add euler problem 38

Author: coryallegory

DIRECTORY.md

@@ -636,6 +636,7 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_037/sol1.py)
   * Problem 038
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_038/sol1.py)
+    * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_038/sol2.py)
   * Problem 039
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_039/sol1.py)
   * Problem 040

project_euler/problem_38/__init__.py

@@ -0,0 +1 @@
+#

project_euler/problem_38/sol1.py

@@ -0,0 +1,81 @@
+"""
+Project Euler Problem 38
+https://projecteuler.net/problem=38
+
+Take the number 192 and multiply it by each of 1, 2, and 3:
+
+192 × 1 = 192
+192 × 2 = 384
+192 × 3 = 576
+
+By concatenating each product we get the 1 to 9 pandigital, 192384576. We will
+call 192384576 the concatenated product of 192 and (1,2,3)
+
+The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5,
+giving the pandigital, 918273645, which is the concatenated product of 9 and
+(1,2,3,4,5).
+
+What is the largest 1 to 9 pandigital 9-digit number that can be formed as the
+concatenated product of an integer with (1,2, ... , n) where n > 1?
+"""
+
+
+def solution() -> int:
+    """
+    Calculates and returns the largest 1-9 pandigital 9-digit number as product of
+    integer set n>1.
+
+    Answer:
+    >>> solution()
+    932718654
+
+    Assumptions:
+    1. Since n>1, we'll concatenate the fixed number at least twice. We only need
+         9 digits, so 2x5=10 would exceed this safely. Thus target number is 4
+         digits or less.
+    2. We know at least one 9-digit pandigital starts with 9 as per the example,
+         so can short circuit search based on that.
+    3. We can start with 90, the next 9-start number following 9 (already calculated)
+    """
+
+    largest_pandigital = 918273645
+
+    for fixed_num in range(90, 10000):  # assumption 1 and 3
+        if not str(fixed_num).startswith("9"):  # assumption 2
+            continue
+
+        pan = fixed_num
+        i = 2
+        while len(str(pan)) < 9:
+            pan = int(str(pan) + str(fixed_num * i))
+
+        if is_pandigital(pan) and pan > largest_pandigital:
+            largest_pandigital = pan
+
+    return largest_pandigital
+
+
+def is_pandigital(num: int) -> bool:
+    """
+    Validates that a number is 9-digit pandigital.
+    i.e. contains all numbers 1-9 without zero.
+
+    Examples:
+    >>> is_pandigital(123456789)
+    True
+    >>> is_pandigital(987654321)
+    True
+    >>> is_pandigital(1234567890)
+    False
+    >>> is_pandigital(111111111)
+    False
+    """
+
+    # check that only digits 1-9 are present with no duplicates
+    numberset = set(str(num))
+    return len(numberset) == 9 and "0" not in numberset
+
+
+if __name__ == "__main__":
+
+    print(solution())