[pull] master from TheAlgorithms:master

backtracking/word_break.py

@@ -0,0 +1,71 @@
+"""
+Word Break Problem is a well-known problem in computer science.
+Given a string and a dictionary of words, the task is to determine if
+the string can be segmented into a sequence of one or more dictionary words.
+
+Wikipedia: https://en.wikipedia.org/wiki/Word_break_problem
+"""
+
+
+def backtrack(input_string: str, word_dict: set[str], start: int) -> bool:
+    """
+    Helper function that uses backtracking to determine if a valid
+    word segmentation is possible starting from index 'start'.
+
+    Parameters:
+    input_string (str): The input string to be segmented.
+    word_dict (set[str]): A set of valid dictionary words.
+    start (int): The starting index of the substring to be checked.
+
+    Returns:
+    bool: True if a valid segmentation is possible, otherwise False.
+
+    Example:
+    >>> backtrack("leetcode", {"leet", "code"}, 0)
+    True
+
+    >>> backtrack("applepenapple", {"apple", "pen"}, 0)
+    True
+
+    >>> backtrack("catsandog", {"cats", "dog", "sand", "and", "cat"}, 0)
+    False
+    """
+
+    # Base case: if the starting index has reached the end of the string
+    if start == len(input_string):
+        return True
+
+    # Try every possible substring from 'start' to 'end'
+    for end in range(start + 1, len(input_string) + 1):
+        if input_string[start:end] in word_dict and backtrack(
+            input_string, word_dict, end
+        ):
+            return True
+
+    return False
+
+
+def word_break(input_string: str, word_dict: set[str]) -> bool:
+    """
+    Determines if the input string can be segmented into a sequence of
+    valid dictionary words using backtracking.
+
+    Parameters:
+    input_string (str): The input string to segment.
+    word_dict (set[str]): The set of valid words.
+
+    Returns:
+    bool: True if the string can be segmented into valid words, otherwise False.
+
+    Example:
+    >>> word_break("leetcode", {"leet", "code"})
+    True
+
+    >>> word_break("applepenapple", {"apple", "pen"})
+    True
+
+    >>> word_break("catsandog", {"cats", "dog", "sand", "and", "cat"})
+    False
+    """
+
+    return backtrack(input_string, word_dict, 0)

data_structures/linked_list/has_loop.py

@@ -14,11 +14,11 @@ def __init__(self, data: Any) -> None:
 
     def __iter__(self):
         node = self
-        visited = []
+        visited = set()
         while node:
             if node in visited:
                 raise ContainsLoopError
-            visited.append(node)
+            visited.add(node)
             yield node.data
             node = node.next_node
 

data_structures/stacks/next_greater_element.py

@@ -6,9 +6,20 @@
 
 def next_greatest_element_slow(arr: list[float]) -> list[float]:
     """
-    Get the Next Greatest Element (NGE) for all elements in a list.
-    Maximum element present after the current one which is also greater than the
-    current one.
+    Get the Next Greatest Element (NGE) for each element in the array
+    by checking all subsequent elements to find the next greater one.
+
+    This is a brute-force implementation, and it has a time complexity
+    of O(n^2), where n is the size of the array.
+
+    Args:
+        arr: List of numbers for which the NGE is calculated.
+
+    Returns:
+        List containing the next greatest elements. If no
+        greater element is found, -1 is placed in the result.
+
+    Example:
     >>> next_greatest_element_slow(arr) == expect
     True
     """
@@ -28,9 +39,21 @@ def next_greatest_element_slow(arr: list[float]) -> list[float]:
 
