input (for that we will see the choice diagram). 

BASE CASE:

By seeing the argument of function we can see only 2 arguments that will change while recursion calls, i.e. lengths of both strings. So for the base case think of the smallest input we can give. You’ll see the smallest input is 0, i.e. empty lengths. hence when n == 0 or m == 0 is the base case. n and m can’t be less than zero. Now we know the condition and we need to return the length of subsequence as per the question. if the length is 0, then means one of the string is empty, and there is no common subsequence possible, so we have to return 0 for the same.

int maxSubsequenceSubstring(string &X,string &Y,int n,int m)
{
    if (n==0 || m==0) return 0;
    ....
    return ans;
}

CHILDREN CALLS: 

Choice Diagram

We can see how to make calls by seeing if the last char of both strings are the same or not. We can see how this is slightly different from the LCS (Longest Common Subsequence) question.

int maxSubsequenceSubstring(string &X,string &Y,int n,int m)
{
    // Base Case
    if (n==0 || m==0) return 0;
    
    // Calls on smaller inputs
    
    // if the last char of both strings are equal
    if(X[n-1] == Y[m-1])
    {
        return  1 + maxSubsequenceSubstring(X,Y,n-1,m-1);
    }
    
    // if the last char of both strings are not equal
    else
    {
        return maxSubsequenceSubstring(X,Y,n-1,m);    
    }
}

Now here is the main crux of the question, we can see we are calling for X[0..n] and Y[0..m], in our recursion function it will return the answer of maximum length of the subsequence of X, and Substring of Y (and the ending char of that substring of Y ends at length m). This is very important as we want to find all intermediary substrings also. hence we need to use a for loop where we will call the above function for all the lengths from 0 to m of Y, and return the maximum of the answer there. Here is the final code in C++ for the same.

Implementation:

Length for maximum possible Subsequence of string X which is Substring of Y -> 3

Time Complexity: O(n*m) (For every call in the recursion function we are decreasing n, hence we will reach the base case exactly after n calls, and we are using for loop for m times for the different lengths of string Y).

Space Complexity: O(n) (For recursion calls we are using stacks for each call).
 

Method 3: (Dynamic Programming):                                                                                                                                                                                                             

Submethod 1:-  Memoization

NOTE: Must see the above recursion solution to understand its optimization as Memoization

From the above recursion solution, there are multiple calls and further, we are using recursion in for loop, there is a high probability we have already solved the answer for a call. hence to optimize our recursion solution we will use? (See we have only 2 arguments that vary throughout the calls, hence the dimension of the array is 2 and the size is  because we need to store answers for all possible calls from 0..n and 0..m). Hence it is a 2D array.

we can use vectors for the same or dynamic allocation of the array.

// initialise a vector like this
vector<vector<int>> dp(n+1,vector<int>(m+1,-1));

// or Dynamic allocation
int **dp = new int*[n+1];
for(int i = 0;i<=n;i++)
{
    dp[i] = new int[m+1];
    for(int j = 0;j<=m;j++)
    {
        dp[i][j] = -1;
    }
}

By initializing the 2D vector we will use this array as the 5th argument in recursion and store our answer. also, we have filled it with -1, which means we have not solved this call hence use traditional recursion for it, if dp[n][m] at any call is not -1, means we have already solved the call, hence use the answer of dp[n][m].

// In recursion calls we will check for if we have solved the answer for the call or not
if(dp[n][m] != -1) return dp[n][m];

// Else we will store the result and return that back from this call
if(X[n-1] == Y[m-1]) 
    return dp[n][m] = 1 + maxSubsequenceSubstring(X,Y,n-1,m-1,dp);
else 
    return dp[n][m] = maxSubsequenceSubstring(X,Y,n-1,m,dp);

Code for memoization is here:

Length for maximum possible Subsequence of string X which is Substring of Y -> 3

Time Complexity: O(n*m) (It will be definitely better than the recursion solution, the worst case is possible only possible when none of the char of string X is there in String Y.)
Space Complexity: O(n*m + n) (the Size of Dp array and stack call size of recursion)

Submethod 2:  Tabulation 

Let n be length of X and m be length of Y. Create a 2D array ‘dp[][]’ of m + 1 rows and n + 1 columns. Value dp[i][j] is maximum length of subsequence of X[0….j] which is substring of Y[0….i]. Now for each cell of dp[][] fill value as : 

for (i = 1 to m)
  for (j = 1 to n)
    if (x[j-1] == y[i - 1])
      dp[i][j] = dp[i-1][j-1] + 1;
    else
      dp[i][j] = dp[i][j-1];

And finally, the length of the longest subsequence of x which is substring of y is max(dp[i][n]) where 1 <= i <= m.

Below is implementing this approach:

3

Time Complexity: O(n*m) (Time required to fill the Dp array)
Space Complexity: O(n*m + n) (the Size of Dp array)

This article is contributed by Anuj Chauhan. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.