[pull] master from TheAlgorithms:master

financial/time_and_half_pay.py

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+"""
+Calculate time and a half pay
+"""
+
+
+def pay(hours_worked: float, pay_rate: float, hours: float = 40) -> float:
+    """
+    hours_worked = The total hours worked
+    pay_rate = Amount of money per hour
+    hours = Number of hours that must be worked before you receive time and a half
+
+    >>> pay(41, 1)
+    41.5
+    >>> pay(65, 19)
+    1472.5
+    >>> pay(10, 1)
+    10.0
+    """
+    # Check that all input parameters are float or integer
+    assert isinstance(hours_worked, (float, int)), (
+        "Parameter 'hours_worked' must be of type 'int' or 'float'"
+    )
+    assert isinstance(pay_rate, (float, int)), (
+        "Parameter 'pay_rate' must be of type 'int' or 'float'"
+    )
+    assert isinstance(hours, (float, int)), (
+        "Parameter 'hours' must be of type 'int' or 'float'"
+    )
+
+    normal_pay = hours_worked * pay_rate
+    over_time = max(0, hours_worked - hours)
+    over_time_pay = over_time * pay_rate / 2
+    return normal_pay + over_time_pay
+
+
+if __name__ == "__main__":
+    # Test
+    import doctest
+
+    doctest.testmod()

project_euler/problem_136/sol1.py

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+"""
+Project Euler Problem 136: https://projecteuler.net/problem=136
+
+Singleton Difference
+
+The positive integers, x, y, and z, are consecutive terms of an arithmetic progression.
+Given that n is a positive integer, the equation, x^2 - y^2 - z^2 = n,
+has exactly one solution when n = 20:
+                              13^2 - 10^2 - 7^2 = 20.
+
+In fact there are twenty-five values of n below one hundred for which
+the equation has a unique solution.
+
+How many values of n less than fifty million have exactly one solution?
+
+By change of variables
+
+x = y + delta
+z = y - delta
+
+The expression can be rewritten:
+
+x^2 - y^2 - z^2 = y * (4 * delta - y) = n
+
+The algorithm loops over delta and y, which is restricted in upper and lower limits,
+to count how many solutions each n has.
+In the end it is counted how many n's have one solution.
+"""
+
+
+def solution(n_limit: int = 50 * 10**6) -> int:
+    """
+    Define n count list and loop over delta, y to get the counts, then check
+    which n has count == 1.
+
+    >>> solution(3)
+    0
+    >>> solution(10)
+    3
+    >>> solution(100)
+    25
+    >>> solution(110)
+    27
+    """
+    n_sol = [0] * n_limit
+
+    for delta in range(1, (n_limit + 1) // 4 + 1):
+        for y in range(4 * delta - 1, delta, -1):
+            n = y * (4 * delta - y)
+            if n >= n_limit:
+                break
+            n_sol[n] += 1
+
+    ans = 0
+    for i in range(n_limit):
+        if n_sol[i] == 1:
+            ans += 1
+
+    return ans
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")