1216. Valid Palindrome III - Detailed Explanation

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Problem Statement

Description:
A string is called a k-palindrome if it can be transformed into a palindrome by removing at most k characters from it. Given a string s and an integer k, return true if s is a k-palindrome, and false otherwise.

Example Inputs & Outputs:

Example 1:
- Input:
  - s = "abcdeca"
  - k = 2
- Process:
  One way to transform "abcdeca" into a palindrome is by removing the characters 'b' and 'e', yielding "acdca", which is a palindrome.
- Output: true
- Explanation: Only 2 removals are needed, which is within the allowed limit.
Example 2:
- Input:
  - s = "acdcb"
  - k = 1
- Process:
  No matter which single character you remove, the remaining string cannot be rearranged into a palindrome.
- Output: false
- Explanation: More than 1 removal would be necessary to form a palindrome.
Example 3:
- Input:
  - s = "aba"
  - k = 0
- Process:
  The string "aba" is already a palindrome.
- Output: true
- Explanation: No removals are needed, which is within the allowed limit.

Constraints:

$(1 \leq s.length \leq 1000)$ (or similar in different variations)
$(0 \leq k \leq s.length)$
s consists of lowercase English letters (and possibly uppercase letters in some variations).

Hints to Approach the Problem

Hint 1:
Notice that if you can transform s into a palindrome by removing at most k characters, then the number of characters you must remove is exactly
$[ \text{removals} = \text{s.length} - \text{(length of the longest palindromic subsequence)} ]$
Hint 2:
Compute the Longest Palindromic Subsequence (LPS) of s. If
$[ \text{s.length} - \text{LPS} \leq k ]$ then s is a k-palindrome.

Approaches

Approach 1: Brute Force (Recursive Backtracking)

Idea:
Try all possible ways to remove up to k characters and check if the resulting string is a palindrome.
Downside:
This approach has exponential time complexity and is impractical for even moderately sized strings.

Approach 2: Dynamic Programming via Longest Palindromic Subsequence (Optimal)

Idea:
Instead of removing characters one by one, compute the length of the longest palindromic subsequence (LPS) of s.
- The minimum number of deletions needed to make s a palindrome is given by: $[ \text{minDeletions} = \text{s.length} - \text{LPS} ]$
- If $(\text{minDeletions} \leq k$ ), then s is a k-palindrome.
How to Compute LPS:
Use dynamic programming with a 2D table where:
- dp[i][j] represents the length of the LPS in the substring s[i...j].
- Transition:
  - If s[i] == s[j], then
    $[ dp[i][j] = dp[i+1][j-1] + 2 ]$
  - Otherwise,
    $[ dp[i][j] = \max(dp[i+1][j], dp[i][j-1]) ]$
- Base Case:
  Every single character is a palindrome of length 1 (i.e. dp[i][i] = 1).

Python Code

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Java Code

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Complexity Analysis

Time Complexity:
- The dynamic programming solution uses two nested loops over the string’s length, leading to a time complexity of $(O(n^2)$ ), where (n) is the length of s.
Space Complexity:
- A 2D array of size (n \times n) is used, resulting in $(O(n^2)$ ) space.
- (Note: Space optimizations are possible but add complexity.)

Step-by-Step Walkthrough and Visual Example

Example: s = "abcdeca", k = 2

Compute LPS via DP:
- For every single character, dp[i][i] = 1.
- For a substring like "ab", if characters differ, dp[i][j] = max(dp[i+1][j], dp[i][j-1]).
- Continuing for longer substrings, eventually you compute dp[0][6], which represents the LPS for "abcdeca".
- In this case, the LPS is "aceca" with a length of 5.
Calculate Minimum Deletions:
- Total characters = 7
- Minimum deletions needed = (7 - 5 = 2)
Compare with k:
- Since $(2 \leq 2$ ), the string is a k-palindrome, and the function returns true.

Visual Summary:

s: a b c d e c a
LPS: a   c   e   c   a  (length = 5)
Min deletions = 7 - 5 = 2  <= k (2)  → Valid k-palindrome

Common Pitfalls

Recursive Brute Force:
Attempting to try all deletion combinations without DP leads to exponential time complexity.
Incorrect DP Table Setup:
Failing to correctly initialize the base cases (e.g., dp[i][i] = 1) or improper handling of substrings of length 2.
Off-by-One Errors:
When iterating over substring lengths and indices, it is easy to make off-by-one mistakes.

Edge Cases

Already a Palindrome:
If s is already a palindrome, then no removals are needed, and the answer is true.
Large k:
If k is greater than or equal to the length of s, then s is trivially a k-palindrome.
Single Character String:
A string with one character is always a palindrome.

Alternative Variation:
Some variations might ask for the minimum number of deletions needed instead of just a boolean answer.
Related Problems for Further Practice:
- Longest Palindromic Subsequence: Directly computing the LPS is a classic problem.
- Minimum Deletions to Make a String Palindrome: Essentially the same DP formulation.
- Edit Distance: Related string transformation problems.