AlgorithmAndLeetCode · pull · May 22, 2025 · May 22, 2025 · May 22, 2025 · May 22, 2025
diff --git a/backtracking/sum_of_subsets.py b/backtracking/sum_of_subsets.py
@@ -1,16 +1,26 @@
 """
-The sum-of-subsetsproblem states that a set of non-negative integers, and a
+The sum-of-subsets problem states that a set of non-negative integers, and a
 value M, determine all possible subsets of the given set whose summation sum
 equal to given M.
 
 Summation of the chosen numbers must be equal to given number M and one number
 can be used only once.
 """
 
-from __future__ import annotations
 
+def generate_sum_of_subsets_solutions(nums: list[int], max_sum: int) -> list[list[int]]:
+    """
+    The main function. For list of numbers 'nums' find the subsets with sum
+    equal to 'max_sum'
+
+    >>> generate_sum_of_subsets_solutions(nums=[3, 34, 4, 12, 5, 2], max_sum=9)
+    [[3, 4, 2], [4, 5]]
+    >>> generate_sum_of_subsets_solutions(nums=[3, 34, 4, 12, 5, 2], max_sum=3)
+    [[3]]
+    >>> generate_sum_of_subsets_solutions(nums=[3, 34, 4, 12, 5, 2], max_sum=1)
+    []
+    """
 
-def generate_sum_of_subsets_soln(nums: list[int], max_sum: int) -> list[list[int]]:
     result: list[list[int]] = []
     path: list[int] = []
     num_index = 0
@@ -34,7 +44,21 @@ def create_state_space_tree(
     This algorithm follows depth-fist-search and backtracks when the node is not
     branchable.
 
+    >>> path = []
+    >>> result = []
+    >>> create_state_space_tree(
+    ...     nums=[1],
+    ...     max_sum=1,
+    ...     num_index=0,
+    ...     path=path,
+    ...     result=result,
+    ...     remaining_nums_sum=1)
+    >>> path
+    []
+    >>> result
+    [[1]]
     """
+
     if sum(path) > max_sum or (remaining_nums_sum + sum(path)) < max_sum:
         return
     if sum(path) == max_sum:
@@ -51,16 +75,7 @@ def create_state_space_tree(
         )
 
 
-"""
-remove the comment to take an input from the user
-
-print("Enter the elements")
-nums = list(map(int, input().split()))
-print("Enter max_sum sum")
-max_sum = int(input())
+if __name__ == "__main__":
+    import doctest
 
-"""
-nums = [3, 34, 4, 12, 5, 2]
-max_sum = 9
-result = generate_sum_of_subsets_soln(nums, max_sum)
-print(*result)
+    doctest.testmod()
diff --git a/data_structures/hashing/hash_map.py b/data_structures/hashing/hash_map.py
@@ -16,7 +16,7 @@
 VAL = TypeVar("VAL")
 
 
-@dataclass(frozen=True, slots=True)
+@dataclass(slots=True)
 class _Item(Generic[KEY, VAL]):
     key: KEY
     val: VAL
@@ -72,16 +72,17 @@ def _try_set(self, ind: int, key: KEY, val: VAL) -> bool:
 
         If bucket is empty or key is the same, does insert and return True.
 
-        If bucket has another key or deleted placeholder,
-        that means that we need to check next bucket.
+        If bucket has another key that means that we need to check next bucket.
         """
         stored = self._buckets[ind]
         if not stored:
+            # A falsy item means that bucket was never used (None)
+            # or was deleted (_deleted).
             self._buckets[ind] = _Item(key, val)
             self._len += 1
             return True
         elif stored.key == key:
-            self._buckets[ind] = _Item(key, val)
+            stored.val = val
             return True
         else:
             return False
@@ -228,6 +229,27 @@ def __delitem__(self, key: KEY) -> None:
         Traceback (most recent call last):
         ...
         KeyError: 4
+
+        # Test resize down when sparse
+        ## Setup: resize up
+        >>> hm = HashMap(initial_block_size=100, capacity_factor=0.75)
+        >>> len(hm._buckets)
+        100
+        >>> for i in range(75):
+        ...     hm[i] = i
+        >>> len(hm._buckets)
+        100
+        >>> hm[75] = 75
+        >>> len(hm._buckets)
+        200
+
+        ## Resize down
+        >>> del hm[75]
+        >>> len(hm._buckets)
+        200
+        >>> del hm[74]
+        >>> len(hm._buckets)
+        100
         """
         for ind in self._iterate_buckets(key):
             item = self._buckets[ind]

diff --git a/dynamic_programming/matrix_chain_order.py b/dynamic_programming/matrix_chain_order.py
@@ -5,13 +5,19 @@
 Implementation of Matrix Chain Multiplication
 Time Complexity: O(n^3)
 Space Complexity: O(n^2)
+
+Reference: https://en.wikipedia.org/wiki/Matrix_chain_multiplication
 """
 
 
-def matrix_chain_order(array):
+def matrix_chain_order(array: list[int]) -> tuple[list[list[int]], list[list[int]]]:
+    """
+    >>> matrix_chain_order([10, 30, 5])
+    ([[0, 0, 0], [0, 0, 1500], [0, 0, 0]], [[0, 0, 0], [0, 0, 1], [0, 0, 0]])
+    """
     n = len(array)
-    matrix = [[0 for x in range(n)] for x in range(n)]
-    sol = [[0 for x in range(n)] for x in range(n)]
+    matrix = [[0 for _ in range(n)] for _ in range(n)]
+    sol = [[0 for _ in range(n)] for _ in range(n)]
 
     for chain_length in range(2, n):
         for a in range(1, n - chain_length + 1):
@@ -28,26 +34,33 @@ def matrix_chain_order(array):
     return matrix, sol
 
 
-# Print order of matrix with Ai as Matrix
-def print_optiomal_solution(optimal_solution, i, j):
+def print_optimal_solution(optimal_solution: list[list[int]], i: int, j: int):
+    """
+    Print order of matrix with Ai as Matrix.
+    """
+
     if i == j:
         print("A" + str(i), end=" ")
     else:
         print("(", end=" ")
-        print_optiomal_solution(optimal_solution, i, optimal_solution[i][j])
-        print_optiomal_solution(optimal_solution, optimal_solution[i][j] + 1, j)
+        print_optimal_solution(optimal_solution, i, optimal_solution[i][j])
+        print_optimal_solution(optimal_solution, optimal_solution[i][j] + 1, j)
         print(")", end=" ")
 
 
 def main():
+    """
+    Size of matrix created from array [30, 35, 15, 5, 10, 20, 25] will be:
+    30*35 35*15 15*5 5*10 10*20 20*25
+    """
+
     array = [30, 35, 15, 5, 10, 20, 25]
     n = len(array)
-    # Size of matrix created from above array will be
-    # 30*35 35*15 15*5 5*10 10*20 20*25
+
     matrix, optimal_solution = matrix_chain_order(array)
 
     print("No. of Operation required: " + str(matrix[1][n - 1]))
-    print_optiomal_solution(optimal_solution, 1, n - 1)
+    print_optimal_solution(optimal_solution, 1, n - 1)
 
 
 if __name__ == "__main__":