132. Palindrome Partitioning II

Problem Description

The problem is centered around a given string s and the goal is to partition the string in such a way that each substring of the partition is a palindrome. A palindrome is defined as a string that reads the same forward and backward. The main objective here is to find the minimum number of cuts necessary to achieve this partitioning. A cut defines a division between two characters in the string, creating separate substrings that must all be palindromes.

For example, if s is "aab", a partition like "aa"|"b" is valid, as both "aa" and "b" are palindromes. The minimum number of cuts in this example is 1.

Intuition

The intuition behind the solution is to first precalculate which substrings are palindromes and then use dynamic programming to find the minimum number of cuts needed.

To determine if substrings are palindromes or not, we use a two-dimensional array g, where g[i][j] will be True if the substring from s[i] to s[j] (inclusive) is a palindrome. This is calculated in a bottom-up manner starting from the end of the string, using the fact that a substring s[i:j] is a palindrome if its outer letters are equal (s[i] == s[j]) and the substring s[i + 1:j - 1] is itself a palindrome.

Then we use a one-dimensional array f, where f[i] represents the minimum number of cuts needed to partition the string s up to index i. The relationship between g and f is crucial; we iterate over all possible end indices i of a palindrome and, for each, iterate over all possible start indices j. If g[j][i] is True (so s[j:i+1] is a palindrome), we know that one possible way to partition the string up to i is to cut right before j and then right after i. Therefore, f[i] can be updated to be the minimum between its current value and f[j-1] + 1 (because we make one cut before j). If j is zero, we do not need a cut, as it means the entire s[0:i+1] is a palindrome.

The base case for f is that f[i] is at most i, because the worst case is cutting the string at every character making each letter a palindrome substring. At the end of the process, f[-1] (last element of f) gives us the minimum number of cuts needed for the entire string s.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses dynamic programming and the concept of palindrome checking to solve the problem in an efficient manner. It involves two phases: precalculation of palindrome substrings and finding the minimum number of cuts using the precalculated data.

Precomputation of Palindrome Substrings

A two-dimensional table g is created to store whether a substring s[i...j] is a palindrome or not. It uses a bottom-up approach to fill this table. The base cases are that all single characters are palindromes (g[i][i] = True for all i), and for two character substrings, they are palindromes if both characters are the same (g[i][i+1] = s[i] == s[i+1]).

For substrings longer than two characters, g[i][j] is True if the first and last characters are the same (s[i] == s[j]) and the substring between them, s[i+1...j-1], is also a palindrome (g[i+1][j-1] == True). This is done by the following nested loop:

1for i in range(n - 1, -1, -1):
2    for j in range(i + 1, n):
3        g[i][j] = s[i] == s[j] and g[i + 1][j - 1]

Dynamic Programming for Minimum Cuts

The dynamic programming array f is initialized to represent the worst-case number of cuts for each index, which is when all characters up to that index are cut separately (f[i] = i). Then the algorithm iterates over each end index i and checks every start index j to see if a palindrome is formed (g[j][i] == True).

To update the minimum cuts for f[i], the code takes the minimum of the current value f[i] or f[j-1] + 1, the latter representing a cut at j producing one more cut than the minimum cuts needed for s[0...j-1]. The if j else 0 part handles the case where the entire substring s[0...i] is a palindrome, so no cut is needed, setting f[i] to 0 in that case.

1f = list(range(n))
2for i in range(1, n):
3    for j in range(i + 1):
4        if g[j][i]:
5            f[i] = min(f[i], 1 + (f[j - 1] if j else 0))

The final answer representing the minimum cuts needed for the whole string s is stored in f[-1].

By using a combination of a palindrome precomputation with dynamic programming, the proposed solution efficiently minimizes the cuts required to partition the string into palindrome substrings.

Example Walkthrough

Let's walk through an example of the solution approach with the input string s = "aab" as per the problem description.

Precomputation of Palindrome Substrings

First, we initialize the two-dimensional table g, which will store True or False values indicating whether the substring s[i...j] is a palindrome.

Given the string s = "aab", our table g starts off with the following base cases filled:

• Single characters are always palindromes: g[0][0] = True, g[1][1] = True, g[2][2] = True.
• For two adjacent characters: g[0][1] = True because s[0] == s[1] ("aa"), g[1][2] = False because s[1] != s[2] ("ab").

Now, we need to check substrings longer than two characters:

• Since our string length n = 3, there is no substring longer than two characters and we skip this step.

So, our precomputation step completes with the table g looking like this:

1g = [ [True, True, False],
2      [False, True, False],
3      [False, False, True] ]

Dynamic Programming for Minimum Cuts

We now initialize the dynamic programming array f with worst-case number of cuts for each index: f = [0, 1, 2].

Next, we iterate over each end index i and check every start index j to see if a palindrome is formed:

1. For i = 1:

   • j = 0: g[0][1] == True, so we check f[1] against 1 + f[j - 1] if j else 0, which translates to 1 + 0 = 1. Therefore, f[1] stays the same.
   • j = 1: We skip this since g[j][i] == g[1][1] == True and is a base case for single characters.

2. For i = 2:

   • j = 0: g[0][2] == False, we do not update f[2].
   • j = 1: g[1][2] == False, we do not update f[2].
   • j = 2: g[2][2] == True, we check f[2] against 1 + f[j - 1] if j else 0, which is 1 + 1 = 2. Therefore, we update f[2] with the minimum, which remains 2.

At the end of the process, f looks like this: f = [0, 1, 2].

The final answer is the minimum number of cuts needed to partition the string into palindrome substrings, and it is stored in f[-1], which is 2 in this case. However, we have overlooked one small detail: in the second iteration for i = 2, when j = 1, since g[1][2] is False, we do not update f[2]. Instead, we should look one character before index 1. Here is how it plays out:

• When j = 0, and since g[0][1] == True, we get f[2] could be updated to 1 + (f[j - 1] if j else 0) = 1 instead of the previous 2.

This correction gives us the final updated f = [0, 1, 1] and f[-1] = 1, which represents the minimum cuts needed for the entire string s.

Therefore, the minimum number of cuts required for "aab" to make every substring a palindrome is 1, with the cut after the first "aa" to isolate the "b".

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the solution is determined by two nested loops to fill the g matrix and two nested loops to compute the f array.

• Filling the g matrix requires a nested loop where i ranges from n - 1 to 0 and j ranges from i + 1 to n - 1. Each element g[i][j] is computed once, resulting in O(n^2) for this part as each cell of an n x n grid is computed through the loops.

• Computing the f array involves another nested loop where i ranges from 1 to n - 1 and for each i, j ranges from 0 to i. Inside this loop, a constant time operation is performed to update f[i]. The number of operations can be represented by the sum of the first n natural numbers minus one (since j starts from 0), which is (n * (n - 1)) / 2, also resulting in O(n^2) time.

Thus, the overall time complexity is O(n^2) + O(n^2) which simplifies to O(n^2).

Space Complexity

The space complexity of the solution is mainly determined by the space required to store the g matrix and the f array.

• The g matrix is an n x n boolean matrix where n is the length of the input string s, taking up O(n^2) space.

• The f array is a one-dimensional array of length n, taking up O(n) space.

So, the overall space complexity of the function is O(n^2) + O(n) which simplifies to O(n^2) as the n^2 term dominates for large n.

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