Repeated square root operations in a given range of numbers
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I was curious to find the solution for the question
'Given a range of numbers, return the maximum no of repeated squares root operations within that range'.
For instance, given the range of numbers are (2,20), the perfect squares are [4,9,16] within that range. Now the maximum no of square root operations is 2
. Reason behind
sqrt(16) -> sqrt(4) -> 2.
I have come with a solution, however, it takes a lot of time to return the results. If you can provide a better version of this code, please let me know.
import math
# For each perfect square, calculate the depth of square root operations
# applied against it.
def square(n):
if n == 2 or n == 1:
return 0
sol = (int(n**0.5))**2
if sol != n : return 0
return 1 + square(int(n**0.5))
A = int(input())
B = int(input())
# Returns the perfect squares within that range
squares = [x for x in range(A,B) if (int(x**0.5))**2 == x]
print(max([square(i) for i in squares]))
python performance algorithm complexity
New contributor
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add a comment |
$begingroup$
I was curious to find the solution for the question
'Given a range of numbers, return the maximum no of repeated squares root operations within that range'.
For instance, given the range of numbers are (2,20), the perfect squares are [4,9,16] within that range. Now the maximum no of square root operations is 2
. Reason behind
sqrt(16) -> sqrt(4) -> 2.
I have come with a solution, however, it takes a lot of time to return the results. If you can provide a better version of this code, please let me know.
import math
# For each perfect square, calculate the depth of square root operations
# applied against it.
def square(n):
if n == 2 or n == 1:
return 0
sol = (int(n**0.5))**2
if sol != n : return 0
return 1 + square(int(n**0.5))
A = int(input())
B = int(input())
# Returns the perfect squares within that range
squares = [x for x in range(A,B) if (int(x**0.5))**2 == x]
print(max([square(i) for i in squares]))
python performance algorithm complexity
New contributor
$endgroup$
add a comment |
$begingroup$
I was curious to find the solution for the question
'Given a range of numbers, return the maximum no of repeated squares root operations within that range'.
For instance, given the range of numbers are (2,20), the perfect squares are [4,9,16] within that range. Now the maximum no of square root operations is 2
. Reason behind
sqrt(16) -> sqrt(4) -> 2.
I have come with a solution, however, it takes a lot of time to return the results. If you can provide a better version of this code, please let me know.
import math
# For each perfect square, calculate the depth of square root operations
# applied against it.
def square(n):
if n == 2 or n == 1:
return 0
sol = (int(n**0.5))**2
if sol != n : return 0
return 1 + square(int(n**0.5))
A = int(input())
B = int(input())
# Returns the perfect squares within that range
squares = [x for x in range(A,B) if (int(x**0.5))**2 == x]
print(max([square(i) for i in squares]))
python performance algorithm complexity
New contributor
$endgroup$
I was curious to find the solution for the question
'Given a range of numbers, return the maximum no of repeated squares root operations within that range'.
For instance, given the range of numbers are (2,20), the perfect squares are [4,9,16] within that range. Now the maximum no of square root operations is 2
. Reason behind
sqrt(16) -> sqrt(4) -> 2.
I have come with a solution, however, it takes a lot of time to return the results. If you can provide a better version of this code, please let me know.
import math
# For each perfect square, calculate the depth of square root operations
# applied against it.
def square(n):
if n == 2 or n == 1:
return 0
sol = (int(n**0.5))**2
if sol != n : return 0
return 1 + square(int(n**0.5))
A = int(input())
B = int(input())
# Returns the perfect squares within that range
squares = [x for x in range(A,B) if (int(x**0.5))**2 == x]
print(max([square(i) for i in squares]))
python performance algorithm complexity
python performance algorithm complexity
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