Problem Description

You are given a string s and a positive integer k. Your task is to select a set of non-overlapping substrings from the string that meet specific criteria and maximize the count of such substrings.

The selected substrings must satisfy two conditions:

1. Each substring must have a length of at least k characters
2. Each substring must be a palindrome (reads the same forwards and backwards)

The substrings you select cannot overlap with each other - meaning they cannot share any character positions in the original string.

Your goal is to find the maximum number of such palindromic substrings you can select from the string s.

For example, if you have a string like "abaccdbbd" and k = 3, you could potentially select palindromes like "aba" and "bdb" (which don't overlap), giving you a count of 2. The challenge is to find the optimal selection that maximizes this count.

A substring is defined as a contiguous sequence of characters within the string, meaning the characters must appear consecutively in the original string.

Intuition

To solve this problem, we need to think about two main challenges: identifying all possible palindromes and then selecting the maximum number of non-overlapping ones.

The first insight is that we need to know which substrings are palindromes before we can make any selection decisions. Checking if a substring is a palindrome repeatedly would be inefficient, so we can precompute this information. We can use dynamic programming to build a 2D table dp[i][j] that tells us whether the substring from index i to index j is a palindrome. The key observation here is that a substring s[i...j] is a palindrome if and only if s[i] == s[j] and the inner substring s[i+1...j-1] is also a palindrome.

Once we know all the palindromes, the second challenge is selecting the maximum number of non-overlapping ones. This is where the problem becomes interesting - it's essentially an optimization problem where we need to make choices. At each position in the string, we have a decision to make: should we include a palindrome starting at this position, or skip it and look for palindromes starting later?

This decision-making process naturally leads to a recursive approach. For any position i in the string, we can either:

1. Skip position i and find the best solution starting from position i+1
2. Choose a palindrome starting at position i (if one exists with length at least k), and then find the best solution starting after that palindrome ends

The key insight is that once we choose a palindrome from position i to position j, the next palindrome can only start from position j+1 or later (to avoid overlapping). This creates a natural recursive structure where each decision leads to a smaller subproblem.

By using memoization with our recursive function dfs(i), we avoid recalculating the same subproblems multiple times. The function dfs(i) represents the maximum number of palindromes we can select starting from position i onwards, considering all the constraints.

Learn more about Greedy, Two Pointers and Dynamic Programming patterns.

Solution Approach

The solution consists of two main phases: preprocessing to identify all palindromes and a memoized recursive search to find the maximum number of non-overlapping palindromes.

Phase 1: Preprocessing Palindromes

We create a 2D boolean table dp[i][j] where dp[i][j] = True if the substring s[i...j] is a palindrome. We build this table using dynamic programming:

• Initialize all single characters as palindromes: dp[i][i] = True
• For each substring, we check from the end of the string backwards (starting from i = n-1 down to 0)
• For each starting position i, we check all ending positions j > i
• A substring s[i...j] is a palindrome if: s[i] == s[j] and dp[i+1][j-1] is True

This preprocessing takes O(n^2) time and space, where n is the length of the string.

Phase 2: Memoized Recursive Search

We define a recursive function dfs(i) that returns the maximum number of non-overlapping palindromes that can be selected from s[i...] (the substring starting at position i).

The recurrence relation is:

dfs(i) = max(
    dfs(i + 1),  // Skip position i
    max{1 + dfs(j + 1) for all j where j >= i + k - 1 and dp[i][j] is True}
)

Breaking this down:

• Base case: If i >= n, return 0 (no more characters to process)
• We have two choices at each position:
  1. Skip the current position and solve for dfs(i + 1)
  2. Choose a palindrome starting at position i with length at least k

For the second choice, we iterate through all possible ending positions j where:

• j >= i + k - 1 (ensuring the palindrome has at least k characters)
• dp[i][j] = True (ensuring the substring is actually a palindrome)

If we choose the palindrome from i to j, we add 1 to our count and continue searching from position j + 1 (the first position after our chosen palindrome).

The @cache decorator (memoization) ensures that each unique state dfs(i) is computed only once, reducing the time complexity from exponential to O(n^2).

