CSEU2 Algorithms Guided Project code
Nabeelah Yousuph 11 minutes ago
# 1
def baz(n): # O(n*sqrt(n) ^ 2)
s = 0 # O(1)
for i in range(n): # O(n)
for j in range(int(math.sqrt(n))): # O(sqrt(n))
s += i * j # O(1)
return s
# 2
def frotz(n): # O(2n ^ 2)
s = 0 # O(1)
for i in range(n): # O(n)
for j in range(2 * n): # O(2n)
s += i * j # O(1)
return s
# 3
def bar(x): # O((1478 + x) ^ 3)
sum = 0 # O(1)
for i in range(0, 1463): # O(1463)
i += sum # O(1)
for _ in range(0, x): # O(x)
for _ in range(x, x + 15): # O(15 + x)
sum += 1 # O(1)
Jakub Maleta 11 minutes ago
# 1
def baz(n): # O(n*sqrt(n))
s = 0 # O(1)
for i in range(n): # O(n)
for j in range(int(math.sqrt(n))): # O(sqrt(n))
s += i * j # O(1)
return s # O(1)
# 2
def frotz(n): # O(2n^2) => O(n^2)
s = 0 # O(1)
for i in range(n): # O(n)
for j in range(2 * n): # O(2n)
s += i * j # O(1)
return s # O(1)
# 3
def bar(x): # O(x^2)
sum = 0 # O(1)
for i in range(0, 1463): # O(1)
i += sum # O(1)
for _ in range(0, x): # O(x)
for _ in range(x, x + 15): # O(x)
sum += 1 # O(1)
Shola Ayeni 11 minutes ago
def baz(n):
s = 0
for i in range(n): # O(n)
for j in range(int(math.sqrt(n))): # 0(sqrt(n))
s += i * j # O(1)
return s # O(1)
# O(n^(3/2) + 1)
# 2
def frotz(n):
s = 0
for i in range(n): # O(n)
for j in range(2 * n): # O(2n)
s += i * j # O(1)
return s # O(1)
# O(n^2 + 1)
# 3
def bar(x):
sum = 0 # O(1)
for i in range(0, 1463): # O(n)
i += sum #O(1)
for _ in range(0, x): # O(n)
for _ in range(x, x + 15): # O(n)
sum += 1 #O(1)
#O(n^3)
Aaron Thompson 11 minutes ago
#Total: O(n sqrt(n))
def baz(n):
s = 0
for i in range(n): #O(n)
for j in range(int(math.sqrt(n))): #O(sqrt(n))
s += i * j #O(1)
return s
# 2
#Total: O(n^2)
def frotz(n):
s = 0
for i in range(n): #O(n)
for j in range(2 * n): #O(2n)
s += i * j #O(1)
return s
# 3
#Total: O(n)
def bar(x):
sum = 0
for i in range(0, 1463): #O(1)
i += sum
for _ in range(0, x): #O(n)
for _ in range(x, x + 15): #O(1)
sum += 1 #O(1)
Pascal Ulor:andela: 11 minutes ago
def baz(n):
s = 0 #O(1)
for i in range(n): #(n)
for j in range(int(sqrt(n))): # n * sqrt(n) => O(sqrt(n)*n)
s += i * j # O(1)
return s #O(1)
"""
3*O(1) + O(n) + O(sqrt(n)*n) => O(sqrt(n)*n)
"""
# 2
def frotz(n):
s = 0 #O(1)
for i in range(n): #O(n)
for j in range(2 * n): # n * 2n => O(2n^2) => O(n^2)
s += i * j #O(1)
return s #O(1)
"""
3*O(1) + O(n) + O(n^2) => O(n^2)
"""
# 3
def bar(x):
sum = 0 #O(1)
for i in range(0, 1463): O(n)
i += sum # n * 1
for _ in range(0, x): # n * n
for _ in range(x, x + 15): # n * n * n
sum += 1 # n * n * n => O(n^3)
Kelechi ogbonna 9 minutes ago
def baz(n):
s = 0 # c
for i in range(n): # n
for j in range(int(sqrt(n))): #sqrt(n)
s += i * j # c
return s # c
# n * sqrt(n)
# 2
def frotz(n):
s = 0 # c
for i in range(n): # n
for j in range(2 * n): n
s += i * j #c
return s # c
# O(n^2)
# 3
def bar(x):
sum = 0 # c
for i in range(0, 1463): # c
i += sum # c
for j in range(0, x): # x
for k in range(x, x + 15): # 2x
sum += 1 # c
# O(2x^2) => O(x^2)
Matt Hardman:bat: 9 minutes ago
O(n)
:fire:
2
Matt Hardman:bat: 8 minutes ago
^third
Jakub Maleta 7 minutes ago
third updated:
# 3
def bar(x): # O(n)
sum = 0 # O(1)
for i in range(0, 1463): # O(1)
i += sum # O(1)
for _ in range(0, x): # O(n)
for _ in range(x, x + 15): # O(1)
sum += 1 # O(1)