Palindrome Partitioning II

The problem of finding the minimum cuts for a palindrome partitioning of a string can be solved using dynamic programming. We can break down the problem into subproblems and use a bottom-up approach to find the final solution.

Here’s a step-by-step explanation of the solution:

1. Create a 2D DP Array for Palindromes: Create a boolean 2D array is_palindrome[i][j], where is_palindrome[i][j] is True if the substring s[i:j+1] is a palindrome. Otherwise, it’s False.

2. Create a 1D DP Array for Minimum Cuts: Create an integer array cuts[i] to hold the minimum number of cuts required to partition the substring s[0:i+1] into palindromes.

3. Iteratively Calculate Palindromes: Iterate through the string and use two pointers to check if substrings are palindromes, updating the is_palindrome array as needed.

4. Calculate Minimum Cuts: For each position i in the string, iterate through all the positions j before i and check if the substring s[j+1:i+1] is a palindrome. If it is, update cuts[i] based on cuts[j].

5. Return the Final Result: The minimum number of cuts for the whole string is stored in cuts[-1].

Here’s the code:

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class Solution:
    def minCut(self, s: str) -> int:
        n = len(s)

        # Initialize is_palindrome table
        is_palindrome = [[False] * n for _ in range(n)]
        for i in range(n):
            is_palindrome[i][i] = True

        # Fill in is_palindrome table
        for length in range(2, n + 1):
            for i in range(n - length + 1):
                j = i + length - 1
                if s[i] == s[j]:
                    if length == 2 or is_palindrome[i + 1][j - 1]:
                        is_palindrome[i][j] = True

        # Initialize cuts array
        cuts = [float('inf')] * n
        for i in range(n):
            if is_palindrome[0][i]:
                cuts[i] = 0
            else:
                for j in range(i):
                    if is_palindrome[j + 1][i]:
                        cuts[i] = min(cuts[i], cuts[j] + 1)

        return cuts[-1]

The is_palindrome table helps us quickly check if any substring is a palindrome. Then we use it to determine the minimum cuts required for each position in the string, leading to the final result. The time complexity of this approach is (O(n^2)), and the space complexity is also (O(n^2)).

10 Prerequisite LeetCode Problems

“132. Palindrome Partitioning II” falls under dynamic programming and string manipulation. The task involves finding the minimum number of cuts needed for a palindrome partitioning of a given string, and it requires knowledge of how to manage state transitions and solve subproblems.

Here are 10 problems to prepare for this problem:

1. Problem 5. Longest Palindromic Substring: This problem provides foundational knowledge of how to detect palindromes within a string.

2. Problem 647. Palindromic Substrings: This problem further drills the concept of finding all palindromic substrings in a string.

3. Problem 516. Longest Palindromic Subsequence: This problem introduces the concept of finding palindromic subsequences, which will be helpful for understanding palindromic partitions.

4. Problem 131. Palindrome Partitioning: This problem is a precursor to the one in question, dealing with finding all possible palindrome partitioning of a string but without the minimum cuts requirement.

5. Problem 139. Word Break: This problem deals with partitioning a string into valid words. It will give you an understanding of how to divide a string into meaningful parts.

6. Problem 322. Coin Change: Though not related to string manipulation, this problem’s concept of finding the minimum number of coins required to make a certain amount can be translated to finding the minimum cuts required in the Palindrome Partitioning II problem.

7. Problem 279. Perfect Squares: This problem requires finding the least number of perfect square numbers that sum to a given number, similar to the minimum cuts in the Palindrome Partitioning II problem.

8. Problem 91. Decode Ways: The ways of decoding a message can be seen as partitions of a string, similar to the Palindrome Partitioning II problem, albeit without the palindrome requirement.

9. Problem 72. Edit Distance: The problem introduces dynamic programming with string editing. Understanding the state transition in this problem helps in similar problems like Palindrome Partitioning II.

10. Problem 120. Triangle: This problem is a basic dynamic programming problem which can help understand the idea of how to break a problem down into subproblems and build up the answer.

These cover dynamic programming, string manipulation, and state transition, which are necessary for solving the Palindrome Partitioning II problem. Starting from understanding how to detect palindromes, these problems gradually increase in complexity, eventually preparing you for the challenge of finding the minimum cuts needed for a palindrome partitioning.

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class Solution(object):
    def minCut(self, s):
        """
        s="abcgcbafj"
        :type s: str
        :rtype: int
        """

        if s == s[::-1]:
            return 0

        for i in range(len(s)):
            if s[:i] == s[:i][::-1] and s[i:] == s[i:][::-1]:
                return 1

        l=len(s)

        x=[[0 for i  in range(l)] for j in range(l)]
        for i in range(l):
          for j in range(i,l):
            st=s[i:j+1]
            x[i][j]= (st==st[::-1])


        p=[0 for c in range(l)]
        for i in range(1,l):
          if x[0][i]:
            p[i]=0
          else:
            m=float("inf")
            for j in range(i,0,-1):
              if x[j][i]: 
                m=min(m,p[j-1])
            p[i]=m+1
        return p[-1]

Problem Classification

The problem asks for the minimum number of cuts needed to partition a text into palindromes. This is about breaking a text into parts with a certain property, so it’s a Text Partitioning Problem.

