Problem Description

You are given a string text that needs to be split into k substrings (subtext₁, subtext₂, ..., textₖ) with the following conditions:

1. Each substring must be non-empty
2. When concatenated together, all substrings must equal the original text (i.e., subtext₁ + subtext₂ + ... + subtextₖ == text)
3. The substrings must form a palindromic pattern where subtextᵢ == subtextₖ₋ᵢ₊₁ for all valid values of i (where 1 ≤ i ≤ k)

The third condition means that the first substring must equal the last substring, the second must equal the second-to-last, and so on. For example, if we split "ghiabcdefhelloadamhelloabcdefghi" into 5 parts: ["ghi", "abcdef", "hello", "adamhello", "abcdef", "ghi"], we can see that:

• 1st substring "ghi" equals 5th substring "ghi"
• 2nd substring "abcdef" equals 4th substring "abcdef"
• 3rd substring "helloadamhello" is in the middle (equals itself when k is odd)

Your task is to find the maximum possible value of k - the largest number of substrings you can split the text into while satisfying all three conditions.

The solution uses a greedy two-pointer approach, starting from both ends of the string and looking for the shortest matching prefix and suffix. When a match is found, it counts as 2 pieces (the prefix and suffix), and the process continues with the remaining middle portion. If no match can be found, the remaining string counts as 1 piece.

Intuition

The key insight is that we want to maximize the number of pieces, which means we should try to create as many matching pairs as possible from the ends of the string.

Think about it this way: if we have a string and we can find a prefix that matches a suffix, we've created 2 pieces. The question becomes - should we take the shortest matching pair or look for longer ones?

Let's consider why choosing the shortest matching prefix-suffix pair is optimal:

Imagine we have two options:

• Option A: Take a short matching pair (like "a" from both ends)
• Option B: Take a longer matching pair (like "abc" from both ends)

If we choose the longer pair (Option B), we're consuming more characters, leaving less room for potential matches in the middle. By choosing the shorter pair (Option A), we leave more characters in the middle, giving us more opportunities to find additional matching pairs.

Here's a visual example: Consider the string "abcXYZabc"

• If we greedily take "a" and "c" as our first matching pair (both single characters 'a'), we're left with "bcXYZab"
• If we instead took "abc" as our matching pair, we're left with just "XYZ"

The greedy choice of shorter matches gives us more flexibility for the remaining string.

This naturally leads to a two-pointer approach:

1. Start with pointers at the beginning and end of the string
2. Try to find the shortest substring from the left that matches a substring from the right
3. When found, count it as 2 pieces and move both pointers inward
4. Repeat until the pointers meet or cross
5. If there's a remaining middle portion that couldn't be matched, it counts as 1 piece

The correctness of this greedy approach stems from the fact that any valid decomposition can be rearranged to have shorter matches on the outside without reducing the total number of pieces.

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

Solution Approach

The implementation uses a greedy approach with two pointers to find the maximum number of palindromic chunks.

We maintain two pointers:

• i starting from the beginning of the string (index 0)
• j starting from the end of the string (index len(text) - 1)

The algorithm works as follows:

1. Initialize variables:

   • ans = 0 to count the number of chunks
   • i = 0 and j = len(text) - 1 as our two pointers

2. Main loop - while i <= j:

   • Set k = 1 to represent the current substring length we're checking
   • Set ok = False to track whether we found a matching pair

3. Inner loop - search for the shortest matching prefix and suffix:

   • While i + k - 1 < j - k + 1 (ensuring substrings don't overlap):
     • Check if text[i : i + k] == text[j - k + 1 : j + 1]
     • If they match:
       • Add 2 to the answer (we found a matching pair)
       • Move i forward by k positions: i += k
       • Move j backward by k positions: j -= k
       • Set ok = True and break from the inner loop
     • If no match, increment k to try a longer substring

4. Handle the middle portion:

   • If ok is still False after the inner loop (no matching pair found):
     • Add 1 to the answer (the remaining middle portion becomes a single chunk)
     • Break from the main loop

5. Return the final answer

Example walkthrough with "ghiabcdefhelloadamhelloabcdefghi":

• First iteration: Find "ghi" matches at both ends, ans += 2, move pointers inward
• Second iteration: Find "abcdef" matches at both ends, ans += 2, move pointers inward
• Third iteration: Remaining string "helloadamhello" has no matching prefix-suffix, ans += 1
• Final answer: 5

The time complexity is O(n²) in the worst case, where we might need to check multiple substring lengths at each step. The space complexity is O(n) for the string slicing operations.

Example Walkthrough

Let's walk through the algorithm with the string "abcab" to find the maximum number of palindromic substrings.

