2484. Count Palindromic Subsequences - Detailed Explanation

Problem Statement

Given a string s, return the number of different non-empty palindromic subsequences in s. Since the answer may be very large, return it modulo 10⁹ + 7.

A subsequence is a sequence that can be derived from the string by deleting some or no characters without changing the order of the remaining characters. A palindrome is a string that reads the same forward and backward.

Important:

• Two palindromic subsequences are considered different if they occur in different positions or have different compositions.
• The string s consists only of the characters 'a', 'b', 'c', and 'd'.

Example Inputs & Outputs

• Example 1:
  Input: "bccb"
  Output: 6
  Explanation:
  The 6 different palindromic subsequences are:

  • "b"
  • "c"
  • "bb"
  • "cc"
  • "bcb"
  • "bccb"

• Example 2:
  Input: "abcd"
  Output: 4
  Explanation:
  Each character itself is a palindrome. There are no longer palindromic subsequences since all characters are distinct.

• Example 3:
  Input: "aaa"
  Output: 3
  Explanation:
  The distinct palindromic subsequences are:

  • "a"
  • "aa"
  • "aaa"

Constraints

• 1 ≤ s.length ≤ 1000
• s[i] is either 'a', 'b', 'c', or 'd'.

Hints

1. Dynamic Programming (DP) on Substrings:
   Think about defining a DP state that represents the count of distinct palindromic subsequences in a substring of s. For example, let dp[i][j] denote the answer for the substring s[i…j].

2. Handling Duplicate Subsequences:
   When the two ends of a substring are the same (say, s[i] == s[j]), you need to carefully count additional palindromes—taking into account whether there are more of the same character between i and j. You might have to consider three cases:

   • No same character exists between i and j.
   • Exactly one occurrence exists.
   • More than one occurrence exists.

3. Recurrence Based on End Characters:
   If s[i] != s[j], the answer for s[i…j] can be expressed in terms of dp[i+1][j], dp[i][j-1], and dp[i+1][j-1].

Approaches

1. Brute Force (Not Feasible)

• Idea: Generate all subsequences and check for palindromes while ensuring uniqueness.
• Drawback: The total number of subsequences is exponential (2ⁿ), which makes this approach impractical for s.length up to 1000.

2. Optimal Dynamic Programming Approach

Idea:
Use a two-dimensional DP array dp[i][j] that stores the number of distinct palindromic subsequences in the substring s[i…j].
The recurrence is built based on whether s[i] and s[j] are the same:

• If s[i] != s[j]: [ dp[i][j] = dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1] ] Here, you add the subsequences from the right and left parts and subtract the overlapping subsequences counted twice.

• If s[i] == s[j]:
  Let char = s[i] and then find the first occurrence l and the last occurrence r of char in the substring s[i+1…j-1].

  • Case 1: No occurrence of char in s[i+1…j-1]
    [ dp[i][j] = dp[i+1][j-1] \times 2 + 2 ] (The extra “+2” counts the subsequences: the single character and the pair "cc".)

  • Case 2: One occurrence of char in s[i+1…j-1]
    [ dp[i][j] = dp[i+1][j-1] \times 2 + 1 ]

  • Case 3: More than one occurrence of char in s[i+1…j-1]
    [ dp[i][j] = dp[i+1][j-1] \times 2 - dp[l+1][r-1] ] (Subtract the palindromic subsequences that are double-counted between the first and last occurrences.)

• Base Case:
  For a single character (i.e. when i == j), there is exactly 1 palindromic subsequence: [ dp[i][i] = 1 ]

Modular Arithmetic:
Remember to take results modulo 10⁹ + 7 at each step.

Step-by-Step Walkthrough

Consider the string s = "bccb":

1. Initialization:
   For all i, set dp[i][i] = 1 since each individual character is a palindrome.

2. Length 2 Substrings:
   Evaluate all substrings of length 2.

   • For s[0…1] = "bc" (b ≠ c):
     [ dp[0][1] = dp[1][1] + dp[0][0] - dp[1][0] \quad \text{(dp[1][0] is 0 by convention)} = 1 + 1 = 2 ]
   • Similar process for other substrings.

3. Length 4 Substring ("bccb"):
   For s[0] == s[3] ('b'):

   • Find the first occurrence of 'b' in s[1…2] → not found (only 'c' exists), so this is Case 1.
   • Then,
     [ dp[0][3] = dp[1][2] \times 2 + 2 ]
   • After computing dp[1][2] (which counts palindromic subsequences in "cc"), you derive the final answer.

Note: The detailed values in each dp cell require careful iteration by increasing substring lengths.

Code Implementations

Python Code