Added solution for Project Euler problem 145

Author: chakeson

project_euler/problem_145/__init__.py

@@ -0,0 +1,89 @@
+"""
+Problem 145: https://projecteuler.net/problem=145
+
+Name: How many reversible numbers are there below one-billion?
+
+Some positive integers n have the property that the 
+sum [ n + reverse(n) ] consists entirely of odd (decimal) digits. 
+For instance, 36 + 63 = 99 and 409 + 904 = 1313. 
+We will call such numbers reversible; so 36, 63, 409, and 904 are reversible. 
+Leading zeroes are not allowed in either n or reverse(n).
+
+There are 120 reversible numbers below one-thousand.
+
+How many reversible numbers are there below one-billion (10^9)?
+
+
+Solution:
+
+Here a brute force solution is used to find and count the reversible numbers.
+
+"""
+from __future__ import annotations
+
+
+def check_if_odd(sum: int) -> int:
+    """
+    Check if the last digit in the sum is even or odd. If even return 0.
+    If odd then floor devision by 10 is used to remove the last number.
+    Process continues until sum becomes 0 because no more numbers.
+    >>> check_if_odd(36)
+    0
+    >>> check_if_odd(33)
+    1
+    """
+    while sum > 0:
+        if (sum % 10) % 2 == 0:
+            return 0
+        sum = sum // 10
+    return 1
+
+
+def find_reverse_nr(nr: int) -> int:
+    """
+    Reverses the given number. Does not work with number that end in zero.
+    >>> find_reverse_nr(36)
+    63
+    >>> find_reverse_nr(409)
+    904
+    """
+    reverse = 0
+
+    while nr > 0:
+        temp = nr % 10
+        reverse = reverse * 10 + temp
+        nr = nr // 10
+
+    return reverse
+
+
+def solution(nr: int) -> int:
+    """
+    Loops over the range of numbers.
+    Checks if they have ending zeros which disqualifies them from being reversable.
+    If that condition is passed it generates the reveresed number.
+    Then sum up n and reverse(n).
+    Then check if all the numbers in the sum are odd. If true add to the answer.
+    >>> solution(1000000000)
+    608720
+    >>> solution(1000000)
+    18720
+    >>> solution(1000000)
+    18720
+    >>> solution(1000)
+    120
+    """
+    answer = 0
+    for x in range(1, nr):
+        if x % 10 != 0:
+            reversed_nr = find_reverse_nr(x)
+            sum = x + reversed_nr
+            answer += check_if_odd(sum)
+
+    return answer
+
+
+nr = 1000000000
+
+if __name__ == "__main__":
+    print(f"{solution(nr) = }")