Methods to solve the problem of constructing backstory strings using Python

Problem definition

The problem of constructing back text strings can be defined into the following two problems:

The longest palindromic subsequence problem: Given a string, find out the length of the longest palindromic subsequence. Pastoral subsequence refers to pastoral strings formed after deleting some characters (or not deleting) from the original string.
Minimum number of deletion issues: Given a string, calculate the minimum number of deletions required to convert it into a palindromic string.

These two questions are actually equivalent. Because the length of the longest palindrome subsequence is equal to the original string length minus the minimum number of deletions. Therefore, we only need to solve one of the problems and we can easily get the answer to the other.

Algorithm selection

Dynamic programming (DP) is an efficient and commonly used algorithm for constructing backstory string problems. Dynamic programming significantly improves algorithm efficiency by decomposing problems into subproblems and storing solutions to subproblems to avoid repeated calculations.

When solving the problem of longest palindromic subsequence, we can define a two-dimensional array dp, where dp[i][j] represents the length of the longest palindromic subsequence of the string from index i to j. By filling this two-dimensional array, we can gradually solve the longest palindromic subsequence length of the entire string.

Python implementation

Next, we will use Python to implement a dynamic programming algorithm to solve the problem of longest palindromic subsequence.

1. Define the problem

Suppose we have a string s and we need to find the length of the longest palindromic subsequence.

2. Dynamic programming status definition

We define a two-dimensional array dp where dp[i][j] represents the length of the longest palindromic subsequence of the string s from index i to j.

3. State transfer equation

According to the properties of palindromic strings, we can obtain the following state transfer equation:

If s[i] == s[j], then dp[i][j] = dp[i+1][j-1] + 2. Because s[i] and s[j] can form both ends of palindrome, the length of the longest palindrome subsequence is equal to the length of the longest palindrome subsequence from s[i+1] to s[j-1] plus 2.
If s[i] != s[j], then dp[i][j] = max(dp[i+1][j], dp[i][j-1]). Because s[i] and s[j] cannot appear in palindrome at the same time, the length of the longest palindrome subsequence is equal to the larger value of the length of the longest palindrome subsequence from s[i+1] to s[j] and s[i] to s[j-1].

4. Initialization

For all cases of i>j, dp[i][j] = 0, because the substring does not exist. For all cases where i == j, dp[i][j] = 1, because a single character is itself a palindrome.

5. Fill order

We need to fill the dp array from small to large according to the length of the substring. Because the value of dp[i][j] depends on dp[i+1][j-1], dp[i+1][j] and dp[i][j-1], we should fill it in the order of rows or columns.

6. Python code implementation

def longest_palindrome_subsequence(s):
    n = len(s)
    # Initialize dp array    dp = [[0] * n for _ in range(n)]
    
    #Fill dp array    for i in range(n-1, -1, -1):
        dp[i][i] = 1  # Single character is a palindrome        for j in range(i+1, n):
            if s[i] == s[j]:
                dp[i][j] = dp[i+1][j-1] + 2
            else:
                dp[i][j] = max(dp[i+1][j], dp[i][j-1])
    
    return dp[0][n-1]

7. Call the algorithm and output the result

s = "bbbab"
length = longest_palindrome_subsequence(s)
print(f"String'{s}'The longest palindromic subsequence length is: {length}")

Run the above code and the output is:

The longest palindromic subsequence length of the string 'bbbab' is: 4
Because "bbbb" is a palindal subsequence of "bbbab" and its length is 4.

Algorithm optimization

Although dynamic programming algorithms have been able to efficiently solve the problem of constructing backstory strings, in practical applications, we may need to optimize the algorithm to improve performance. Here are some possible optimization methods:

1. Space optimization

In the dynamic programming algorithm, we use the two-dimensional array dp to store the solutions to the subproblem. However, we can find that when filling the dp array, we only need the data of the current row and the previous row. Therefore, we can optimize the two-dimensional array into one-dimensional array, thereby reducing the spatial complexity from O(n^2) to O(n).

2. Scrolling array optimization

Scrolling array optimization is a commonly used spatial optimization method. For dynamic programming problems, if we only need the data of the current row and the previous row, we can use two one-dimensional arrays to store the data alternately, thereby reducing the spatial complexity to O(n).

