Hacktoberfest: added sol for P104 Project Euler

project_euler/problem_104/sol.py

@@ -0,0 +1,138 @@
+"""
+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(a : 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
+
+  '''
+
+  checkLast=[0]*11 
+  checkFront=[0]*11
+
+  #mark last 9 numbers
+  for x in range(9):
+      checkLast[int(a%10)]=1
+      a=a//10
+  #flag
+  f=True
+
+  #check last 9 numbers for pandigitality
+
+  for x in range(9):
+      if( not checkLast[x+1]):
+          f=False
+  if(not f):
+    return f
+
+
+  #mark first 9 numbers
+  a= int(str(a)[:9])
+
+  for x in range(9):
+      checkFront[int(a%10)]=1
+      a=a//10
+
+  #check first 9 numbers for pandigitality
+
+  for x in range(9):
+      if( not checkFront[x+1]):
+          f=False
+  return f
+
+
+
+
+def check1(a : int) ->bool:
+  '''
+  Takes a number and checks if it is pandigital from END
+
+
+  >>> check1(123456789987654321)
+  True
+
+  >>> check1(120000987654321)
+  True
+
+  >>> check1(12345678957656779870004321)
+  False
+
+  '''
+
+  checkLast=[0]*11 
+
+  #mark last 9 numbers
+  for x in range(9):
+      checkLast[int(a%10)]=1
+      a=a//10
+  #flag
+  f=True
+
+  #check last 9 numbers for pandigitality
+
+  for x in range(9):
+      if( not checkLast[x+1]):
+          f=False
+  return f
+
+
+def solution()->int:
+    """
+    Outputs the answer ie the least fib 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
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")