Solution Approach

The solution implements a 3D dynamic programming approach to count palindromic subsequences. Let's walk through the implementation step by step:

1. Initialization:

dp = [[[0] * 4 for _ in range(n)] for _ in range(n)]

We create a 3D DP table where dp[i][j][k] stores the count of distinct palindromic subsequences in substring s[i...j] that start and end with character k (where k maps to 'a', 'b', 'c', 'd' as 0, 1, 2, 3).

2. Base Case:

for i, c in enumerate(s):  dp[i][i][ord(c) - ord('a')] = 1

Single characters are palindromes themselves. For each position i, we mark that there's exactly 1 palindrome of length 1 ending with that character.

3. Building Up Solutions:

for l in range(2, n + 1): # substring length  for i in range(n - l + 1): # starting position  j = i + l - 1 # ending position

We iterate through all possible substring lengths from 2 to n, and for each length, we consider all possible starting positions.

4. Core DP Transition: For each substring s[i...j] and each possible character c in 'abcd':

• Case 1: s[i] == s[j] == c

  dp[i][j][k] = 2 + sum(dp[i + 1][j - 1])

  When both boundaries match character c, we can form:

  • The single character c (count = 1)
  • The double character cc (count = 1)
  • Any palindrome from inner substring wrapped with c on both sides

• Case 2: s[i] == c but s[j] != c

  dp[i][j][k] = dp[i][j - 1][k]

  Palindromes with character c must start at position i, so we exclude the last character.

• Case 3: s[i] != c but s[j] == c

  dp[i][j][k] = dp[i + 1][j][k]

  Palindromes with character c must end at position j, so we exclude the first character.

• Case 4: Neither boundary is c

  dp[i][j][k] = dp[i + 1][j - 1][k]

  Any palindrome with character c must be completely within the inner substring.

5. Final Result:

return sum(dp[0][-1]) % mod

The total count is the sum of all palindromic subsequences in the entire string s[0...n-1] for all possible characters, taken modulo 10^9 + 7.

Time Complexity: O(n² × 4) = O(n²) where n is the length of the string Space Complexity: O(n² × 4) = O(n²) for the 3D DP table

Example Walkthrough

Let's walk through a small example with s = "bccb" to illustrate the solution approach.

Step 1: Initialize the DP table We create a 3D DP table dp[i][j][k] where indices represent:

• i, j: substring boundaries (0 to 3 for our string of length 4)
• k: character type (0='a', 1='b', 2='c', 3='d')

Step 2: Base case - single characters

s = "b c c b"  0 1 2 3 (indices)

For each single character:

• Position 0: 'b' → dp[0][0][1] = 1 (one palindrome "b")
• Position 1: 'c' → dp[1][1][2] = 1 (one palindrome "c")
• Position 2: 'c' → dp[2][2][2] = 1 (one palindrome "c")
• Position 3: 'b' → dp[3][3][1] = 1 (one palindrome "b")

Step 3: Build substrings of length 2

For substring "bc" (i=0, j=1):

• For character 'b': s[0]='b', s[1]='c' → only left boundary matches
  • dp[0][1][1] = dp[0][0][1] = 1 (palindrome "b")
• For character 'c': s[0]='b', s[1]='c' → only right boundary matches
  • dp[0][1][2] = dp[1][1][2] = 1 (palindrome "c")

For substring "cc" (i=1, j=2):

• For character 'c': s[1]='c', s[2]='c' → both boundaries match!
  • dp[1][2][2] = 2 + sum(dp[2][1])
  • Since i+1 > j-1, inner substring is empty, sum = 0
  • dp[1][2][2] = 2 (palindromes "c" and "cc")

For substring "cb" (i=2, j=3):

• For character 'c': s[2]='c', s[3]='b' → only left boundary matches
  • dp[2][3][2] = dp[2][2][2] = 1 (palindrome "c")
• For character 'b': s[2]='c', s[3]='b' → only right boundary matches
  • dp[2][3][1] = dp[3][3][1] = 1 (palindrome "b")

Step 4: Build substrings of length 3

For substring "bcc" (i=0, j=2):

• For character 'b': s[0]='b', s[2]='c' → only left boundary matches
  • dp[0][2][1] = dp[0][1][1] = 1 (palindrome "b")
• For character 'c': s[0]='b', s[2]='c' → only right boundary matches
  • dp[0][2][2] = dp[1][2][2] = 2 (palindromes "c", "cc")

For substring "ccb" (i=1, j=3):

• For character 'c': s[1]='c', s[3]='b' → only left boundary matches
  • dp[1][3][2] = dp[1][2][2] = 2 (palindromes "c", "cc")
• For character 'b': s[1]='c', s[3]='b' → only right boundary matches
  • dp[1][3][1] = dp[2][3][1] = 1 (palindrome "b")

Step 5: Build substring of length 4 (entire string)

For substring "bccb" (i=0, j=3):

• For character 'b': s[0]='b', s[3]='b' → both boundaries match!
  • Inner substring is "cc" (i+1=1, j-1=2)
  • Sum of inner palindromes = dp[1][2][2] = 2
  • dp[0][3][1] = 2 + 2 = 4 (palindromes "b", "bb", "bcb", "bccb")
• For character 'c': s[0]='b', s[3]='b' → neither boundary matches
  • dp[0][3][2] = dp[1][2][2] = 2 (palindromes "c", "cc")

