Fix sphinx/build_docs warnings for dynamic_programming

dynamic_programming/all_construct.py

@@ -8,9 +8,10 @@
 
 def all_construct(target: str, word_bank: list[str] | None = None) -> list[list[str]]:
     """
-        returns the list containing all the possible
-        combinations a string(target) can be constructed from
-        the given list of substrings(word_bank)
+    returns the list containing all the possible
+    combinations a string(`target`) can be constructed from
+    the given list of substrings(`word_bank`)
+
     >>> all_construct("hello", ["he", "l", "o"])
     [['he', 'l', 'l', 'o']]
     >>> all_construct("purple",["purp","p","ur","le","purpl"])

dynamic_programming/combination_sum_iv.py

@@ -1,24 +1,25 @@
 """
 Question:
-You are given an array of distinct integers and you have to tell how many
-different ways of selecting the elements from the array are there such that
-the sum of chosen elements is equal to the target number tar.
+    You are given an array of distinct integers and you have to tell how many
+    different ways of selecting the elements from the array are there such that
+    the sum of chosen elements is equal to the target number tar.
 
 Example
 
 Input:
-N = 3
-target = 5
-array = [1, 2, 5]
+    * N = 3
+    * target = 5
+    * array = [1, 2, 5]
 
 Output:
-9
+    9
 
 Approach:
-The basic idea is to go over recursively to find the way such that the sum
-of chosen elements is “tar”. For every element, we have two choices
-    1. Include the element in our set of chosen elements.
-    2. Don't include the element in our set of chosen elements.
+    The basic idea is to go over recursively to find the way such that the sum
+    of chosen elements is `target`. For every element, we have two choices
+
+        1. Include the element in our set of chosen elements.
+        2. Don't include the element in our set of chosen elements.
 """
 
 

dynamic_programming/fizz_buzz.py

@@ -3,11 +3,12 @@
 
 def fizz_buzz(number: int, iterations: int) -> str:
     """
-    Plays FizzBuzz.
-    Prints Fizz if number is a multiple of 3.
-    Prints Buzz if its a multiple of 5.
-    Prints FizzBuzz if its a multiple of both 3 and 5 or 15.
-    Else Prints The Number Itself.
+    | Plays FizzBuzz.
+    | Prints Fizz if number is a multiple of ``3``.
+    | Prints Buzz if its a multiple of ``5``.
+    | Prints FizzBuzz if its a multiple of both ``3`` and ``5`` or ``15``.
+    | Else Prints The Number Itself.
+
     >>> fizz_buzz(1,7)
     '1 2 Fizz 4 Buzz Fizz 7 '
     >>> fizz_buzz(1,0)

dynamic_programming/knapsack.py

@@ -11,7 +11,7 @@ def mf_knapsack(i, wt, val, j):
     """
     This code involves the concept of memory functions. Here we solve the subproblems
     which are needed unlike the below example
-    F is a 2D array with -1s filled up
+    F is a 2D array with ``-1`` s filled up
     """
     global f  # a global dp table for knapsack
     if f[i][j] < 0:
@@ -45,22 +45,24 @@ def knapsack_with_example_solution(w: int, wt: list, val: list):
     the several possible optimal subsets.
 
     Parameters
-    ---------
+    ----------
 
-    W: int, the total maximum weight for the given knapsack problem.
-    wt: list, the vector of weights for all items where wt[i] is the weight
-    of the i-th item.
-    val: list, the vector of values for all items where val[i] is the value
-    of the i-th item
+    * `w`: int, the total maximum weight for the given knapsack problem.
+    * `wt`: list, the vector of weights for all items where ``wt[i]`` is the weight
+       of the ``i``-th item.
+    * `val`: list, the vector of values for all items where ``val[i]`` is the value
+      of the ``i``-th item
 
     Returns
     -------
-    optimal_val: float, the optimal value for the given knapsack problem
-    example_optional_set: set, the indices of one of the optimal subsets
-    which gave rise to the optimal value.
+
+    * `optimal_val`: float, the optimal value for the given knapsack problem
+    * `example_optional_set`: set, the indices of one of the optimal subsets
+      which gave rise to the optimal value.
 
