1216. Valid Palindrome III

Problem Description

The task is to determine if a given string s can be transformed into a palindrome by removing at most k characters. A palindrome is a sequence that reads the same backward as forward, such as "radar" or "level." A k-palindrome is an extended version of this concept, where some flexibility is allowed by permitting the removal of up to k characters to achieve a palindrome. To solve this problem, we need to establish a method for deciding whether or not the string, after the allowed modifications, can be considered a palindrome.

Intuition

The solution leverages a dynamic programming approach. We use a 2D array f with dimensions equivalent to the length of the string s, where f[i][j] represents the length of the longest palindromic subsequence within the substring s[i..j]. The underlying intuition is that if we can identify such a subsequence that is sufficiently long, then the characters that are not part of this subsequence are the ones we can consider removing. If the number of these characters is less than or equal to k, then s is a k-palindrome.

We start by recognizing that any single character is trivially a palindrome, which gives us the initial condition for our dynamic programming table: f[i][i] = 1 for all indices i. From there, we move on to consider substrings of increasing length. When the characters at both ends of the current substring are the same, it contributes to a longer palindromic subsequence by simply framing any palindromic subsequence found within the interior of the substring (i.e., without the matched characters). When they do not match, we have a choice to make: we either consider the longest subsequence possible without the first character (s[i]) or without the last character (s[j]), which is embodied by taking the maximum of f[i+1][j] and f[i][j-1].

Finally, if we find a f[i][j] value such that f[i][j] + k is greater than or equal to the length of the string n, we have established that s can be considered a k-palindrome and can return true. If no such pair of indices (i, j) yields a suitable longest palindromic subsequence, the function will ultimately return false.

The intuition emerges from recognizing that palindromicity is, fundamentally, about symmetry and that dynamic programming allows us to explore and remember the outcomes of smaller problems (substrings) to solve larger ones.

Learn more about Dynamic Programming patterns.

Solution Approach

The provided code employs a dynamic programming technique to solve the problem. The algorithm involves the following steps:

1. Initialize a 2D table f with dimensions n x n, where n is the length of the string s. This table will store the length of the longest palindromic subsequence for every substring s[i..j].

2. Set f[i][i] = 1 for all i, representing the fact that each individual character is a palindrome of length 1.

3. Fill the table f in a bottom-up manner. Start by considering all substrings of length 2 and increase the length gradually. The outer loop starts from the second to last character and moves towards the first one (in reverse order).

4. For each pair (i, j) where i < j, two cases are possible:

   • If s[i] == s[j], the characters at both ends are the same, and they can contribute to the palindromic subsequence. The current cell f[i][j] is updated to f[i+1][j-1] + 2, adding the two matching characters to the length of the longest palindrome within the inner substring s[i+1..j-1].

   • If s[i] != s[j], the characters are different and cannot be part of the same palindrome so we must choose whether to exclude the character at the start or at the end of the string. The cell f[i][j] is updated to the maximum between f[i+1][j] and f[i][j-1], effectively discarding either s[i] or s[j] and considering the longer of the two resulting subsequences.

5. As we fill the table, we check after each update if the condition f[i][j] + k >= n is met. This condition checks whether the length of the longest palindromic subsequence plus the maximum allowed number of removals k covers the entire string length n. If it does, we know that the string s can be transformed into a palindrome by removing at most k characters, and we can return true.

6. If we go through the entire table without the condition being met, we return false, as it is not possible to transform s into a palindrome within the constraints of the problem.

The use of dynamic programming here is crucial as it optimizes the solution by avoiding redundant computation of the longest palindromic subsequence through memoization in the table f. This optimization reduces the time complexity of the algorithm compared to checking all possible subsequences individually.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach:

Suppose the given string s is "abxcyba" and k is 1. We want to determine if "abxcyba" can be turned into a palindrome by removing at most one character.

We'll initialize a 2D array f with dimensions 7x7 (n = 7 for our example, since the string length is 7). Initially, every entry f[i][i] is set to 1, which corresponds to each individual character being a palindrome of length 1.

1f = [1, 0, 0, 0, 0, 0, 0]
2    [0, 1, 0, 0, 0, 0, 0]
3    [0, 0, 1, 0, 0, 0, 0]
4    [0, 0, 0, 1, 0, 0, 0]
5    [0, 0, 0, 0, 1, 0, 0]
6    [0, 0, 0, 0, 0, 1, 0]
7    [0, 0, 0, 0, 0, 0, 1]

We start populating f for substrings of length 2:

• For i = 5 and j = 6 (s[i] is 'b' and s[j] is 'a'), they don't match, so we take the maximum of f[i+1][j] and f[i][j-1], both of which are 0, so f[i][j] remains 0.

• For i = 4 and j = 5 (s[i] is 'y' and s[j] is 'b'), they don't match, we do the same and f[4][5] remains 0.

• ... (We continue this for all substrings of length 2)

Then we move to substrings of length 3 and so on, up to the length of the whole string:

• During this process, for i = 0 and j = 6 (the whole string), if s[i] and s[j] (s[0] = 'a' and s[6] = 'a') are the same, so f[i][j] is f[1][5] + 2.

In the end, we check if f[0][n-1] (which represents the length of the longest palindromic subsequence of the whole string) plus k (which is 1 in this case) is greater than or equal to the length of the string n (which is 7). For the example, we'll assume that the longest palindromic subsequence we found has a length of 5.

1f[0][n-1] = 5

Since f[0][n-1] + k = 5 + 1 = 6 is not greater than or equal to 7, we determine that we cannot transform the string "abxcyba" into a palindrome by removing at most one character, and we return false.

This walkthrough provides a step-by-step illustration of how the dynamic programming table is populated and how the decision is made based on the values computed.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n^2). This arises because the algorithm contains two nested for-loops each running from 0 to n-1 for the outer loop and from i+1 to n for the inner loop, which performs a number of operations proportional to the square of the length of the input string s.

The space complexity of the algorithm is also O(n^2) due to the allocation of a 2D array f with n * n elements, where n is the length of string s. This array stores the length of the longest palindrome sequence found at every start and end index pair in the input string.

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