Repeated square root operations in a given range of numbers
$begingroup$
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
asked 11 mins ago
s326280s326280
1
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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
asked 11 mins ago
s326280s326280
1
New contributor
s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
The code doesn't solve the stated problem. It is not asking how many perfect squares are in the range. It asks how many times a square root operation can be iteratively applied starting from some number.
$endgroup$
– vnp
8 secs ago
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
asked 11 mins ago
s326280s326280
1
New contributor
s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
asked 11 mins ago
s326280s326280
1
New contributor
s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 mins ago
s326280s326280
1
New contributor
s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 mins ago
s326280s326280
1
New contributor
s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 mins ago
s326280s326280
1
asked 11 mins ago
s326280s326280
1
1
New contributor
s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
The code doesn't solve the stated problem. It is not asking how many perfect squares are in the range. It asks how many times a square root operation can be iteratively applied starting from some number.
$endgroup$
– vnp
8 secs ago
add a comment |
$begingroup$
The code doesn't solve the stated problem. It is not asking how many perfect squares are in the range. It asks how many times a square root operation can be iteratively applied starting from some number.
$endgroup$
– vnp
8 secs ago
$begingroup$
The code doesn't solve the stated problem. It is not asking how many perfect squares are in the range. It asks how many times a square root operation can be iteratively applied starting from some number.
$endgroup$
– vnp
8 secs ago
$begingroup$
The code doesn't solve the stated problem. It is not asking how many perfect squares are in the range. It asks how many times a square root operation can be iteratively applied starting from some number.
$endgroup$
– vnp
8 secs ago
add a comment |
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$begingroup$
The code doesn't solve the stated problem. It is not asking how many perfect squares are in the range. It asks how many times a square root operation can be iteratively applied starting from some number.
$endgroup$
– vnp
8 secs ago