     Examples
-    -------
+    --------
+
     >>> knapsack_with_example_solution(10, [1, 3, 5, 2], [10, 20, 100, 22])
     (142, {2, 3, 4})
     >>> knapsack_with_example_solution(6, [4, 3, 2, 3], [3, 2, 4, 4])
@@ -104,19 +106,19 @@ def _construct_solution(dp: list, wt: list, i: int, j: int, optimal_set: set):
     a filled DP table and the vector of weights
 
     Parameters
-    ---------
-
-    dp: list of list, the table of a solved integer weight dynamic programming problem
+    ----------
 
-    wt: list or tuple, the vector of weights of the items
-    i: int, the index of the item under consideration
-    j: int, the current possible maximum weight
-    optimal_set: set, the optimal subset so far. This gets modified by the function.
+    * `dp`: list of list, the table of a solved integer weight dynamic programming
+      problem
+    * `wt`: list or tuple, the vector of weights of the items
+    * `i`: int, the index of the item under consideration
+    * `j`: int, the current possible maximum weight
+    * `optimal_set`: set, the optimal subset so far. This gets modified by the function.
 
     Returns
     -------
-    None
 
+    ``None``
     """
     # for the current item i at a maximum weight j to be part of an optimal subset,
     # the optimal value at (i, j) must be greater than the optimal value at (i-1, j).

dynamic_programming/longest_common_substring.py

@@ -1,15 +1,19 @@
 """
-Longest Common Substring Problem Statement: Given two sequences, find the
-longest common substring present in both of them. A substring is
-necessarily continuous.
-Example: "abcdef" and "xabded" have two longest common substrings, "ab" or "de".
-Therefore, algorithm should return any one of them.
+Longest Common Substring Problem Statement:
+    Given two sequences, find the
+    longest common substring present in both of them. A substring is
+    necessarily continuous.
+
+Example:
+    ``abcdef`` and ``xabded`` have two longest common substrings, ``ab`` or ``de``.
+    Therefore, algorithm should return any one of them.
 """
 
 
 def longest_common_substring(text1: str, text2: str) -> str:
     """
     Finds the longest common substring between two strings.
+
     >>> longest_common_substring("", "")
     ''
     >>> longest_common_substring("a","")

dynamic_programming/longest_increasing_subsequence.py

@@ -4,11 +4,13 @@
 This is a pure Python implementation of Dynamic Programming solution to the longest
 increasing subsequence of a given sequence.
 
-The problem is  :
-Given an array, to find the longest and increasing sub-array in that given array and
-return it.
-Example: [10, 22, 9, 33, 21, 50, 41, 60, 80] as input will return
-         [10, 22, 33, 41, 60, 80] as output
+The problem is:
+    Given an array, to find the longest and increasing sub-array in that given array and
+    return it.
+
+Example:
+    ``[10, 22, 9, 33, 21, 50, 41, 60, 80]`` as input will return
+    ``[10, 22, 33, 41, 60, 80]`` as output
 """
 
 from __future__ import annotations
@@ -17,6 +19,7 @@
 def longest_subsequence(array: list[int]) -> list[int]:  # This function is recursive
     """
     Some examples
+
     >>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
     [10, 22, 33, 41, 60, 80]
     >>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])

dynamic_programming/matrix_chain_multiplication.py

@@ -1,42 +1,48 @@
 """
-Find the minimum number of multiplications needed to multiply chain of matrices.
-Reference: https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/
+| Find the minimum number of multiplications needed to multiply chain of matrices.
+| Reference: https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/
 
-The algorithm has interesting real-world applications. Example:
-1. Image transformations in Computer Graphics as images are composed of matrix.
-2. Solve complex polynomial equations in the field of algebra using least processing
-   power.
-3. Calculate overall impact of macroeconomic decisions as economic equations involve a
-   number of variables.
-4. Self-driving car navigation can be made more accurate as matrix multiplication can
-   accurately determine position and orientation of obstacles in short time.
+The algorithm has interesting real-world applications.
 