Language Agnostic Coding Drills

There are several key concepts that must be understood in order to fully grasp the solution. Each of these can be studied individually as a coding drill, which can then be combined into the final solution. Here are the drills in increasing order of difficulty:

1. String Operations: Learn how to create, access, and manipulate strings. Practice iterating through a string, accessing characters in a string, and reversing a string.

   Drill: Create a string and perform operations such as accessing individual characters, slicing substrings, and reversing the string.

2. If-Else Statements and Loops: Understand how to use if-else statements and for loops. Practice creating conditional statements and loops that iterate over a range of numbers or through a string.

   Drill: Write a program that iterates over a string and performs a certain operation (e.g. counting a specific character) based on a conditional statement.

3. List Comprehensions: Learn about list comprehensions, a concise way to create lists based on existing lists.

   Drill: Given a list of numbers, create a new list that contains the squares of all the numbers using a list comprehension.

4. 2D Lists (or Matrices): Understand how to create, access, and manipulate 2D lists. Practice creating a 2D list with given dimensions, accessing elements, and updating elements.

   Drill: Create a 2D list with a certain dimension, access specific elements, and update elements.

5. Dynamic Programming: Understand how to solve complex problems by breaking them down into simpler subproblems. Learn about memoization and how to fill up a DP table.

   Drill: Solve classical DP problems such as the Fibonacci sequence or the Knapsack problem.

Here is how the final problem is solved:

The problem is asking to partition a string into as few parts as possible so that each part is a palindrome. The solution utilizes dynamic programming to solve the problem.

1. The function first checks if the entire string is a palindrome. If it is, then the function returns 0 because no cuts are necessary.

2. Then the function checks if it can cut the string into two palindromes. If it can, then the function returns 1 because only one cut is necessary.

3. If the above two conditions are not met, then the function builds a 2D boolean DP table, x, where x[i][j] is True if the substring s[i:j+1] is a palindrome.

4. The function then builds a list p where p[i] is the minimum number of cuts required to partition the substring s[0:i+1] into palindromes. It iteratively checks each possible partition point and keeps track of the minimum number of cuts needed.

5. Finally, the function returns p[-1], the minimum number of cuts needed for the entire string.

Targeted Drills in Python

1. String Operations:

   Drill: Create a string and perform operations such as accessing individual characters, slicing substrings, and reversing the string.

   1 2 3 4 5

   s = "Hello, World!" print(s[0]) # Accessing the first character print(s[-1]) # Accessing the last character print(s[7:12]) # Slicing a substring print(s[::-1]) # Reversing the string

2. If-Else Statements and Loops:

   Drill: Write a program that iterates over a string and performs a certain operation (e.g. counting a specific character) based on a conditional statement.

   1 2 3 4 5 6

   s = "Hello, World!" count = 0 for char in s: if char.lower() == 'o': count += 1 print(count)

3. List Comprehensions:

   Drill: Given a list of numbers, create a new list that contains the squares of all the numbers using a list comprehension.

   1 2 3

   numbers = [1, 2, 3, 4, 5] squares = [num ** 2 for num in numbers] print(squares)

4. 2D Lists (or Matrices):

   Drill: Create a 2D list with a certain dimension, access specific elements, and update elements.

   1 2 3 4

   matrix = [[0 for _ in range(5)] for _ in range(5)] # Create a 5x5 2D list print(matrix[0][0]) # Access the first element matrix[0][0] = 1 # Update the first element print(matrix)

5. Dynamic Programming:

   Drill: Solve classical DP problems such as the Fibonacci sequence or the Knapsack problem.

   Fibonacci Sequence:

   1 2 3 4 5 6 7

   def fibonacci(n): dp = [0, 1] + [0] * (n - 1) for i in range(2, n + 1): dp[i] = dp[i - 1] + dp[i - 2] return dp[n] print(fibonacci(10))

Each of these drills addresses a concept that is critical for understanding and implementing the provided solution. To create a solution for the provided problem, you would need to apply and combine these individual concepts.

title: Palindrome Partitioning tags: palindrome partitioning two-pointers-moving-towards-each-other

Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. A palindrome string is a string that reads the same backward as forward.

1. Define the Interface

Input: string Output: Array of arrays (each array consisting of palindrome)

2. Define the Terms

Partition: A palindrome is a word or a phrase that reads the same forwards and backwards. Example: ‘racecar’

Palindrome: A palindrome is a word or a phrase that reads the same forwards and backwards – ‘racecar’, ‘Step on no pets’.

3. Problem Decomposition

• How do we reduce the size of the problem?
• Should we reduce the input size by one, two, half, by some constant factor or different varying size?

