Refactor kth_permutation and fix linter errors

Author: umutKaracelebi

machine_learning/linear_discriminant_analysis.py

@@ -249,16 +249,16 @@ def accuracy(actual_y: list, predicted_y: list) -> float:
     return (correct / len(actual_y)) * 100
 
 
-num = TypeVar("num")
+T = TypeVar("T")
 
 
 def valid_input(
-    input_type: Callable[[object], num],  # Usually float or int
+    input_type: Callable[[object], T],  # Usually float or int
     input_msg: str,
     err_msg: str,
-    condition: Callable[[num], bool] = lambda _: True,
+    condition: Callable[[T], bool] = lambda _: True,
     default: str | None = None,
-) -> num:
+) -> T:
     """
     Ask for user value and validate that it fulfill a condition.
 

maths/kth_lexicographic_permutation.py

@@ -1,24 +1,50 @@
-def kth_permutation(k, n):
+def kth_permutation(k: int, n: int) -> list[int]:
     """
     Finds k'th lexicographic permutation (in increasing order) of
     0,1,2,...n-1 in O(n^2) time.
 
+    :param k: The index of the permutation (0-based)
+    :param n: The number of elements in the permutation
+    :return: The k-th lexicographic permutation of size n
+
     Examples:
-    First permutation is always 0,1,2,...n
-    >>> kth_permutation(0,5)
+    First permutation is always 0,1,2,...n-1
+    >>> kth_permutation(0, 5)
     [0, 1, 2, 3, 4]
 
     The order of permutation of 0,1,2,3 is [0,1,2,3], [0,1,3,2], [0,2,1,3],
     [0,2,3,1], [0,3,1,2], [0,3,2,1], [1,0,2,3], [1,0,3,2], [1,2,0,3],
     [1,2,3,0], [1,3,0,2]
-    >>> kth_permutation(10,4)
+    >>> kth_permutation(10, 4)
     [1, 3, 0, 2]
+
+    >>> kth_permutation(10, 0)
+    Traceback (most recent call last):
+        ...
+    ValueError: n must be positive
+
+    >>> kth_permutation(-1, 5)
+    Traceback (most recent call last):
+        ...
+    IndexError: k must be non-negative
+
+    >>> kth_permutation(120, 5)
+    Traceback (most recent call last):
+        ...
+    IndexError: k out of bounds
     """
-    # Factorails from 1! to (n-1)!
+    if n <= 0:
+        raise ValueError("n must be positive")
+    if k < 0:
+        raise IndexError("k must be non-negative")
+
+    # Factorials from 1! to (n-1)!
     factorials = [1]
     for i in range(2, n):
         factorials.append(factorials[-1] * i)
-    assert 0 <= k < factorials[-1] * n, "k out of bounds"
+
+    if k >= factorials[-1] * n:
+        raise IndexError("k out of bounds")
 
     permutation = []
     elements = list(range(n))
@@ -28,7 +54,7 @@ def kth_permutation(k, n):
         factorial = factorials.pop()
         number, k = divmod(k, factorial)
         permutation.append(elements[number])
-        elements.remove(elements[number])
+        elements.pop(number)  # elements.remove(elements[number]) is slower and redundant
     permutation.append(elements[0])
 
     return permutation