Make a beautiful binary string

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This is the "Beautiful Binary String" problem at HackerRank:

Alice has a binary string. She thinks a binary string is beautiful if and only if it doesn't contain the substring 010.

In one step, Alice can change a 0 to a 1 or vice versa. Count and print the minimum number of steps needed to make Alice see the string as beautiful.

For example, if Alice's string is 010 she can change any one element and have a beautiful string.

This is my Python code:

def beautifulBinaryString(b):
 temp = list(b)
 count,i = 0,0
 if len(b) == 3 and b == "010": count += 1
 elif len(b) == 3 and b != "010": count = count
 else:
 while (i+3 <= len(temp)):
 if temp[i:i+3] == ['0','1','0']: 
 count += 1
 del temp[i:i+3]
 else: i += 1
 return count

It seems to me as having too many conditionals (though readable, I guess). Is there a more concise way to accomplish this?

Some test cases:

0101010 
01100

2
0

python strings programming-challenge

share|improve this question

edited 1 hour ago

Gareth Rees

43k394172

asked 2 hours ago

db18

614

add a comment | 

up vote
3
down vote

favorite

This is the "Beautiful Binary String" problem at HackerRank:

Alice has a binary string. She thinks a binary string is beautiful if and only if it doesn't contain the substring 010.

In one step, Alice can change a 0 to a 1 or vice versa. Count and print the minimum number of steps needed to make Alice see the string as beautiful.

For example, if Alice's string is 010 she can change any one element and have a beautiful string.

This is my Python code:

def beautifulBinaryString(b):
 temp = list(b)
 count,i = 0,0
 if len(b) == 3 and b == "010": count += 1
 elif len(b) == 3 and b != "010": count = count
 else:
 while (i+3 <= len(temp)):
 if temp[i:i+3] == ['0','1','0']: 
 count += 1
 del temp[i:i+3]
 else: i += 1
 return count

It seems to me as having too many conditionals (though readable, I guess). Is there a more concise way to accomplish this?

Some test cases:

0101010 
01100

2
0

python strings programming-challenge

share|improve this question

edited 1 hour ago

Gareth Rees

43k394172

asked 2 hours ago

db18

614

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

This is the "Beautiful Binary String" problem at HackerRank:

Alice has a binary string. She thinks a binary string is beautiful if and only if it doesn't contain the substring 010.

In one step, Alice can change a 0 to a 1 or vice versa. Count and print the minimum number of steps needed to make Alice see the string as beautiful.

For example, if Alice's string is 010 she can change any one element and have a beautiful string.

This is my Python code:

def beautifulBinaryString(b):
 temp = list(b)
 count,i = 0,0
 if len(b) == 3 and b == "010": count += 1
 elif len(b) == 3 and b != "010": count = count
 else:
 while (i+3 <= len(temp)):
 if temp[i:i+3] == ['0','1','0']: 
 count += 1
 del temp[i:i+3]
 else: i += 1
 return count

It seems to me as having too many conditionals (though readable, I guess). Is there a more concise way to accomplish this?

Some test cases:

0101010 
01100

2
0

python strings programming-challenge

share|improve this question

edited 1 hour ago

Gareth Rees

43k394172

asked 2 hours ago

db18

614

This is the "Beautiful Binary String" problem at HackerRank:

Alice has a binary string. She thinks a binary string is beautiful if and only if it doesn't contain the substring 010.

In one step, Alice can change a 0 to a 1 or vice versa. Count and print the minimum number of steps needed to make Alice see the string as beautiful.

For example, if Alice's string is 010 she can change any one element and have a beautiful string.

This is my Python code:

def beautifulBinaryString(b):
 temp = list(b)
 count,i = 0,0
 if len(b) == 3 and b == "010": count += 1
 elif len(b) == 3 and b != "010": count = count
 else:
 while (i+3 <= len(temp)):
 if temp[i:i+3] == ['0','1','0']: 
 count += 1
 del temp[i:i+3]
 else: i += 1
 return count

It seems to me as having too many conditionals (though readable, I guess). Is there a more concise way to accomplish this?

Some test cases:

0101010 
01100

2
0

python strings programming-challenge

python strings programming-challenge

share|improve this question

edited 1 hour ago

Gareth Rees

43k394172

asked 2 hours ago

db18

614

share|improve this question

edited 1 hour ago

Gareth Rees

43k394172

asked 2 hours ago

db18

614

share|improve this question

share|improve this question

edited 1 hour ago

Gareth Rees

43k394172

edited 1 hour ago

Gareth Rees

43k394172

edited 1 hour ago

Gareth Rees

43k394172

43k394172

asked 2 hours ago

db18

614

asked 2 hours ago

db18

614

asked 2 hours ago

db18

614

614

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

Style

• 
  

  Read the PEP8 style guide!

