Problem Description

You have a string s that contains only the letters 'a' and 'b'. Your goal is to make this string empty by repeatedly removing palindromic subsequences from it.

In each step, you can:

• Choose any palindromic subsequence from the string
• Remove it from the string

A subsequence is formed by deleting some (possibly zero) characters from the original string without changing the order of the remaining characters. The characters don't need to be consecutive.

A palindrome is a string that reads the same forwards and backwards (like "aba" or "aa").

You need to find the minimum number of steps required to make the string completely empty.

For example:

• If s = "ababa", this is already a palindrome, so you can remove the entire string in 1 step
• If s = "abb", you might remove "aa" first (taking the two 'a's as a subsequence), then remove "bb" in the second step, for a total of 2 steps

The key insight is that since the string only contains two different characters ('a' and 'b'), you can always remove all 'a's as one palindromic subsequence (since "aaa...a" is a palindrome) and all 'b's as another palindromic subsequence (since "bbb...b" is a palindrome). This means the answer will never be more than 2. If the entire string is already a palindrome, the answer is 1. If the string is not empty but not a palindrome, the answer is 2.

Intuition

The first observation is that the string only contains two distinct characters: 'a' and 'b'. This is a crucial constraint that drastically simplifies the problem.

Let's think about what palindromic subsequences we can form:

• Any sequence of identical characters forms a palindrome: "aaa...a" or "bbb...b"
• We can pick all 'a's from the string (regardless of their positions) to form a palindromic subsequence
• Similarly, we can pick all 'b's to form another palindromic subsequence

This means we can always empty the string in at most 2 steps:

1. Remove all 'a's as one palindromic subsequence
2. Remove all 'b's as another palindromic subsequence

But can we do better? Yes, if the entire string itself is already a palindrome! In that case, we can remove the whole string in just 1 step.

So the problem reduces to:

• If the string is empty (edge case), return 0
• If the entire string is a palindrome, return 1
• Otherwise, return 2

Since the problem guarantees a non-empty string, we only need to check if s is a palindrome. If it is, we need 1 step. If not, we need 2 steps.

The elegant solution return 1 if s[::-1] == s else 2 captures this logic perfectly - it checks if the string equals its reverse (palindrome check) and returns the appropriate number of steps.

Learn more about Two Pointers patterns.

Solution Approach

The implementation is remarkably simple due to the problem's constraints. Here's how the solution works:

Step 1: Check if the string is a palindrome

We use Python's string slicing to reverse the string: s[::-1]

• The slice notation [::-1] creates a reversed copy of the string
• We compare this with the original string s

Step 2: Return the appropriate value

return 1 if s[::-1] == s else 2

This conditional expression handles both cases:

• If s[::-1] == s is True (the string is a palindrome), return 1
• Otherwise, return 2

Why this works:

1. Case 1 - String is a palindrome: If the entire string reads the same forwards and backwards, we can remove it all at once as a single palindromic subsequence. Result: 1 step.

2. Case 2 - String is not a palindrome: Since we only have characters 'a' and 'b', we can:

   • First remove all 'a's (they form a palindromic subsequence like "aaa...a")
   • Then remove all 'b's (they form a palindromic subsequence like "bbb...b")
   • Result: 2 steps maximum.

Time Complexity: O(n) where n is the length of the string, as we need to create a reversed string and compare it.

Space Complexity: O(n) for storing the reversed string during comparison.

The beauty of this solution lies in recognizing that with only two distinct characters, the answer can only be 1 or 2, eliminating the need for complex dynamic programming or greedy algorithms.

Example Walkthrough

Let's walk through two examples to illustrate the solution approach:

Example 1: s = "abba"

Step 1: Check if the string is a palindrome

• Original string: "abba"
• Reversed string (s[::-1]): "abba"
• Are they equal? Yes! "abba" == "abba" is True

Step 2: Since the string is a palindrome, we return 1

We can remove the entire string "abba" in one step because it's already a palindromic subsequence.

