1278. Palindrome Partitioning III

Problem Description

The given problem presents a scenario where we have a string s comprising lowercase English letters and an integer k. Our objective is to carry out two tasks. Firstly, we are to modify some of the characters in s into other lowercase English letters. Secondly, we must divide the modified string into k non-empty, disjoint substrings with the condition that each of these substrings must be a palindrome.

A palindrome is a sequence of characters which reads the same backward as forward (for example, "radar" or "level"). The goal is to achieve this division by making the minimal number of character changes in the original string s.

Intuition

To solve this problem, we need to consider two main aspects:

1. The cost of making a substring palindrome - This is the minimum number of character changes required to make any given substring a palindrome.

2. The optimal way to partition the string into k palindromes - Once we understand how to compute the cost for single substrings, we need to figure out an optimal way to partition the string into k segments such that the total cost (the sum of the costs of making all substrings palindromic) is minimized.

Here is the thought process to arrive at the solution approach:

• First, we need to calculate the cost of converting every possible substring into a palindrome. We do this by using dynamic programming to precompute and store the costs in a 2D array.

• Next, we use another 2D dynamic programming array, where f[i][j] represents the minimum number of changes needed to split the first i characters of s into j palindromes.

• The dynamic step progresses by considering all possible splits of the string where the rightmost palindrome ends just before the i-th character and the rest of the string is split into j-1 palindromes. We update our f[i][j] value by finding the minimum among all these possibilities.

• The final solution, f[n][k], would be the minimal number of changes to divide the complete string s of length n into k palindromic substrings.

By carefully computing these costs and properly combining them while maintaining the imposed conditions, we can incrementally construct the solution, allowing us to find the minimum total cost requested.

In the solution code given, g[i][j] is used to store the cost of converting the substring s[i:j+1] into a palindrome. The array f[i][j] keeps track of the minimum number of changes for the first i characters of s into j palindromes. The nested loops are used to calculate these values, using previously computed ones, which underline the dynamic programming approach.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution utilizes Dynamic Programming (DP), a method for solving complex problems by breaking them down into simpler subproblems. The approach uses two DP arrays, g and f, to store intermediate results and build up the solution progressively. Here is a breakdown of how the solution approach is implemented:

• Initialize a 2D array g, where g[i][j] will hold the cost for converting the substring from s[i] to s[j] into a palindrome. The cost is the number of characters that need to be changed.

• Fill the g array with the necessary costs. This involves iterating over all possible substrings in s:

  • Starting from the end of the string and moving backwards ensures that we calculate the costs for smaller substrings first. This is essential since the cost of a larger substring might depend on the cost of its smaller substrings.
  • For each pair of indices (i, j), we start by setting g[i][j] to be one if the characters at these indices are different (int(s[i] != s[j])), signifying a single necessary change to help form a palindrome.
  • If we have a substring longer than two characters (i + 1 < j), we add the cost of making the inner substring s[i+1:j-1] a palindrome, which we have already calculated (g[i + 1][j - 1]).

• Initialize a 2D array f, where f[i][j] will store the minimum number of changes required to divide the first i characters of s into j palindromes.

• Populate the f array using the previously computed costs in the g array:

  • Iterate over each position i in the string and for each possible number of palindromes j.
  • If we are looking for only one palindrome (j == 1), the cost is straight from the g array, since the whole substring s[0:i] needs to be palindromic.
  • For more than one palindrome (j > 1), we consider all possible previous positions h where the last palindrome could have started after (f[h][j-1] is the cost for the first h characters into j-1 palindromes). We then add the cost of making s[h:i] a palindrome (g[h][i - 1]).
  • We take the minimum of all these possible previous positions h to ensure we have the least number of changes.

• The minimum number of changes required to divide the entire string s into k palindromes is then found in f[n][k], where n is the length of s.

The variables i, j, and h represent indices in the string s and the loops over these indices address the subproblems in a bottom-up manner, allowing the solution to be assembled incrementally. The inf value serves as an initial high cost to ensure that any actual cost calculated will be lower, enabling the min function to work correctly.

Through this approach, all possible ways to form k palindromes out of the string s by making the fewest changes are evaluated efficiently. The nested loops ensure that every necessary substring and partition count scenario is considered, while previous calculations are reused, following the typical dynamic programming pattern of solving overlapping subproblems and optimizing the decision at each step.

