[Dynamic Programming] Distinct Subsequences - Medium

The Dynamic Programming Distinct Subsequences problem is about finding how many unique subsequences we can form from string T using the characters in another string S. This problem is common in combinatorial algorithms. The solution counts the ways to create one sequence from another. We can use dynamic programming to find the number of distinct subsequences. We build a table that shows the connections between the characters of S and T.

In this article, we will look at different ways to solve the Distinct Subsequences problem. We will talk about a recursive method, memoization techniques, and three bottom-up methods in Java, Python, and C++. We will also discuss how to save space and analyze the complexity to see how efficient our solutions are. Here is what we will cover:

Dynamic Programming Distinct Subsequences Problem Overview
Dynamic Programming Distinct Subsequences Recursive Approach
Dynamic Programming Distinct Subsequences Memoization Technique
Dynamic Programming Distinct Subsequences Bottom Up Approach in Java
Dynamic Programming Distinct Subsequences Bottom Up Approach in Python
Dynamic Programming Distinct Subsequences Bottom Up Approach in C++
Dynamic Programming Distinct Subsequences Space Optimization Techniques
Dynamic Programming Distinct Subsequences Complexity Analysis
Frequently Asked Questions

For more reading on similar dynamic programming problems, we can check out articles about Dynamic Programming Fibonacci Number and Dynamic Programming Longest Common Subsequence.

Dynamic Programming Distinct Subsequences Recursive Approach

The Distinct Subsequences problem is to find how many different subsequences of string S equal string T. We can solve this problem by breaking it down into smaller problems using recursion.

Recursive Function Definition

We define a recursive function called countDistinctSubsequences(S, T, m, n). Here: - m is the length of S - n is the length of T

Base Cases

If n is 0, which means T is an empty string, there is one subsequence of S that matches T. This subsequence is the empty subsequence. So, we return 1.
If m is 0 and n is more than 0, we return 0. This is because we cannot make a non-empty T from an empty S.

Recursive Cases

If the last characters of S and T match (if S[m-1] == T[n-1]):
- We count the subsequences by including the last character: countDistinctSubsequences(S, T, m-1, n-1)
- We also count the subsequences by not including the last character of S: countDistinctSubsequences(S, T, m-1, n)

The total count is the sum of these two counts.

If the last characters do not match (if S[m-1] != T[n-1]):
- We only count the subsequences by not including the last character of S: countDistinctSubsequences(S, T, m-1, n)

Recursive Implementation

Here is a simple code in Python:

def countDistinctSubsequences(S, T, m, n):
    if n == 0:
        return 1
    if m == 0:
        return 0
    
    if S[m-1] == T[n-1]:
        return (countDistinctSubsequences(S, T, m-1, n-1) + 
                countDistinctSubsequences(S, T, m-1, n))
    else:
        return countDistinctSubsequences(S, T, m-1, n)

# Example Usage
S = "rabbbit"
T = "rabbit"
count = countDistinctSubsequences(S, T, len(S), len(T))
print(count)  # Output: 3

This recursive way works but can be slow because of overlapping subproblems. For big strings, we can think about using memoization or a bottom-up dynamic programming way to make it faster.

Dynamic Programming Distinct Subsequences Memoization Technique

We use the memoization technique to solve the “Distinct Subsequences” problem in dynamic programming. This method helps us improve the recursive approach. We store results of smaller problems so we do not have to do the same calculations again.

Problem Definition

We have two strings S and T. Our goal is to count how many distinct subsequences of T we can make by deleting some characters from S.

Recursive Approach

In the recursive way, we create a function countDistinctSubsequences(i, j). This function gives us the number of distinct subsequences of T[0..j-1] in S[0..i-1].

Memoization Implementation

To use memoization, we need a 2D array dp. In this array, dp[i][j] saves the result of countDistinctSubsequences(i, j). If we already calculated dp[i][j], we can just return its value.