Step 6: Calculate final result Total palindromic subsequences = sum of all characters for entire string:

• dp[0][3][0] (character 'a') = 0
• dp[0][3][1] (character 'b') = 4
• dp[0][3][2] (character 'c') = 2
• dp[0][3][3] (character 'd') = 0

Total = 0 + 4 + 2 + 0 = 6

The 6 distinct palindromic subsequences are: "b", "c", "bb", "cc", "bcb", "bccb"

Time and Space Complexity

Time Complexity: O(n² × 4) = O(n²)

The algorithm uses dynamic programming with three nested loops:

• The outer loop iterates through all possible substring lengths from 2 to n: O(n) iterations
• The middle loop iterates through all possible starting positions for substrings of length l: O(n) iterations
• The inner loop iterates through all 4 possible characters ('a', 'b', 'c', 'd'): O(4) = O(1) iterations

Within the innermost loop, the operations are mostly O(1), except for one case where sum(dp[i + 1][j - 1]) is called, which takes O(4) = O(1) time since it sums over 4 elements.

Therefore, the total time complexity is O(n × n × 4) = O(n²).

Space Complexity: O(n² × 4) = O(n²)

The algorithm uses a 3-dimensional DP table:

• First dimension represents the starting index: n positions
• Second dimension represents the ending index: n positions
• Third dimension represents the character type (a, b, c, or d): 4 values

The total space required is n × n × 4 = O(n²) since 4 is a constant.

No additional significant auxiliary space is used beyond the DP table and a few variables, so the overall space complexity remains O(n²).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Modulo Application

A critical pitfall in this solution is applying the modulo operation only at the final return statement. During the DP computation, intermediate values can exceed the integer limit, potentially causing overflow issues in languages with fixed integer sizes or leading to incorrect results.

Problem Code:

dp[start][end][char_index] = 2 + sum(dp[start + 1][end - 1]) # ... other assignments without modulo return sum(dp[0][n - 1]) % MOD

Solution: Apply modulo operation during each DP update to prevent overflow:

if s[start] == s[end] == char:  dp[start][end][char_index] = (2 + sum(dp[start + 1][end - 1])) % MOD elif s[start] == char:  dp[start][end][char_index] = dp[start][end - 1][char_index] % MOD elif s[end] == char:  dp[start][end][char_index] = dp[start + 1][end][char_index] % MOD else:  dp[start][end][char_index] = dp[start + 1][end - 1][char_index] % MOD

2. Index Out of Bounds When Accessing Inner Substring

When computing dp[start + 1][end - 1] for substrings of length 2, we get start + 1 > end - 1, which creates an invalid range. The current code doesn't handle this edge case explicitly, which could lead to accessing uninitialized values or incorrect results.

Problem Scenario: When length = 2, we have end = start + 1, so start + 1 = end and end - 1 = start, making dp[start + 1][end - 1] reference dp[end][start] which represents an invalid substring.

Solution: Handle the edge case explicitly:

if s[start] == s[end] == char:  if length == 2:  # For length 2, we have two palindromes: single char and double char  dp[start][end][char_index] = 2  else:  # For longer strings, include inner palindromes  inner_sum = sum(dp[start + 1][end - 1]) % MOD  dp[start][end][char_index] = (2 + inner_sum) % MOD

3. Assumption About Character Set

The code assumes the input string contains only characters 'a', 'b', 'c', 'd'. If the string contains other characters, the solution will fail or produce incorrect results.

Problem:

for char in 'abcd': # Hard-coded character set  char_index = ord(char) - ord('a')

Solution: Either validate the input or dynamically determine the character set:

# Option 1: Validate input unique_chars = set(s) if not unique_chars.issubset({'a', 'b', 'c', 'd'}):  raise ValueError("String must contain only characters a, b, c, d")  # Option 2: Dynamic character set unique_chars = sorted(set(s)) char_to_index = {char: i for i, char in enumerate(unique_chars)} dp = [[[0] * len(unique_chars) for _ in range(n)] for _ in range(n)]

Complete Corrected Solution:

class Solution:  def countPalindromicSubsequences(self, s: str) -> int:  MOD = 10**9 + 7  n = len(s)   dp = [[[0] * 4 for _ in range(n)] for _ in range(n)]   # Base case  for i, char in enumerate(s):  char_index = ord(char) - ord('a')  dp[i][i][char_index] = 1   # Process substrings  for length in range(2, n + 1):  for start in range(n - length + 1):  end = start + length - 1   for char in 'abcd':  char_index = ord(char) - ord('a')   if s[start] == s[end] == char:  if length == 2:  dp[start][end][char_index] = 2  else:  inner_sum = sum(dp[start + 1][end - 1]) % MOD  dp[start][end][char_index] = (2 + inner_sum) % MOD  elif s[start] == char:  dp[start][end][char_index] = dp[start][end - 1][char_index]  elif s[end] == char:  dp[start][end][char_index] = dp[start + 1][end][char_index]  else:  if length > 2:  dp[start][end][char_index] = dp[start + 1][end - 1][char_index]  # else: remains 0 for length 2 when neither end matches   return sum(dp[0][n - 1]) % MOD