Practicing Coding Interview Questions in Python
Kirill Smirnov
Data Science Consultant, Altran
$n! = n\cdot(n-1)\cdot(n-2)\cdot...\cdot1$
$n = 4$:
$4! = 4\cdot3\cdot2\cdot1$
4! = 24
$n! = n\cdot(n-1)\cdot(n-2)\cdot...\cdot1 = $
$ = 1\cdot2\cdot3\cdot...\cdot n$
Iterative solution:
# iterative factorial
def fact_iter(n):
result = 1
# looping over numbers from 1 to n
for num in range(1, n+1)
result = num * result
return result
$n = 4:$
result
= 1
result
= 1 * result
(1) = 1result
= 2 * result
(1) = 2result
= 3 * result
(2) = 6result
= 4 * result
(4) = 24$4! = 1 \cdot 2 \cdot 3 \cdot 4 = 24$
$n!$ $=n\cdot(n-1)!$
def fact_rec(n):
return n * fact_rec(n-1)
What's wrong with that code?
fact_rec(4)
RecursionError
We must define a base case!
$n! = n\cdot(n-1)\cdot(n-2)\cdot...\cdot1$
A stopping criterion / base case: $1! = 1$
def fact_rec(n):
if n == 1:
return 1
return n * fact_rec(n-1)
fact_rec(4)
24
Recursive functions have two main components:
x - a new sample $(x_1, x_2)$
# Pseudo algorithm for finding out the category:
category = pred(node, x):
# Check if there is a split if node.hasSplitting:
# Check which child node to take if node.goToLeftChild(x): return pred(node.leftChild, x) if node.goToRightChild(x): return pred(node.rightChild, x)
x - a new sample $(x_1, x_2)$
# Pseudo algorithm for finding out the category:
category = pred(node, x):
# Check if there is a split if node.hasSplitting:
# Check which child node to take if node.goToLeftChild(x): return pred(node.leftChild, x) if node.goToRightChild(x): return pred(node.rightChild, x)
# Returning the category return node.category
Practicing Coding Interview Questions in Python