 def next_greatest_element_fast(arr: list[float]) -> list[float]:
     """
-    Like next_greatest_element_slow() but changes the loops to use
-    enumerate() instead of range(len()) for the outer loop and
-    for in a slice of arr for the inner loop.
+    Find the Next Greatest Element (NGE) for each element in the array
+    using a more readable approach. This implementation utilizes
+    enumerate() for the outer loop and slicing for the inner loop.
+
+    While this improves readability over next_greatest_element_slow(),
+    it still has a time complexity of O(n^2).
+
+    Args:
+        arr: List of numbers for which the NGE is calculated.
+
+    Returns:
+        List containing the next greatest elements. If no
+        greater element is found, -1 is placed in the result.
+
+    Example:
     >>> next_greatest_element_fast(arr) == expect
     True
     """
@@ -47,14 +70,23 @@ def next_greatest_element_fast(arr: list[float]) -> list[float]:
 
 def next_greatest_element(arr: list[float]) -> list[float]:
     """
-    Get the Next Greatest Element (NGE) for all elements in a list.
-    Maximum element present after the current one which is also greater than the
-    current one.
-
-    A naive way to solve this is to take two loops and check for the next bigger
-    number but that will make the time complexity as O(n^2). The better way to solve
-    this would be to use a stack to keep track of maximum number giving a linear time
-    solution.
+    Efficient solution to find the Next Greatest Element (NGE) for all elements
+    using a stack. The time complexity is reduced to O(n), making it suitable
+    for larger arrays.
+
+    The stack keeps track of elements for which the next greater element hasn't
+    been found yet. By iterating through the array in reverse (from the last
+    element to the first), the stack is used to efficiently determine the next
+    greatest element for each element.
+
+    Args:
+        arr: List of numbers for which the NGE is calculated.
+
+    Returns:
+        List containing the next greatest elements. If no
+        greater element is found, -1 is placed in the result.
+
+    Example:
     >>> next_greatest_element(arr) == expect
     True
     """

dynamic_programming/floyd_warshall.py

@@ -12,19 +12,58 @@ def __init__(self, n=0):  # a graph with Node 0,1,...,N-1
         ]  # dp[i][j] stores minimum distance from i to j
 
     def add_edge(self, u, v, w):
+        """
+        Adds a directed edge from node u
+        to node v with weight w.
+
+        >>> g = Graph(3)
+        >>> g.add_edge(0, 1, 5)
+        >>> g.dp[0][1]
+        5
+        """
         self.dp[u][v] = w
 
     def floyd_warshall(self):
+        """
+        Computes the shortest paths between all pairs of
+        nodes using the Floyd-Warshall algorithm.
+
+        >>> g = Graph(3)
+        >>> g.add_edge(0, 1, 1)
+        >>> g.add_edge(1, 2, 2)
+        >>> g.floyd_warshall()
+        >>> g.show_min(0, 2)
+        3
+        >>> g.show_min(2, 0)
+        inf
+        """
         for k in range(self.n):
             for i in range(self.n):
                 for j in range(self.n):
                     self.dp[i][j] = min(self.dp[i][j], self.dp[i][k] + self.dp[k][j])
 
     def show_min(self, u, v):
+        """
+        Returns the minimum distance from node u to node v.
+
+        >>> g = Graph(3)
+        >>> g.add_edge(0, 1, 3)
+        >>> g.add_edge(1, 2, 4)
+        >>> g.floyd_warshall()
+        >>> g.show_min(0, 2)
+        7
+        >>> g.show_min(1, 0)
+        inf
+        """
         return self.dp[u][v]
 
 
 if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+
+    # Example usage
     graph = Graph(5)
     graph.add_edge(0, 2, 9)
     graph.add_edge(0, 4, 10)
@@ -38,5 +77,9 @@ def show_min(self, u, v):
     graph.add_edge(4, 2, 4)
     graph.add_edge(4, 3, 9)
     graph.floyd_warshall()
-    graph.show_min(1, 4)
-    graph.show_min(0, 3)
+    print(
+        graph.show_min(1, 4)
+    )  # Should output the minimum distance from node 1 to node 4
+    print(
+        graph.show_min(0, 3)
+    )  # Should output the minimum distance from node 0 to node 3