The final answer is dfs(0), which gives us the maximum number of non-overlapping palindromes starting from the beginning of the string.

Example Walkthrough

Let's walk through the solution with the string s = "abaccdbbd" and k = 3.

Phase 1: Building the Palindrome Table

First, we build our dp table to identify all palindromes. We'll create a 9×9 table (string length is 9).

Starting with single characters (all are palindromes):

• dp[0][0] = True ('a')
• dp[1][1] = True ('b')
• ... and so on for all positions

Then we check longer substrings, working backwards from position 8 to 0:

For position 7 (character 'b'):

• Check s[7...8]: "bd" → 'b' ≠ 'd', so dp[7][8] = False

For position 6 (character 'b'):

• Check s[6...7]: "bb" → 'b' = 'b', so dp[6][7] = True
• Check s[6...8]: "bbd" → 'b' ≠ 'd', so dp[6][8] = False

Continuing this process, we identify key palindromes of length ≥ 3:

• dp[0][2] = True: "aba" (positions 0-2)
• dp[2][4] = True: "acc" (positions 2-4) - Wait, this is False ('a' ≠ 'c')
• dp[3][5] = True: "ccd" - False ('c' ≠ 'd')
• dp[5][7] = True: "dbb" - False ('d' ≠ 'b')
• dp[6][8] = True: "bbd" - False ('b' ≠ 'd')

Actually, let me recalculate the palindromes correctly:

• "aba" (0-2): 'a' = 'a' and 'b' is palindrome → dp[0][2] = True
• "bac" (1-3): 'b' ≠ 'c' → dp[1][3] = False
• "acc" (2-4): 'a' ≠ 'c' → dp[2][4] = False
• "ccd" (3-5): 'c' ≠ 'd' → dp[3][5] = False
• "cdb" (4-6): 'c' ≠ 'b' → dp[4][6] = False
• "dbb" (5-7): 'd' ≠ 'b' → dp[5][7] = False
• "bbd" (6-8): 'b' ≠ 'd' → dp[6][8] = False

Let's check for "bdb" (positions 1,5,6): This isn't contiguous! We need contiguous substrings.

Actually, looking more carefully at "abaccdbbd":

• Position 5-7: "dbb" → 'd' ≠ 'b', not a palindrome
• Position 4-6: "cdb" → 'c' ≠ 'b', not a palindrome

The valid palindromes of length ≥ 3 are:

• "aba" (positions 0-2): dp[0][2] = True

Phase 2: Finding Maximum Non-overlapping Palindromes

Now we use our recursive function dfs(i) to find the maximum count:

dfs(0): Starting from position 0

• Option 1: Skip position 0 → dfs(1)
• Option 2: Take palindrome "aba" (0-2) → 1 + dfs(3)

Let's evaluate dfs(1): Starting from position 1

• No palindromes of length ≥ 3 start at position 1
• Return dfs(2)

dfs(2): Starting from position 2

• No palindromes of length ≥ 3 start at position 2
• Continue similarly...

dfs(3): Starting from position 3

• No palindromes of length ≥ 3 start here
• Continue checking...

Since we don't find any other palindromes of length ≥ 3 in the remaining string, most dfs calls return 0.

Therefore:

• dfs(3) = 0 (no more palindromes found)
• dfs(0) = max(dfs(1), 1 + dfs(3)) = max(0, 1 + 0) = 1

The maximum number of non-overlapping palindromes of length at least 3 is 1 (just "aba").

This example shows how the algorithm:

1. Precomputes all palindromes efficiently
2. Uses memoized recursion to avoid overlapping selections
3. Maximizes the count by trying all valid combinations

Solution Implementation

Time and Space Complexity

The time complexity is O(n^2) and the space complexity is O(n^2), where n is the length of the string s.

Time Complexity Analysis:

• Building the palindrome DP table: The nested loops iterate through all pairs (i, j) where i < j, which takes O(n^2) time.
• The DFS function with memoization: Each state i (from 0 to n-1) is computed at most once due to the @cache decorator. For each state, we iterate through positions from i + k - 1 to n - 1, which is at most O(n) operations. Therefore, the DFS contributes O(n) * O(n) = O(n^2) time.
• Overall time complexity: O(n^2) + O(n^2) = O(n^2).

