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LCS variant using python and DP

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0

The code below barfs for some reason that I cannot understand. I'm currently honing my DP skills and one of the assignments is to make sure a string s2 can be constructed using a given string s1 whereby the only limitations is uppercase characters cannot be converted to lowercase. So we can delete and convert chars in s1 to upper case. The script returns True or False depending on whether the conversion is possible.

The basic idea is:

1) test len(s1) >= len(s2)

2) check no uppercase in s1 that aren't in s2

3) check that removal/uppercasing s1 yields s2

I am testing the code and some answers show up as 'YES' where they are supposed to be 'NO'. Is something wrong with the code?

def abbreviation(a, b):
 borig = b
 a = deque(a[::-1])
 b = deque(b[::-1])
 flag = True

 if len(a) < len(b):
 return 'NO'

 bc = 0
 ac = 0

 arr = []
 test = [i for i in a if i.isupper() and i not in b]
 if len(test) > 0:
 return 'NO'
 while bc < len(b) and len(a) >= len(b) and len(a) - ac >= len(b) - bc:
 flag = True

 if a[ac] == b[bc] or a[ac].upper() == b[bc]:
 flag = False
 arr.append(a[ac])
 bc += 1
 ac +=1
 #print 'increment valid'
 if flag:
 ac +=1
 #print 'icrement invalid'

 print arr
 if ''.join([i.upper() for i in arr[::-1]]) == borig:
 return 'YES'
 else:
 return 'NO'

for i in range(0,20,2):
 print abbreviation(f[i],f[i+1])

python

share|improve this question

asked Mar 8 at 12:25

S. L.S. L.

118112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Could you add an example of a test case where your code fails?
  
  – Valentino
  Mar 8 at 12:53
  
  

  
  
  
  

add a comment | 

0

The code below barfs for some reason that I cannot understand. I'm currently honing my DP skills and one of the assignments is to make sure a string s2 can be constructed using a given string s1 whereby the only limitations is uppercase characters cannot be converted to lowercase. So we can delete and convert chars in s1 to upper case. The script returns True or False depending on whether the conversion is possible.

The basic idea is:

1) test len(s1) >= len(s2)

2) check no uppercase in s1 that aren't in s2

3) check that removal/uppercasing s1 yields s2

I am testing the code and some answers show up as 'YES' where they are supposed to be 'NO'. Is something wrong with the code?

def abbreviation(a, b):
 borig = b
 a = deque(a[::-1])
 b = deque(b[::-1])
 flag = True

 if len(a) < len(b):
 return 'NO'

 bc = 0
 ac = 0

 arr = []
 test = [i for i in a if i.isupper() and i not in b]
 if len(test) > 0:
 return 'NO'
 while bc < len(b) and len(a) >= len(b) and len(a) - ac >= len(b) - bc:
 flag = True

 if a[ac] == b[bc] or a[ac].upper() == b[bc]:
 flag = False
 arr.append(a[ac])
 bc += 1
 ac +=1
 #print 'increment valid'
 if flag:
 ac +=1
 #print 'icrement invalid'

 print arr
 if ''.join([i.upper() for i in arr[::-1]]) == borig:
 return 'YES'
 else:
 return 'NO'

for i in range(0,20,2):
 print abbreviation(f[i],f[i+1])

python

share|improve this question

asked Mar 8 at 12:25

S. L.S. L.

118112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Could you add an example of a test case where your code fails?
  
  – Valentino
  Mar 8 at 12:53
  
  

  
  
  
  

add a comment | 

The code below barfs for some reason that I cannot understand. I'm currently honing my DP skills and one of the assignments is to make sure a string s2 can be constructed using a given string s1 whereby the only limitations is uppercase characters cannot be converted to lowercase. So we can delete and convert chars in s1 to upper case. The script returns True or False depending on whether the conversion is possible.

The basic idea is:

1) test len(s1) >= len(s2)

2) check no uppercase in s1 that aren't in s2

3) check that removal/uppercasing s1 yields s2

I am testing the code and some answers show up as 'YES' where they are supposed to be 'NO'. Is something wrong with the code?

def abbreviation(a, b):
 borig = b
 a = deque(a[::-1])
 b = deque(b[::-1])
 flag = True

 if len(a) < len(b):
 return 'NO'

 bc = 0
 ac = 0

 arr = []
 test = [i for i in a if i.isupper() and i not in b]
 if len(test) > 0:
 return 'NO'
 while bc < len(b) and len(a) >= len(b) and len(a) - ac >= len(b) - bc:
 flag = True

 if a[ac] == b[bc] or a[ac].upper() == b[bc]:
 flag = False
 arr.append(a[ac])
 bc += 1
 ac +=1
 #print 'increment valid'
 if flag:
 ac +=1
 #print 'icrement invalid'

 print arr
 if ''.join([i.upper() for i in arr[::-1]]) == borig:
 return 'YES'
 else:
 return 'NO'

for i in range(0,20,2):
 print abbreviation(f[i],f[i+1])

python

share|improve this question

asked Mar 8 at 12:25

S. L.S. L.

118112

def abbreviation(a, b):
 borig = b
 a = deque(a[::-1])
 b = deque(b[::-1])
 flag = True

 if len(a) < len(b):
 return 'NO'

 bc = 0
 ac = 0

 arr = []
 test = [i for i in a if i.isupper() and i not in b]
 if len(test) > 0:
 return 'NO'
 while bc < len(b) and len(a) >= len(b) and len(a) - ac >= len(b) - bc:
 flag = True

 if a[ac] == b[bc] or a[ac].upper() == b[bc]:
 flag = False
 arr.append(a[ac])
 bc += 1
 ac +=1
 #print 'increment valid'
 if flag:
 ac +=1
 #print 'icrement invalid'

 print arr
 if ''.join([i.upper() for i in arr[::-1]]) == borig:
 return 'YES'
 else:
 return 'NO'

for i in range(0,20,2):
 print abbreviation(f[i],f[i+1])

share|improve this question

asked Mar 8 at 12:25

S. L.S. L.

118112

share|improve this question

asked Mar 8 at 12:25

S. L.S. L.

118112

asked Mar 8 at 12:25

S. L.S. L.

118112

asked Mar 8 at 12:25

S. L.S. L.

118112