Count Different Palindromic Subsequences

“Count Different Palindromic Subsequences” involves dynamic programming, strings and palindromic sequences. Here are some simpler problems to prepare for this problem:

  1. 125. Valid Palindrome: This problem can help understand basic checks for palindromic strings.

  2. 647. Palindromic Substrings: This problem practices finding all palindromic substrings in a string.

  3. 516. Longest Palindromic Subsequence: This problem uses dynamic programming to find the longest palindromic subsequence.

  4. 680. Valid Palindrome II: This problem introduces the concept of making modifications to create a palindrome.

  5. 5. Longest Palindromic Substring: More practice on finding palindromes in a string.

  6. 300. Longest Increasing Subsequence: Although not directly related, this dynamic programming problem gives experience with building solutions on previous results.

  7. 1143. Longest Common Subsequence: Another dynamic programming problem to practice on.

  8. 718. Maximum Length of Repeated Subarray: This problem is about finding common subsequences, but can still help practice with handling subsequences.

  9. 131. Palindrome Partitioning: This problem involves finding all possible ways of partitioning a string into palindromic substrings.

  10. 279. Perfect Squares: This dynamic programming problem provides a different context, but the thought process is useful for complex problems.

Clarification Questions

What are the clarification questions we can ask about this problem?

Identifying Problem Isomorphism

“Count Different Palindromic Subsequences” can be approximately mapped to “Distinct Subsequences II”.

Reasoning:

Both problems involve counting distinct subsequences within a given string, and share a common core challenge, which is to account for the repetition of characters.

In “Count Different Palindromic Subsequences”, we count the distinct palindromic subsequences. A dynamic programming approach could be used, where the string is scanned from both ends and counts are made for palindromic sequences, with repetitions subtracted.

In “Distinct Subsequences II”, we count all distinct subsequences of the string. A dynamic programming approach is used here as well, where for each character we either choose to include it in our subsequence or not, and subtract repetitions.

“Count Different Palindromic Subsequences” is more complex because it has an added constraint that the subsequences must be palindromic. “Distinct Subsequences II” only requires that the subsequences be distinct, which makes it a simpler problem.

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class Solution:
    def countPalindromicSubsequences(self, s: str) -> int:
        n = len(s)
        mod = 10**9 + 7
        dp = [[0] * n for _ in range(n)]
        for i in range(n):
            dp[i][i] = 1
        for i in range(n - 1, -1, -1):
            for j in range(i + 1, n):
                if s[i] == s[j]:
                    left, right = i + 1, j - 1
                    while left <= right and s[left] != s[i]:
                        left += 1
                    while left <= right and s[right] != s[i]:
                        right -= 1
                    if left > right:
                        dp[i][j] = dp[i + 1][j - 1] * 2 + 2
                    elif left == right:
                        dp[i][j] = dp[i + 1][j - 1] * 2 + 1
                    else:
                        dp[i][j] = dp[i + 1][j - 1] * 2 - dp[left + 1][right - 1]
                else:
                    dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1]
                dp[i][j] = (dp[i][j] + mod) % mod
        return dp[0][n - 1]

Problem Classification

This problem can be classified as a Combinatorial and Dynamic Programming problem.

The problem is about string manipulation, subsequence generation, and counting distinct palindromic subsequences.

  • Combinatorial: The task involves calculating the total number of unique palindromic subsequences in a given string. This requires an understanding of combinatorics to navigate the possible combinations of characters and identify those that form palindromes.

  • Dynamic Programming: The problem can be efficiently solved using a dynamic programming approach. This involves breaking down the problem into smaller subproblems and solving each subproblem only once, reusing their solutions for larger problems.

‘What’ Components:

  1. A string ’s’: The input for this problem is a string ’s’ of length between 1 and 1000. Each character in the string is either ‘a’, ‘b’, ‘c’, or ’d’.

  2. Non-empty palindromic subsequences: We are tasked with identifying the number of unique palindromic subsequences. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. A sequence is palindromic if it reads the same backward as forward.

  3. Output number: The output should be the total number of unique non-empty palindromic subsequences in ’s’, modulo 10^9 + 7, to handle the potential for very large outputs. The modulo operation is used to keep the result within a manageable range.

  4. Constraints: The given string will only contain the characters ‘a’, ‘b’, ‘c’, and ’d’, and its length will not exceed 1000 characters. This ensures that the problem remains within computationally feasible bounds.

The problem requires both combinatorial logic to understand all the different ways characters in the string can be combined to form subsequences and dynamic programming techniques to efficiently calculate and store those combinations that form unique palindromes.