3. Central extension method

For the problem of constructing backstory strings, we can also use the central extension method to solve it. The basic idea of the central extension method is to start between each character and every two characters and expand to both sides until the palindrome cannot be formed. This method has a time complexity of O(n^2), the same as the dynamic programming algorithm, but it may be easier to implement.

Summarize

This article details how to use Python and dynamic programming algorithms to solve the problem of constructing backstory strings. Dynamic programming algorithms significantly improve algorithm efficiency by decomposing problems into subproblems and storing solutions to subproblems to avoid repeated calculations. Through the study of this article, readers can master the basic principles and implementation methods of dynamic programming algorithms, and can apply them to solve various constructed backstory string problems. In practical applications, we can also optimize and improve the algorithm according to specific needs to improve performance and efficiency.

Expansion: How to judge palindrome using Python

1. Basic method: double pointer method

The double pointer method is one of the most intuitive methods. We can use two pointers, one starts at the beginning of the string and the other starts at the end, gradually moving to the middle and comparing the corresponding characters.

Sample code：

def is_palindrome(s):
    # Remove all non-alphanumeric characters and convert them to lowercase    cleaned = ''.join(() for c in s if ())
    
    left, right = 0, len(cleaned) - 1
    
    while left &lt; right:
        if cleaned[left] != cleaned[right]:
            return False
        left += 1
        right -= 1
    return True
 
# testprint(is_palindrome("A man, a plan, a canal: Panama"))  # Output: Trueprint(is_palindrome("race a car"))  # Output: False

2. Concise method: string inversion method

Python provides a very concise way to reverse strings. We can tell whether it is a palindrome by inverting the string and comparing it to the original string.

Sample code：

def is_palindrome(s):
    # Remove all non-alphanumeric characters and convert them to lowercase    cleaned = ''.join(() for c in s if ())
    
    # Comparison of original string with inverted string    return cleaned == cleaned[::-1]
 
# testprint(is_palindrome("A man, a plan, a canal: Panama"))  # Output: Trueprint(is_palindrome("race a car"))  # Output: False

3. Use built-in functions all and generator expressions

We can use Python'sallFunctions and generator expressions to simplify the code. This method can also efficiently judge palindrome.

Sample code：

def is_palindrome(s):
    # Remove all non-alphanumeric characters and convert them to lowercase    cleaned = ''.join(() for c in s if ())
    
    # Use all functions to compare with generator expressions    return all(cleaned[i] == cleaned[~i] for i in range(len(cleaned) // 2))
 
# testprint(is_palindrome("A man, a plan, a canal: Panama"))  # Output: Trueprint(is_palindrome("race a car"))  # Output: False

4. Regular expression methods that ignore case and non-alphanumeric characters

If stricter handling is required, such as ignoring case and non-alphanumeric characters, you can use regular expressions to clean up the input string.

Sample code：

import re
 
def is_palindrome(s):
    # Use regular expression to remove all non-alphanumeric characters and convert them to lowercase    cleaned = (r'[^A-Za-z0-9]', '', s).lower()
    
    # Comparison of original string with inverted string    return cleaned == cleaned[::-1]
 
# testprint(is_palindrome("A man, a plan, a canal: Panama"))  # Output: Trueprint(is_palindrome("race a car"))  # Output: False

5. Recursive method

While not the most efficient, the recursive approach provides an elegant way to solve the problem. We can recursively check whether the first and last characters of the string are the same, and then repeat the process for the substring.

Sample code：

def is_palindrome_recursive(s):
    # Basic situation: an empty string or a single character is a palindrome    if len(s) &lt;= 1:
        return True
    
    # Remove all non-alphanumeric characters and convert them to lowercase    cleaned = ''.join(() for c in s if ())
    
    # Recursively check the first and last characters    if not cleaned or len(cleaned) == 1:
        return True
    elif cleaned[0] != cleaned[-1]:
        return False
    else:
        return is_palindrome_recursive(cleaned[1:-1])
 
# testprint(is_palindrome_recursive("A man, a plan, a canal: Panama"))  # Output: Trueprint(is_palindrome_recursive("race a car"))  # Output: False

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