4. Consider the Brute Force approach if you cannot think of anything else.

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def palindrome?(string, low, high)
  while low < high
    if string[low] != string[high]
      return false
    end
    
    low += 1
    high -= 1
  end

  return true
end

def wrapper(start, result, solution, s)
  if start >= s.size
    result << solution.dup
  end
    
  for stop in (start..s.size-1)
    if palindrome?(s, start, stop)
      solution << s[start..stop]
      wrapper(stop+1, result, solution, s)
      solution.pop
    end
  end
end

def partition(s)
  result = []
  solution = []
  start = 0

  wrapper(start, result, solution, s)
  result
end

Applying Recursion

To apply recursion to a problem you have to:

• Find a sub-problem which is self-similar to the large problem
• Make the size of the problem a parameter to your recursive function
• Solve the base case of the problem and explicitly write the answer into recursive method
• Assume that recursive_method(size-1) returns the correct answer to a sub-problem of size-1.
• Use the output of recursive_method(size-1) to write the solution to problem recursive_method(size)

Example:

Problem: factorial(n) Subproblem: factorial(n-1) Takes a smaller integer as an argument

Problem: Counting paths through some network Subproblem: Count paths through a smaller network

Problem: Palindrome Subproblem: Parameters to control the size of the problem: start and stop pointers

Each recursive call should be on a smaller instance of the same problem, that is, a smaller subproblem.The recursive calls must eventually reach a base case, which is solved without further recursion.

Is the number of subproblems the same as the recursive calls made during the runtime? No. They are different.

Viewing the Program Execution as a Series of State Transitions

• Initial State
• Intermediate State
• Final State

We make a series of state transitions from the beginning to the end of the program execution.

The additional parameter in the wrapper method fills a similar role to that of a local variable in an iterative implementation. It holds an intermediate state during the execution of the program.

An optimization problem is a problem of finding the best solution from all feasible solutions. And combinatorial problems expect you to figure out the number of ways to do something or the probability of some event happening.

The normal dfs backtracking will need to check each substring for palindrome, but a dp array can be used to record the possible break for palindrome before we start recursion.

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def palindrome?(string, low, high)
  while low < high
    if string[low] != string[high]
      return false
    end
    low += 1
    high -= 1
  end
	
  return true
end

def wrapper(start, result, solution, s)
  if start >= s.size
    result << solution.dup
  end
    
  for stop in (start..s.size-1)
    if palindrome?(s, start, stop)
      solution << s[start..stop]
      wrapper(stop+1, result, solution, s)
      solution.pop
    end
  end
end
# @param {String} s
# @return {String[][]}
def partition(s)
  result = []
  solution = []
  start = 0
  wrapper(start, result, solution, s)
  result
end

s = "aab"

p partition(s)

All backtracking problems are composed by three steps: choose, explore, unchoose. So for each problem, you need to know:

• Choose what? For this problem, we choose each substring.
• How to explore? For this problem, we do the same thing to the remaining string.
• Unchoose: Do the opposite operation of choose.

Let’s take this problem as an example:

1. Define wrapper(): Usually we need a wrapper function in the backtracking problem, to accept more parameters.
2. Parameters: Usually we need the following parameters
   • The object you are working on: For this problem the String s.
3. A start index or an end index which indicate which part you are working on: For this problem, we use substring to indicate the start index.
4. A step result, to remember the current choice and then do unchoose: for this problem, we use an array.
5. A final result, to remember the final result. Usually when we add, we use ‘result.add(partial.dup)’ instead of ‘result.add(partial)’, since partial is passed by reference. We will modify partial later, so we need to copy it and add the copy to the result.
6. Base case: The base case defines when to add the partial into result and when to return.
7. Use for-loop: Usually we need a for loop to iterate through the input strings, so that we can choose all the options.
8. Choose: In this problem, if the substring of s is palindrome, we add it into the partial, which means we choose this substring.
9. Explore: In this problem, we want to do the same thing to the remaining substring. So we recursively call our function.
10. Un-Choose: After the recursive call, remove the chosen substring, in order to try other options.

Time and Space Complexity

The copying of each palindrome takes O(N) time and the decision tree will result in exponential time. So the total runtime is O(N * 2N).

Space: O(N). The space used for storing all palindromes is N.

Variation

Complete the generate_palindromic_decompositions function below. The palindromes are separated by |.

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def palindrome?(string, low, high)
  while low < high
    if string[low] != string[high]
      return false
    end
    
    low += 1
    high -= 1
  end

  return true
end

def wrapper(start, result, solution, s)
  if start >= s.size
    array = solution.dup
    result << array.join('|')
  end
    
  for stop in (start..s.size-1)
    if palindrome?(s, start, stop)
      solution << s[start..stop]
      wrapper(stop+1, result, solution, s)
      solution.pop
    end
  end
end

def generate_palindromic_decompositions(s)
  result = []
  solution = []
  start = 0

  wrapper(start, result, solution, s)
  result
end