-Python doctests can be run with the following command:
-python -m doctest -v matrix_chain_multiply.py
+Example:
+  1. Image transformations in Computer Graphics as images are composed of matrix.
+  2. Solve complex polynomial equations in the field of algebra using least processing
+     power.
+  3. Calculate overall impact of macroeconomic decisions as economic equations involve a
+     number of variables.
+  4. Self-driving car navigation can be made more accurate as matrix multiplication can
+     accurately determine position and orientation of obstacles in short time.
 
-Given a sequence arr[] that represents chain of 2D matrices such that the dimension of
-the ith matrix is arr[i-1]*arr[i].
-So suppose arr = [40, 20, 30, 10, 30] means we have 4 matrices of dimensions
-40*20, 20*30, 30*10 and 10*30.
+Python doctests can be run with the following command::
 
-matrix_chain_multiply() returns an integer denoting minimum number of multiplications to
-multiply the chain.
+  python -m doctest -v matrix_chain_multiply.py
+
+Given a sequence ``arr[]`` that represents chain of 2D matrices such that the dimension
+of the ``i`` th matrix is ``arr[i-1]*arr[i]``.
+So suppose ``arr = [40, 20, 30, 10, 30]`` means we have ``4`` matrices of dimensions
+``40*20``, ``20*30``, ``30*10`` and ``10*30``.
+
+``matrix_chain_multiply()`` returns an integer denoting minimum number of
+multiplications to multiply the chain.
 
 We do not need to perform actual multiplication here.
 We only need to decide the order in which to perform the multiplication.
 
 Hints:
-1. Number of multiplications (ie cost) to multiply 2 matrices
-of size m*p and p*n is m*p*n.
-2. Cost of matrix multiplication is associative ie (M1*M2)*M3 != M1*(M2*M3)
-3. Matrix multiplication is not commutative. So, M1*M2 does not mean M2*M1 can be done.
-4. To determine the required order, we can try different combinations.
+  1. Number of multiplications (ie cost) to multiply ``2`` matrices
+     of size ``m*p`` and ``p*n`` is ``m*p*n``.
+  2. Cost of matrix multiplication is not associative ie ``(M1*M2)*M3 != M1*(M2*M3)``
+  3. Matrix multiplication is not commutative. So, ``M1*M2`` does not mean ``M2*M1``
+     can be done.
+  4. To determine the required order, we can try different combinations.
+
 So, this problem has overlapping sub-problems and can be solved using recursion.
 We use Dynamic Programming for optimal time complexity.
 
 Example input:
-arr = [40, 20, 30, 10, 30]
-output: 26000
+    ``arr = [40, 20, 30, 10, 30]``
+output:
+    ``26000``
 """
 
 from collections.abc import Iterator
@@ -50,25 +56,25 @@ def matrix_chain_multiply(arr: list[int]) -> int:
     Find the minimum number of multiplcations required to multiply the chain of matrices
 
     Args:
-        arr: The input array of integers.
+        `arr`: The input array of integers.
 