  
  
  
  1. Functions and variables should be snake_case
     
  
  
  2. Conditions should be on the next line if a: ... is bad style
  
  
  3. Conditions don't need parenthesis while (a) is the same as while a:
     
  
  
  4. Avoid temp variables
  
  

Algorithm

• 
  

  Your first 2 guard clauses seem very unnecessary

  
  
  

  When I remove them, the code still works.

  
  



• Consider writing docstring/tests or both with the doctest module

Alternative Code

You could use re.findall(substring, string) for counting the occurrence,

OR string.count(substring) making this practically a one-line

import doctest

def beautiful_binary_string(b):
 """
 Returns the steps to make a binary string beautiful

 >>> beautiful_binary_string("0101010")
 2

 >>> beautiful_binary_string("01100")
 0

 >>> beautiful_binary_string("0100101010100010110100100110110100011100111110101001011001110111110000101011011111011001111100011101")
 10
 """
 return b.count("010")

if __name__ == '__main__':
 doctest.testmod()

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Ludisposed

6,41021858

add a comment | 

up vote
1
down vote

Is there a more concise way to accomplish this?

Certainly.

For a start, the special cases are unnecessary. (They make me think that the code has been refactored from a recursive version).

Secondly, the expensive del temp[i:i+3] could be replaced with i += 3, and since the processing is no longer destructive temp is unnecessary. This simplifies the code to

def beautifulBinaryString(b):
 count, i = 0, 0
 while (i+3 <= len(b)):
 if b[i:i+3] == "010":
 count, i = count+1, i+3
 else: i += 1
 return count

share|improve this answer

answered 1 hour ago

Peter Taylor

14.5k2454

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Style

• 
  

  Read the PEP8 style guide!

  
  
  
  1. Functions and variables should be snake_case
     
  
  
  2. Conditions should be on the next line if a: ... is bad style
  
  
  3. Conditions don't need parenthesis while (a) is the same as while a:
     
  
  
  4. Avoid temp variables
  
  

Algorithm

• 
  

  Your first 2 guard clauses seem very unnecessary

  
  
  

  When I remove them, the code still works.

  
  



• Consider writing docstring/tests or both with the doctest module

Alternative Code

You could use re.findall(substring, string) for counting the occurrence,

OR string.count(substring) making this practically a one-line

import doctest

def beautiful_binary_string(b):
 """
 Returns the steps to make a binary string beautiful

 >>> beautiful_binary_string("0101010")
 2

 >>> beautiful_binary_string("01100")
 0

 >>> beautiful_binary_string("0100101010100010110100100110110100011100111110101001011001110111110000101011011111011001111100011101")
 10
 """
 return b.count("010")

if __name__ == '__main__':
 doctest.testmod()

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Ludisposed

6,41021858

add a comment | 

up vote
2
down vote



accepted

Style

• 
  

  Read the PEP8 style guide!

  
  
  
  1. Functions and variables should be snake_case
     
  
  
  2. Conditions should be on the next line if a: ... is bad style
  
  
  3. Conditions don't need parenthesis while (a) is the same as while a:
     
  
  
  4. Avoid temp variables
  
  

Algorithm

• 
  

  Your first 2 guard clauses seem very unnecessary

  
  
  

  When I remove them, the code still works.

  
  



• Consider writing docstring/tests or both with the doctest module

Alternative Code

You could use re.findall(substring, string) for counting the occurrence,

OR string.count(substring) making this practically a one-line

import doctest

def beautiful_binary_string(b):
 """
 Returns the steps to make a binary string beautiful

 >>> beautiful_binary_string("0101010")
 2

 >>> beautiful_binary_string("01100")
 0

 >>> beautiful_binary_string("0100101010100010110100100110110100011100111110101001011001110111110000101011011111011001111100011101")
 10
 """
 return b.count("010")

if __name__ == '__main__':
 doctest.testmod()

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Ludisposed

6,41021858

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Style

• 
  

  Read the PEP8 style guide!

  
  
  
  1. Functions and variables should be snake_case
     
  
  
  2. Conditions should be on the next line if a: ... is bad style
  
  
  3. Conditions don't need parenthesis while (a) is the same as while a:
     
  
  
  4. Avoid temp variables
  
  

Algorithm

• 
  

  Your first 2 guard clauses seem very unnecessary

  
  
  

  When I remove them, the code still works.