Here is how we implement the memoization technique:

public class DistinctSubsequences {
    public int numDistinct(String s, String t) {
        int[][] dp = new int[s.length() + 1][t.length() + 1];

        // Base case: An empty string T has one subsequence in any string S
        for (int i = 0; i <= s.length(); i++) {
            dp[i][0] = 1;
        }

        // Fill dp array
        for (int i = 1; i <= s.length(); i++) {
            for (int j = 1; j <= t.length(); j++) {
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }

        return dp[s.length()][t.length()];
    }
}

Explanation of the Code

Base Case: If T is empty, there is one subsequence of T in S. This is the empty subsequence.
Filling the DP Table:
- If the characters match (s.charAt(i - 1) == t.charAt(j - 1)), we add the count of subsequences that include both characters and those that do not include the current character of S.
- If they do not match, we take the count from the previous character of S.

Complexity

Time Complexity: O(m * n) where m is the length of S and n is the length of T.
Space Complexity: O(m * n) because of the dp array.

This memoization method makes the recursive solution much faster. It avoids repeating calculations for the same smaller problems. This makes it good for larger inputs. For more about similar problems, we can look at Dynamic Programming: Unique Paths in a Grid.

Dynamic Programming Distinct Subsequences Bottom Up Approach in Java

We can solve the Distinct Subsequences problem using the Bottom Up Approach. This method uses a 2D array to build the solutions to smaller problems step by step. It avoids the extra cost of recursion. So, it gives us a faster way to find the answer.

Problem Statement

We have two strings s and t. Our goal is to find how many distinct subsequences of t exist in s. A subsequence means we can remove some characters from s but we must keep the order of the remaining characters.

Dynamic Programming Table Setup

We create a 2D array dp. Here, dp[i][j] represents the number of distinct subsequences of t[0..j-1] in s[0..i-1].
We initialize:
- dp[0][0] = 1: An empty string is a subsequence of another empty string.
- dp[i][0] = 1 for all i: An empty t is a subsequence of any part of s.

Transition Formula

If s[i-1] == t[j-1]:
- dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
If s[i-1] != t[j-1]:
- dp[i][j] = dp[i-1][j]

Java Implementation

public class DistinctSubsequences {
    public int numDistinct(String s, String t) {
        int m = s.length();
        int n = t.length();
        int[][] dp = new int[m + 1][n + 1];

        // Initialize dp table
        dp[0][0] = 1;
        for (int i = 1; i <= m; i++) {
            dp[i][0] = 1; // Empty t is a subsequence of any s
        }

        // Fill dp table
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }

        return dp[m][n]; // Result is in the bottom-right cell
    }

    public static void main(String[] args) {
        DistinctSubsequences ds = new DistinctSubsequences();
        String s = "rabbbit";
        String t = "rabbit";
        System.out.println("Number of distinct subsequences: " + ds.numDistinct(s, t)); // Output: 3
    }
}

Explanation of the Code

The numDistinct function finds the number of distinct subsequences of t in s. It uses the dynamic programming table we created before.

In the main method, we show an example. We take s = "rabbbit" and t = "rabbit". The output is 3.

This Java code uses the Bottom Up Approach for the Distinct Subsequences problem. It quickly finds the number of distinct subsequences using dynamic programming. If you want to learn more about dynamic programming, check topics like Dynamic Programming Fibonacci Number.

Dynamic Programming Distinct Subsequences Bottom Up Approach in Python

We can solve the distinct subsequences problem using a bottom-up approach. This means we build a dynamic programming table to keep track of results. This way, we do not need to use recursion or memoization. Instead, we fill the DP table step by step using values we have already calculated.

Problem Definition

We have two strings, s and t. Our goal is to find how many distinct subsequences of t can be made from s.

Dynamic Programming Table

We create a 2D DP array dp. Here, dp[i][j] shows the number of distinct subsequences of t[0:j] in s[0:i].
We set dp[0][0] to 1. This means there is one way to make an empty string.
For each character in s and t, we update the table using these conditions:
- If the characters match (s[i-1] == t[j-1]), then:
```
dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
```
- If the characters do not match:
```
dp[i][j] = dp[i-1][j]
```

Python Implementation

def numDistinct(s: str, t: str) -> int:
    m, n = len(s), len(t)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    
    # Base case
    for i in range(m + 1):
        dp[i][0] = 1  # An empty string t has one subsequence in any prefix of s
    
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if s[i - 1] == t[j - 1]:
                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
            else:
                dp[i][j] = dp[i - 1][j]
    
    return dp[m][n]

# Example usage
s = "rabbbit"
t = "rabbit"
print(numDistinct(s, t))  # Output: 3

Explanation of the Code

We start with the function numDistinct. It makes a DP table with size (m + 1) x (n + 1). Here m is the length of s, and n is the length of t.