Initial Setup:

• String: "abcab"
• i = 0 (pointing to 'a' at the start)
• j = 4 (pointing to 'b' at the end)
• ans = 0

Iteration 1:

• We're looking for matching prefix and suffix
• Try k = 1: Check if text[0:1] ("a") equals text[4:5] ("b")
  • "a" ≠ "b", no match
• Try k = 2: Check if text[0:2] ("ab") equals text[3:5] ("ab")
  • "ab" = "ab", match found!
• Add 2 to answer: ans = 2
• Move pointers: i = 2, j = 2

Iteration 2:

• Now i = 2 and j = 2 (both pointing to the middle 'c')
• Since i <= j is still true, we continue
• The remaining substring is just "c"
• Try k = 1: But i + k - 1 = 2 is not less than j - k + 1 = 2
• No matching pair can be found (we're at a single character)
• Add 1 to answer: ans = 3
• Break from loop

Result: The string "abcab" can be split into maximum 3 palindromic substrings: ["ab", "c", "ab"]

Let's verify:

• Concatenation: "ab" + "c" + "ab" = "abcab" ✓
• Palindromic pattern: 1st substring "ab" = 3rd substring "ab", 2nd substring "c" = itself ✓

This example demonstrates how the greedy approach finds the shortest matching pairs ("ab" instead of trying longer substrings first) to maximize the total number of pieces.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n²)

The algorithm uses two pointers i and j starting from both ends of the string. In the worst case:

• The outer while loop runs until i <= j, which can iterate up to n/2 times where n is the length of the string
• For each iteration of the outer loop, the inner while loop increments k to find matching substrings
• In the worst case, k can grow up to n/2 (when no matching prefix-suffix pairs are found until the middle)
• String slicing text[i : i + k] and text[j - k + 1 : j + 1] takes O(k) time
• String comparison also takes O(k) time

Therefore, the worst-case scenario involves O(n/2) iterations of the outer loop, with each iteration potentially doing O(n/2) work in the inner loop for substring operations, resulting in O(n²) time complexity.

Space Complexity: O(n)

The space complexity is O(n) due to the string slicing operations. When Python performs slicing like text[i : i + k] and text[j - k + 1 : j + 1], it creates new string objects that require additional memory proportional to the slice length. In the worst case, these slices can be up to O(n/2) in length.

If the implementation were modified to compare characters directly without creating substring copies (using index-based character comparison), the space complexity could be reduced to O(1).

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

Common Pitfalls

Pitfall 1: Incorrect Handling of the Middle Element

The Problem: A common mistake is forgetting to count the middle portion when no matching prefix-suffix pair can be found. When the pointers meet or cross without finding any matches, the remaining string still counts as one valid chunk.

Incorrect Implementation:

while left_pointer <= right_pointer:
    substring_length = 1
    match_found = False
  
    while left_pointer + substring_length - 1 < right_pointer - substring_length + 1:
        if left_substring == right_substring:
            chunk_count += 2
            match_found = True
            break
        substring_length += 1
  
    # WRONG: Forgetting to handle the case when no match is found
    if not match_found:
        break  # This loses the middle chunk!

Correct Implementation:

if not match_found:
    chunk_count += 1  # Count the remaining middle portion
    break

Pitfall 2: Off-by-One Errors in Pointer Boundaries

The Problem: The condition for checking substring overlap is critical. Using <= instead of < in the inner loop condition can cause the algorithm to check overlapping substrings, leading to incorrect results.

Incorrect Implementation:

# WRONG: This allows overlapping substrings
while left_pointer + substring_length - 1 <= right_pointer - substring_length + 1:
    # This might check overlapping regions

Correct Implementation:

# Ensures substrings don't overlap
while left_pointer + substring_length - 1 < right_pointer - substring_length + 1:
    # Proper boundary checking

Pitfall 3: Greedy Algorithm Correctness Assumption

The Problem: Some might think we need to try all possible decompositions to find the maximum k, not trusting that the greedy approach (finding shortest matching prefix-suffix) always yields the optimal solution.

Why Greedy Works: The key insight is that if we have a matching prefix and suffix of length L, splitting them off immediately is always optimal. Any alternative decomposition that keeps these L characters as part of larger chunks would result in fewer total chunks. The greedy choice of smallest matching chunks maximizes the number of pieces.

Example to Illustrate: For string "abcabc":

• Greedy approach: Split as ["a", "b", "c", "a", "b", "c"] → 6 chunks (though this would actually be ["a","bc","a"] → 3 chunks following the algorithm)
• Non-greedy might try: ["abc", "abc"] → 2 chunks
• The greedy approach correctly identifies ["a", "bc", "a"] by finding the shortest match first

Pitfall 4: String Slicing Performance

The Problem: Creating new substring copies with slicing operations in Python can be inefficient for very long strings, potentially causing performance issues.

Alternative Approach Using Indices:

def longestDecomposition(self, text: str) -> int:
    def matches(start1, end1, start2, end2):
        # Compare characters without creating substrings
        if end1 - start1 != end2 - start2:
            return False
        for i in range(end1 - start1):
            if text[start1 + i] != text[start2 + i]:
                return False
        return True
  
    chunk_count = 0
    left = 0
    right = len(text) - 1
  
    while left <= right:
        length = 1
        found = False
      
        while left + length - 1 < right - length + 1:
            if matches(left, left + length, right - length + 1, right + 1):
                chunk_count += 2
                left += length
                right -= length
                found = True
                break
            length += 1
      
        if not found:
            chunk_count += 1
            break
  
    return chunk_count

This approach avoids creating substring copies and can be more memory-efficient for large inputs.