Space Complexity Analysis:

• The 2D DP table dp[i][j] stores palindrome information for all substring pairs, requiring O(n^2) space.
• The memoization cache for the DFS function stores at most n states, each taking O(1) space, contributing O(n) space.
• The recursion stack depth is at most O(n) in the worst case.
• Overall space complexity: O(n^2) + O(n) + O(n) = O(n^2).

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

Common Pitfalls

1. Greedy Selection of Palindromes

The Pitfall: A common mistake is attempting to solve this problem greedily by always selecting the first valid palindrome encountered. For example, given the string "aabaa" with k=2, a greedy approach might select "aa" at position 0-1, then "aa" at position 3-4, yielding 2 palindromes. However, the optimal solution is to select the single palindrome "aabaa" (positions 0-4), then potentially find more palindromes afterward.

Why This Fails: Selecting shorter palindromes early can block the selection of longer palindromes that might lead to better overall solutions. The problem requires global optimization, not local optimization.

Solution: The recursive approach with memoization correctly explores all possibilities - both taking and skipping palindromes at each position - ensuring we find the globally optimal solution.

2. Incorrect Palindrome DP Table Initialization

The Pitfall: When building the palindrome checking table, incorrectly handling the base cases can lead to wrong results. Specifically:

• Forgetting that single characters are palindromes
• Not properly handling adjacent characters (length 2 substrings)
• Accessing dp[i+1][j-1] when i+1 > j-1, which can give incorrect results

Example of Incorrect Code:

# WRONG: This doesn't handle edge cases properly
for start in range(string_length):
    for end in range(start + 1, string_length):
        is_palindrome[start][end] = (s[start] == s[end] and 
                                    is_palindrome[start + 1][end - 1])

Solution: Initialize the diagonal (single characters) as True and process the table in the correct order (bottom-up from the end of the string):

# Initialize all positions as True (handles single characters and edge cases)
is_palindrome = [[True] * string_length for _ in range(string_length)]

# Process from end to beginning to ensure dp[i+1][j-1] is computed before dp[i][j]
for start in range(string_length - 1, -1, -1):
    for end in range(start + 1, string_length):
        is_palindrome[start][end] = (s[start] == s[end] and 
                                    is_palindrome[start + 1][end - 1])

3. Off-by-One Errors in Index Handling

The Pitfall: Confusion between inclusive and exclusive indices when:

• Calculating substring length (should be end - start + 1)
• Checking minimum length constraint (should be end >= start + k - 1)
• Continuing recursion after selecting a palindrome (should continue from end + 1)

Example of Incorrect Code:

# WRONG: Off-by-one error in length check
for end_index in range(start_index + k, string_length):  # Should be start_index + k - 1
    if is_palindrome[start_index][end_index]:
        max_count = max(max_count, 1 + find_max_palindromes(end_index))  # Should be end_index + 1

Solution: Be explicit about index calculations:

# Correct: end_index is inclusive, so length = end - start + 1
# For length >= k, we need end >= start + k - 1
for end_index in range(start_index + k - 1, string_length):
    if is_palindrome[start_index][end_index]:
        # Continue from the position after the selected palindrome
        max_count = max(max_count, 1 + find_max_palindromes(end_index + 1))

4. Memory Optimization Opportunities Missed

The Pitfall: While the solution works correctly, it might use more memory than necessary. The palindrome table uses O(n²) space, but we could potentially optimize this.

Solution: For very large strings, consider using a set to store only valid palindrome ranges instead of a full 2D table:

# Memory-optimized approach for sparse palindromes
valid_palindromes = set()
for start in range(string_length):
    for end in range(start + k - 1, string_length):
        if is_palindrome_check(s, start, end):  # Check palindrome on-the-fly
            valid_palindromes.add((start, end))

This is particularly useful when the number of valid palindromes of length ≥ k is much smaller than n².