We set the base case for when t is empty by filling the first column with 1.

The loops go through each character in both strings. They fill the DP table based on whether the characters match or not.

In the end, we return dp[m][n]. This value gives us the total number of distinct subsequences of t in s.

This bottom-up approach works well and does not use much memory. It is good for bigger input sizes. If you want to learn more about dynamic programming, we can check articles on Dynamic Programming Fibonacci Number and Dynamic Programming Longest Common Subsequence.

Dynamic Programming Distinct Subsequences Bottom Up Approach in C++

The bottom-up way to solve the Distinct Subsequences problem in dynamic programming is to build the solution step by step. We use a table to keep the results of smaller problems.

Problem Statement

We have two strings S and T. Our job is to find how many distinct subsequences of S match T.

Approach

Initialization: We create a 2D array dp with size (m+1) x (n+1). Here m is the length of S and n is the length of T. We set dp[0][0] to 1. This is because an empty string is a subsequence of another empty string.
Filling the DP Table:
- For each character in S (from 1 to m):
  - For each character in T (from 1 to n):
    - If S[i-1] == T[j-1], we do:
      - dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
    - If not, we do:
      - dp[i][j] = dp[i-1][j]
Result: We will find the answer in dp[m][n].

C++ Implementation

#include <vector>
#include <string>
#include <iostream>

class Solution {
public:
    int numDistinct(std::string S, std::string T) {
        int m = S.size();
        int n = T.size();
        std::vector<std::vector<int>> dp(m + 1, std::vector<int>(n + 1, 0));

        // An empty T is a subsequence of any S
        for (int i = 0; i <= m; i++) {
            dp[i][0] = 1;
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (S[i - 1] == T[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[m][n];
    }
};

int main() {
    Solution solution;
    std::string S = "rabbbit";
    std::string T = "rabbit";
    std::cout << "Number of distinct subsequences: " << solution.numDistinct(S, T) << std::endl;
    return 0;
}

In the code: - We have a Solution class. It has a method numDistinct that takes strings S and T. - We use a dynamic programming table dp to count distinct subsequences step by step. - The main function prints the result.

This method is good. It avoids problems that come with recursion and memoization. The bottom-up dynamic programming technique works well for bigger inputs. For more on dynamic programming, we can read about related topics like longest common subsequence and unique paths.

Dynamic Programming Distinct Subsequences Space Optimization Techniques

In the Dynamic Programming Distinct Subsequences problem, we can improve performance by saving memory. The usual way uses a 2D DP array. But we can make it better by using a 1D array. The solution at each step only needs the previous step.

Space Optimization Using a 1D Array

Instead of using a full 2D array, we can keep a single array. This array will store results for the current and previous rows. Let’s see how to do this in different programming languages.

Java Implementation

public class DistinctSubsequences {
    public int numDistinct(String s, String t) {
        int m = s.length(), n = t.length();
        if (n == 0) return 1;
        if (m == 0) return 0;

        int[] dp = new int[n + 1];
        dp[0] = 1; // Base case

        for (int i = 1; i <= m; i++) {
            for (int j = n; j >= 1; j--) {
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    dp[j] += dp[j - 1];
                }
            }
        }
        return dp[n];
    }
}

Python Implementation

class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        m, n = len(s), len(t)
        if n == 0: return 1
        if m == 0: return 0

        dp = [0] * (n + 1)
        dp[0] = 1  # Base case

        for i in range(1, m + 1):
            for j in range(n, 0, -1):
                if s[i - 1] == t[j - 1]:
                    dp[j] += dp[j - 1]
        return dp[n]

C++ Implementation

class Solution {
public:
    int numDistinct(string s, string t) {
        int m = s.size(), n = t.size();
        if (n == 0) return 1;
        if (m == 0) return 0;

        vector<int> dp(n + 1, 0);
        dp[0] = 1; // Base case

        for (int i = 1; i <= m; i++) {
            for (int j = n; j >= 1; j--) {
                if (s[i - 1] == t[j - 1]) {
                    dp[j] += dp[j - 1];
                }
            }
        }
        return dp[n];
    }
};

Key Points

Memory Reduction: Using the 1D array cuts space needed from O(m * n) to O(n).
Bottom-Up Dynamic Programming: This way fits with the bottom-up method in dynamic programming. It uses memory well.
Performance: The time needed is O(m * n), but the space needed is now O(n).