Example Walkthrough

Let's illustrate the solution approach using a small example. Suppose we have the string s = "abracadabra" and we want to split it into k = 3 palindromic substrings, making the minimum number of character changes.

1. Initialize the 2D array g to store the cost of converting substrings to palindromes. Here's how to fill g:

   • For a single character substring, the cost is 0 as one letter is always a palindrome (e.g., g[0][0], g[1][1], ..., g[10][10] should all be 0).
   • For two character substrings, if the characters are the same, the cost is 0. If different, the cost is 1 (e.g., g[0][1] is 1 as 'a' != 'b', g[2][3] is 1 as 'r' != 'a').
   • For three or more character substrings, the cost is dependent on the first and last characters and the cost of the inner substring (e.g., g[0][2] would depend on g[1][1], which is 0, plus int(s[0] != s[2]), which is 0, thus g[0][2] = 0).

2. Repeat the process by expanding the substring length until g contains the cost for all possible substrings.

3. Initialize the 2D array f. f[i][j] will keep track of the minimum number of changes required to split the first i characters into j palindromes.

4. Begin populating f. For j=1, use the values directly from g:

   • f[10][1] = g[0][10]: the cost to change s[0:11] into one palindrome.

5. For j>1, consider where the last palindrome might start and calculate the cost of j-1 palindromes before that plus the cost of the last palindrome:

   • For f[10][3], we check each possible position h for the last palindrome to start, calculate f[h][2] + g[h+1][10] and find the minimum. For instance:
     • f[6][2] + g[7][10]: The cost to change the first 7 characters into 2 palindromes and the cost to change s[7:11] into a palindrome.
     • We do this for all possible h, and f[10][3] will be the minimum of all these values.

6. By the end of this process, f[n][k] (with n being the length of s) will store the minimum number of changes we need to split s into k palindromes.

Now, applying the approach to our example:

• g might look something like this (assuming we computed all necessary costs):

1   0 1 2 3 4 5 6 7 8  9 10
20 [0,1,0,...]
31    [0,1,...]
42       [0,...]
5...
610          [0] 

• Let's say that f[6][2] (min changes to make first 7 characters into 2 palindromes) is 2 and the cost to change s[7:11] ("dabra") into a palindrome is 2 (g[7][10] = 2), then one scenario for f[10][3] would be 2 + 2 = 4. We would compute all other scenarios similarly and take the minimum.

• Eventually, f[10][3] gives us the answer for the entire string s.

This example walkthrough demonstrates how dynamic programming allows us to efficiently compute and combine the costs of smaller problems to solve the larger problem at hand.

Solution Implementation

Time and Space Complexity

The given code snippet is designed to find the minimum number of characters that need to be changed to create k palindromic substrings from the given string s.

Time Complexity

The time complexity of the algorithm can be analyzed as follows:

1. The first loop calculates the minimum number of changes required to make all substrings palindromic. This is a double loop where i ranges from n-1 to 0 and j ranges from i+1 to n. For each i, j pair, it does up to O(1) work if i + 1 >= j or it adds the result of a precomputed value of g[i+1][j-1] which itself is O(1) operation. So this loop runs in O(n^2) time.

2. The second loop computes the minimum number of changes to make exactly k palindromic substrings by building up a solution using dynamic programming. This involves triple nested loops: i ranges from 1 to n, j ranges from 1 to min(i, k), and h ranges from j-1 to i. The innermost loop runs in O(i) time because it ranges from j-1 to i. Since j is at most k and k can be as large as n, in the worst case, the innermost loop is O(n). Out of these three loops, the first two contribute to O(nk) since for the j loop, although it could be as large as n, realistically it is bounded by k.

Thus, the overall time complexity of the second loop is O(n^2k) since for each of the O(nk) outer loop iterations, the innermost loop might run in O(n) time.

Combining both parts, the time complexity of the algorithm can be described as O(n^2 + n^2k), which simplifies to O(n^2k) due to the dominating term when k is not constant.

The final expression to represent the time complexity is:

O(n^2k)

Space Complexity

The space complexity is determined by the space used by the dynamic programming table f and the table g.

1. The space required for the table g is O(n^2) as it is a 2-dimensional square matrix of size n.

2. The space required for the table f is O(n(k+1)) since it has n rows and k+1 columns.

Therefore, the total space complexity is the sum of the two space requirements which is:

O(n^2) + O(nk)

However, the space complexity can be bounded by the larger term, which is O(n^2) when k is not constant.

The final expression to represent the space complexity is:

O(n^2)

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