Example 2: s = "abbab"

Step 1: Check if the string is a palindrome

• Original string: "abbab"
• Reversed string (s[::-1]): "babba"
• Are they equal? No! "abbab" != "babba"

Step 2: Since the string is NOT a palindrome, we return 2

Here's how we could remove it in 2 steps:

• Step 1: Remove all 'a's as a subsequence → removes positions 0 and 3 → "a_a" forms palindrome "aa"
  • Remaining string: "bbb"
• Step 2: Remove all 'b's as a subsequence → removes "bbb" which is already a palindrome
  • Remaining string: empty

Alternative removal (also 2 steps):

• Step 1: Remove all 'b's → removes positions 1, 2, and 4 → subsequence "bbb"
  • Remaining string: "aa"
• Step 2: Remove all 'a's → removes "aa" which is a palindrome
  • Remaining string: empty

The key insight: with only two characters 'a' and 'b', we can always group all identical characters together as palindromic subsequences, guaranteeing we never need more than 2 steps!

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the string s. The code performs a string reversal operation s[::-1] which takes O(n) time to create a reversed copy of the string, and then compares it with the original string using == which also takes O(n) time in the worst case to compare all characters.

Space Complexity: O(n) where n is the length of the string s. The string slicing operation s[::-1] creates a new string that is a reversed copy of the original string, which requires O(n) additional space to store the reversed string temporarily during the comparison.

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

Common Pitfalls

1. Misunderstanding Subsequence vs Substring

A frequent mistake is confusing subsequences with substrings. Many developers initially think they need to find consecutive palindromic substrings to remove, leading to unnecessarily complex solutions involving dynamic programming or sliding windows.

Wrong Thinking:

# Trying to find consecutive palindromic substrings
def removePalindromeSub(self, s: str) -> int:
    count = 0
    while s:
        # Looking for longest palindromic substring from start
        for i in range(len(s), 0, -1):
            if s[:i] == s[:i][::-1]:
                s = s[i:]
                count += 1
                break
    return count

Correct Understanding:

• A subsequence allows you to pick any characters (maintaining order) even if they're not consecutive
• This means all 'a's in "ababa" form the subsequence "aaa" which is a palindrome
• All 'b's form "bb" which is also a palindrome

2. Overcomplicating Due to Not Recognizing the Two-Character Constraint

Many developers miss the crucial insight that having only two distinct characters severely limits the possible answers to just 1 or 2.

Overcomplicated Approach:

def removePalindromeSub(self, s: str) -> int:
    # Trying to find optimal subsequences using DP
    dp = [[0] * len(s) for _ in range(len(s))]
    # Complex DP logic...

Simple Solution: The constraint that the string contains only 'a' and 'b' means:

• If it's already a palindrome → 1 removal
• If it's not a palindrome → 2 removals (all 'a's, then all 'b's)

3. Edge Case: Empty String Handling

While the problem states the string contains 'a' and 'b', some might forget to handle the edge case properly if the string could be empty.

Potential Issue:

def removePalindromeSub(self, s: str) -> int:
    # This would return 2 for empty string!
    return 1 if s[::-1] == s else 2

Robust Solution:

def removePalindromeSub(self, s: str) -> int:
    if not s:  # Handle empty string explicitly
        return 0
    return 1 if s[::-1] == s else 2

Note: Most LeetCode problems guarantee non-empty input, but it's good practice to consider this case.

4. Inefficient Palindrome Check

Some might implement palindrome checking inefficiently:

Inefficient:

def removePalindromeSub(self, s: str) -> int:
    # Manual two-pointer approach (unnecessary here)
    left, right = 0, len(s) - 1
    is_palindrome = True
    while left < right:
        if s[left] != s[right]:
            is_palindrome = False
            break
        left += 1
        right -= 1
    return 1 if is_palindrome else 2

Better for this problem:

def removePalindromeSub(self, s: str) -> int:
    # Python's string slicing is optimized and cleaner
    return 1 if s == s[::-1] else 2

While the two-pointer approach has better space complexity theoretically (O(1) vs O(n)), for this specific problem with guaranteed small inputs, the cleaner slicing approach is preferred for readability and simplicity.