Add sol for P104 Project Euler (#5257)

* Hacktoberfest: added sol for P104 Project Euler

* bot requests resolved

* pre-commit

* Update sol.py

* Update sol.py

* remove trailing zeroes

* Update sol.py

* Update sol.py

* Update sol.py

Co-authored-by: John Law <johnlaw.po@gmail.com>

project_euler/problem_104/sol.py

@@ -0,0 +1,137 @@
+"""
+Project Euler Problem 104 : https://projecteuler.net/problem=104
+
+The Fibonacci sequence is defined by the recurrence relation:
+
+Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
+It turns out that F541, which contains 113 digits, is the first Fibonacci number
+for which the last nine digits are 1-9 pandigital (contain all the digits 1 to 9,
+but not necessarily in order). And F2749, which contains 575 digits, is the first
+Fibonacci number for which the first nine digits are 1-9 pandigital.
+
+Given that Fk is the first Fibonacci number for which the first nine digits AND
+the last nine digits are 1-9 pandigital, find k.
+"""
+
+
+def check(number: int) -> bool:
+    """
+    Takes a number and checks if it is pandigital both from start and end
+
+
+    >>> check(123456789987654321)
+    True
+
+    >>> check(120000987654321)
+    False
+
+    >>> check(1234567895765677987654321)
+    True
+
+    """
+
+    check_last = [0] * 11
+    check_front = [0] * 11
+
+    # mark last 9 numbers
+    for x in range(9):
+        check_last[int(number % 10)] = 1
+        number = number // 10
+    # flag
+    f = True
+
+    # check last 9 numbers for pandigitality
+
+    for x in range(9):
+        if not check_last[x + 1]:
+            f = False
+    if not f:
+        return f
+
+    # mark first 9 numbers
+    number = int(str(number)[:9])
+
+    for x in range(9):
+        check_front[int(number % 10)] = 1
+        number = number // 10
+
+    # check first 9 numbers for pandigitality
+
+    for x in range(9):
+        if not check_front[x + 1]:
+            f = False
+    return f
+
+
+def check1(number: int) -> bool:
+    """
+    Takes a number and checks if it is pandigital from END
+
+    >>> check1(123456789987654321)
+    True
+
+    >>> check1(120000987654321)
+    True
+
+    >>> check1(12345678957656779870004321)
+    False
+
+    """
+
+    check_last = [0] * 11
+
+    # mark last 9 numbers
+    for x in range(9):
+        check_last[int(number % 10)] = 1
+        number = number // 10
+    # flag
+    f = True
+
+    # check last 9 numbers for pandigitality
+
+    for x in range(9):
+        if not check_last[x + 1]:
+            f = False
+    return f
+
+
+def solution() -> int:
+    """
+    Outputs the answer is the least Fibonacci number pandigital from both sides.
+    >>> solution()
+    329468
+    """
+
+    a = 1
+    b = 1
+    c = 2
+    # temporary Fibonacci numbers
+
+    a1 = 1
+    b1 = 1
+    c1 = 2
+    # temporary Fibonacci numbers mod 1e9
+
+    # mod m=1e9, done for fast optimisation
+    tocheck = [0] * 1000000
+    m = 1000000000
+
+    for x in range(1000000):
+        c1 = (a1 + b1) % m
+        a1 = b1 % m
+        b1 = c1 % m
+        if check1(b1):
+            tocheck[x + 3] = 1
+
+    for x in range(1000000):
+        c = a + b
+        a = b
+        b = c
+        # perform check only if in tocheck
+        if tocheck[x + 3] and check(b):
+            return x + 3  # first 2 already done
+    return -1
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")