By using these space optimization techniques, we can solve the Distinct Subsequences problem better. We save memory and can handle larger inputs more easily.

Dynamic Programming Distinct Subsequences Complexity Analysis

The Distinct Subsequences problem is about finding how many distinct subsequences of a string S match a target string T. We need to look at the complexity analysis of different ways to solve this problem. This helps us understand how efficient the algorithms are.

Time Complexity

Recursive Approach:
- The recursive solution takes a lot of time. It has an exponential time complexity of O(2^(m + n)). Here, m is the length of string T and n is the length of string S. This happens because recursion looks at all possible subsequences.
Memoization Technique:
- With memoization, we store results of previous calculations. This reduces the time complexity to O(m * n). Each subproblem gets solved just once, which is a big improvement over the basic recursive way.
Bottom-Up Approach:
- The bottom-up dynamic programming method also works in O(m * n) time. It builds the solution step by step in a table. Each subproblem is solved only one time.

Space Complexity

Recursive Approach:
- The space complexity is O(m + n). This is because of the recursion stack, which can go deep up to the total length of S and T.
Memoization Technique:
- In memoization, the space complexity stays O(m * n). This is due to the extra space for the memoization table.
Bottom-Up Approach:
- For the bottom-up method, we can improve space complexity to O(n). We only keep the last row of the DP table instead of the whole table.

Example of Time Complexity Calculation

Let’s take an example where S = "rabbbit" and T = "rabbit":

With the recursive approach, we check many combinations. This leads to many repeated calculations.
Using memoization, we calculate each unique state only once. This gives a linear relationship based on the lengths of S and T.

Conclusion of Complexity Analysis

In conclusion, the dynamic programming methods really help us solve the Distinct Subsequences problem better than the simple recursive way. By using memoization or a table method, we handle both time and space complexity well. This makes them good for larger input sizes.

For more information on similar dynamic programming topics, you can read about the Dynamic Programming Fibonacci Number or Longest Common Subsequence.

Frequently Asked Questions

1. What is the distinct subsequences problem in dynamic programming?

The distinct subsequences problem is a well-known challenge in dynamic programming. We need to find how many different subsequences of a string S match a target string T. This problem is important in many areas, like string matching and bioinformatics. To solve it well, we usually use methods like recursion, memoization, or bottom-up approaches in programming languages like Java, Python, or C++.

2. How do I approach the distinct subsequences problem recursively?

To solve this problem using recursion, we can make a function that looks at two options for each character in the string S. We can either include the character in the match against T or leave it out. We keep exploring these options until we finish checking the strings or find a complete match. This method is easy to understand but can take a long time to run. So, we often use memoization to make it faster.

3. What is the memoization technique for distinct subsequences?

Memoization is a way to make dynamic programming faster. We save the results we have already calculated. This helps us avoid doing the same work again, which is good for the distinct subsequences problem. We can store results in a 2D array or a dictionary, using the positions in S and T. This method helps reduce the time it takes to solve the problem to O(m*n), where m and n are the lengths of S and T.

4. Can you explain the bottom-up approach for distinct subsequences in Python?

In the bottom-up approach for this problem, we fill a 2D DP table step by step, starting with the simplest cases. We create a table where dp[i][j] shows how many distinct subsequences of the first i characters of S match the first j characters of T. By updating this table based on whether we include or exclude characters, we can find the answer without using recursion. This makes the solution faster and clearer.

5. What are the space optimization techniques for the distinct subsequences problem?

Space optimization techniques for this problem usually cut down the 2D DP table to a 1D array. Instead of keeping a full table, we update the array directly, using only the current and previous values. This method greatly decreases the space we need from O(m*n) to O(n) while keeping the time complexity the same. It is a good way to handle large inputs.

For more insights on dynamic programming techniques, we can look at related topics like Dynamic Programming: Unique Paths in a Grid or Dynamic Programming: Longest Common Subsequence. These articles give more examples and help us understand dynamic programming better.