Language Agnostic Coding Drills

  1. Dissecting the code:

a) Arrays and Matrix Initialization: This concept involves understanding how to create and initialize arrays or matrices with default values. In this code, it is used to create a 2D array dp of size n*n initialized with zeroes and single diagonal elements as one.

b) Loops and iteration: The code uses several nested loops to iterate over the elements in the string, and to fill up the 2D dp array. This requires understanding how to write nested loops, and how to control their iteration.

c) Condition checking (if-else statements): The code uses conditional checks to decide what calculations to perform based on the characters being compared in the string. This is a fundamental programming concept.

d) Dynamic Programming: This is an advanced concept that involves solving a problem by breaking it down into smaller subproblems and storing the results of these subproblems to avoid solving them multiple times. The 2D dp array is a classic example of a DP table.

e) Modulo arithmetic: The code uses modulo arithmetic to keep the results within a certain range. This is a common technique used in problems where the output can be very large.

  1. Order of increasing difficulty:

a) Condition checking (if-else statements): This is a basic concept in most programming languages.

b) Loops and iteration: While still a basic concept, loops require a slightly higher level of understanding than simple condition checking, especially when nested loops are involved.

c) Arrays and Matrix Initialization: This concept requires an understanding of data structures, specifically arrays, and how to create and initialize them.

d) Modulo arithmetic: Requires understanding of mathematical concepts, but still more straightforward than dynamic programming.

e) Dynamic Programming: This is typically considered an advanced topic in computer science due to the level of problem-solving and understanding of recursion it requires.

  1. Problem-solving approach:

The problem is to find the number of distinct non-empty palindromic subsequences in a given string. The strategy used in the solution is a dynamic programming approach where we store the results of smaller subproblems to avoid re-computation.

To solve this, the code first initializes a 2D DP table. It then iterates through the string from end to start, checking for every pair of elements (i, j) whether they are equal or not. If they are equal, the code finds out if there are any characters equal to the current one between indices i and j and based on that, it calculates the number of distinct palindromic subsequences. If the characters at i and j are not equal, the number of palindromes is calculated as the sum of the palindromes calculated for (i+1, j) and (i, j-1) minus the palindromes calculated for (i+1, j-1).

Finally, the function returns the number of distinct non-empty palindromic subsequences for the entire string, which is stored in dp[0][n - 1].

Each of the coding drills mentioned above contribute to the solution in the following way: array initialization is used to set up the dp table; loops are used to fill up this table; condition checking is used to decide how to update the dp table; dynamic programming is the overall strategy used to solve the problem; and modulo arithmetic is used to handle large outputs.

Targeted Drills in Python

  1. Python-based coding drills for each identified concept:

a) Condition checking (if-else statements)

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x = 10
if x > 5:
    print("x is greater than 5")
else:
    print("x is not greater than 5")

b) Loops and iteration

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for i in range(10):
    print(i)

c) Arrays and Matrix Initialization

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# Initialize a list (array in Python)
my_list = [0] * 5  # Creates a list of 5 zeros

# Initialize a 2D list (matrix in Python)
my_matrix = [[0]*5 for _ in range(5)]  # Creates a 5x5 matrix filled with zeros

d) Modulo arithmetic

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x = 10000000000
mod = 10**9 + 7
x_mod = x % mod
print(x_mod)  # Prints x modulo mod

e) Dynamic Programming

Let’s consider the Fibonacci series, which is a simpler problem that can illustrate dynamic programming. In this problem, we store the calculated Fibonacci numbers in an array to avoid re-calculating them.

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def fibonacci(n):
    dp = [0, 1] + [0] * (n - 1)  # initializing the DP table
    for i in range(2, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]
    return dp[n]

  1. Problem-specific concepts: The main problem-specific concept in this problem is the dynamic programming approach to count palindromic subsequences. This involves understanding how to create a 2D DP table, and how to use it to count the palindromic subsequences in a specific way as per the problem’s requirements. A drill for this concept would involve practicing similar dynamic programming problems, particularly those involving strings and subsequences.

  2. Assembling the pieces:

All the pieces can be assembled together in the order of their complexity, starting from the basic concepts like condition checking and looping to more complex ones like dynamic programming.

Firstly, we initialize our 2D DP array. Then we iterate through our string in reverse order using a nested loop. For each pair of characters, we perform different operations based on their equality and position in the string. This involves both condition checking and looping.

Within these conditional checks, we perform additional iterations and calculations to account for various scenarios in the problem (e.g., if the characters are equal but there are more of the same character between them). These operations are then used to update the DP array.

Finally, after filling up the DP array, we return the value in the top right corner of the array (which represents the count of distinct palindromic subsequences for the entire string). The final answer is also taken modulo 10^9 + 7 to keep the number within the range of a 32-bit integer.

This way, each drill is integrated into the final solution, contributing to the overall dynamic programming approach to solve the problem.