  
  



• Consider writing docstring/tests or both with the doctest module

Alternative Code

You could use re.findall(substring, string) for counting the occurrence,

OR string.count(substring) making this practically a one-line

import doctest

def beautiful_binary_string(b):
 """
 Returns the steps to make a binary string beautiful

 >>> beautiful_binary_string("0101010")
 2

 >>> beautiful_binary_string("01100")
 0

 >>> beautiful_binary_string("0100101010100010110100100110110100011100111110101001011001110111110000101011011111011001111100011101")
 10
 """
 return b.count("010")

if __name__ == '__main__':
 doctest.testmod()

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Ludisposed

6,41021858

Style

• 
  

  Read the PEP8 style guide!

  
  
  
  1. Functions and variables should be snake_case
     
  
  
  2. Conditions should be on the next line if a: ... is bad style
  
  
  3. Conditions don't need parenthesis while (a) is the same as while a:
     
  
  
  4. Avoid temp variables
  
  

Algorithm

• 
  

  Your first 2 guard clauses seem very unnecessary

  
  
  

  When I remove them, the code still works.

  
  



• Consider writing docstring/tests or both with the doctest module

Alternative Code

You could use re.findall(substring, string) for counting the occurrence,

OR string.count(substring) making this practically a one-line

import doctest

def beautiful_binary_string(b):
 """
 Returns the steps to make a binary string beautiful

 >>> beautiful_binary_string("0101010")
 2

 >>> beautiful_binary_string("01100")
 0

 >>> beautiful_binary_string("0100101010100010110100100110110100011100111110101001011001110111110000101011011111011001111100011101")
 10
 """
 return b.count("010")

if __name__ == '__main__':
 doctest.testmod()

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Ludisposed

6,41021858

share|improve this answer

share|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 1 hour ago

Ludisposed

6,41021858

answered 1 hour ago

Ludisposed

6,41021858

answered 1 hour ago

Ludisposed

6,41021858

6,41021858

add a comment | 

add a comment | 

up vote
1
down vote

Is there a more concise way to accomplish this?

Certainly.

For a start, the special cases are unnecessary. (They make me think that the code has been refactored from a recursive version).

Secondly, the expensive del temp[i:i+3] could be replaced with i += 3, and since the processing is no longer destructive temp is unnecessary. This simplifies the code to

def beautifulBinaryString(b):
 count, i = 0, 0
 while (i+3 <= len(b)):
 if b[i:i+3] == "010":
 count, i = count+1, i+3
 else: i += 1
 return count

share|improve this answer

answered 1 hour ago

Peter Taylor

14.5k2454

add a comment | 

up vote
1
down vote

Is there a more concise way to accomplish this?

Certainly.

For a start, the special cases are unnecessary. (They make me think that the code has been refactored from a recursive version).

Secondly, the expensive del temp[i:i+3] could be replaced with i += 3, and since the processing is no longer destructive temp is unnecessary. This simplifies the code to

def beautifulBinaryString(b):
 count, i = 0, 0
 while (i+3 <= len(b)):
 if b[i:i+3] == "010":
 count, i = count+1, i+3
 else: i += 1
 return count

share|improve this answer

answered 1 hour ago

Peter Taylor

14.5k2454

add a comment | 

up vote
1
down vote

up vote
1
down vote

Is there a more concise way to accomplish this?

Certainly.

For a start, the special cases are unnecessary. (They make me think that the code has been refactored from a recursive version).

Secondly, the expensive del temp[i:i+3] could be replaced with i += 3, and since the processing is no longer destructive temp is unnecessary. This simplifies the code to

def beautifulBinaryString(b):
 count, i = 0, 0
 while (i+3 <= len(b)):
 if b[i:i+3] == "010":
 count, i = count+1, i+3
 else: i += 1
 return count

share|improve this answer

answered 1 hour ago

Peter Taylor

14.5k2454

Is there a more concise way to accomplish this?

Certainly.

For a start, the special cases are unnecessary. (They make me think that the code has been refactored from a recursive version).

Secondly, the expensive del temp[i:i+3] could be replaced with i += 3, and since the processing is no longer destructive temp is unnecessary. This simplifies the code to

def beautifulBinaryString(b):
 count, i = 0, 0
 while (i+3 <= len(b)):
 if b[i:i+3] == "010":
 count, i = count+1, i+3
 else: i += 1
 return count

share|improve this answer

answered 1 hour ago

Peter Taylor

14.5k2454

share|improve this answer

share|improve this answer

answered 1 hour ago

Peter Taylor

14.5k2454

answered 1 hour ago

Peter Taylor

14.5k2454

answered 1 hour ago

Peter Taylor

14.5k2454

14.5k2454

add a comment | 

add a comment | 

 

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