     Returns:
         Minimum number of multiplications needed to multiply the chain
 
     Examples:
-        >>> matrix_chain_multiply([1, 2, 3, 4, 3])
-        30
-        >>> matrix_chain_multiply([10])
-        0
-        >>> matrix_chain_multiply([10, 20])
-        0
-        >>> matrix_chain_multiply([19, 2, 19])
-        722
-        >>> matrix_chain_multiply(list(range(1, 100)))
-        323398
-
-        # >>> matrix_chain_multiply(list(range(1, 251)))
-        # 2626798
+
+    >>> matrix_chain_multiply([1, 2, 3, 4, 3])
+    30
+    >>> matrix_chain_multiply([10])
+    0
+    >>> matrix_chain_multiply([10, 20])
+    0
+    >>> matrix_chain_multiply([19, 2, 19])
+    722
+    >>> matrix_chain_multiply(list(range(1, 100)))
+    323398
+    >>> # matrix_chain_multiply(list(range(1, 251)))
+    # 2626798
     """
     if len(arr) < 2:
         return 0
@@ -93,8 +99,10 @@ def matrix_chain_multiply(arr: list[int]) -> int:
 def matrix_chain_order(dims: list[int]) -> int:
     """
     Source: https://en.wikipedia.org/wiki/Matrix_chain_multiplication
+
     The dynamic programming solution is faster than cached the recursive solution and
     can handle larger inputs.
+
     >>> matrix_chain_order([1, 2, 3, 4, 3])
     30
     >>> matrix_chain_order([10])
@@ -105,8 +113,7 @@ def matrix_chain_order(dims: list[int]) -> int:
     722
     >>> matrix_chain_order(list(range(1, 100)))
     323398
-
-    # >>> matrix_chain_order(list(range(1, 251)))  # Max before RecursionError is raised
+    >>> # matrix_chain_order(list(range(1, 251)))  # Max before RecursionError is raised
     # 2626798
     """
 

dynamic_programming/max_product_subarray.py

@@ -1,9 +1,10 @@
 def max_product_subarray(numbers: list[int]) -> int:
     """
     Returns the maximum product that can be obtained by multiplying a
-    contiguous subarray of the given integer list `nums`.
+    contiguous subarray of the given integer list `numbers`.
 
     Example:
+
     >>> max_product_subarray([2, 3, -2, 4])
     6
     >>> max_product_subarray((-2, 0, -1))

dynamic_programming/minimum_squares_to_represent_a_number.py

@@ -5,6 +5,7 @@
 def minimum_squares_to_represent_a_number(number: int) -> int:
     """
     Count the number of minimum squares to represent a number
+
     >>> minimum_squares_to_represent_a_number(25)
     1
     >>> minimum_squares_to_represent_a_number(37)

dynamic_programming/regex_match.py

@@ -1,23 +1,25 @@
 """
 Regex matching check if a text matches pattern or not.
 Pattern:
-    '.' Matches any single character.
-    '*' Matches zero or more of the preceding element.
+
+    1. ``.`` Matches any single character.
+    2. ``*`` Matches zero or more of the preceding element.
+
 More info:
     https://medium.com/trick-the-interviwer/regular-expression-matching-9972eb74c03
 """
 
 
 def recursive_match(text: str, pattern: str) -> bool:
-    """
+    r"""
     Recursive matching algorithm.
 
-    Time complexity: O(2 ^ (|text| + |pattern|))
-    Space complexity: Recursion depth is O(|text| + |pattern|).
+    | Time complexity: O(2^(\|text\| + \|pattern\|))
+    | Space complexity: Recursion depth is O(\|text\| + \|pattern\|).
 
     :param text: Text to match.
     :param pattern: Pattern to match.
-    :return: True if text matches pattern, False otherwise.
+    :return: ``True`` if `text` matches `pattern`, ``False`` otherwise.
 
     >>> recursive_match('abc', 'a.c')
     True
@@ -48,15 +50,15 @@ def recursive_match(text: str, pattern: str) -> bool:
 
 
 def dp_match(text: str, pattern: str) -> bool:
-    """
+    r"""
     Dynamic programming matching algorithm.
 
-    Time complexity: O(|text| * |pattern|)
-    Space complexity: O(|text| * |pattern|)
+    | Time complexity: O(\|text\| * \|pattern\|)
+    | Space complexity: O(\|text\| * \|pattern\|)
 
     :param text: Text to match.
     :param pattern: Pattern to match.
-    :return: True if text matches pattern, False otherwise.
+    :return: ``True`` if `text` matches `pattern`, ``False`` otherwise.
 
     >>> dp_match('abc', 'a.c')
     True