Generate next bigger combination in lexicographic order in python
$begingroup$
I am new in python.I am developing my python skill by solving problems in Hackerrank. I am currently experiencing the problem with finding solution one interesting challenges. The task of the problem is the following:
I need to find a bigger string (within lexicographical order) that greater than given string.
Condition:
Given a word, create a new word by swapping some or all of its characters. This new word must meet two criteria:
It must be greater than the original word
It must be the smallest word that meets the first condition
For example, given the w="hega", the next largest word is "gahe" or w="abcd", the next largest word is "abdc".
I first try to do it through permutations methods in python. for some reason, that code did not pass all test cases.
Then, I wrote the following code.
This code works for majority cases (i believe), however, when the given string is too long, it takes a lot of time to converge (times out) and therefore it fails the test.
import string
w="dcba"
number=[n for n in range(1,27)]
letter=[x for x in string.ascii_lowercase]
dict_storage=dict(zip(string.ascii_lowercase,number))
def biggerIsGreater(w):
total_len=len(w)
count1=0
for _ in range(1,len(w)):
if count1==1:
break
else:
total_len-=1
count=total_len
for _ in range(0,len(w[:total_len])):
move_letter=dict_storage.get(w[total_len])
s1_max=dict_storage.get(max(set(w[:total_len])))
if move_letter>s1_max:
break
else:
count-=1
iter_letter=dict_storage.get(w[count])
if move_letter>iter_letter and count1==0:
move_index=total_len
loc_index=count
count1+=1
break
if count1==0:
return "no answer"
else:
left_str=w[:loc_index]
right_str=w[loc_index:move_index]
move_str=w[move_index:]
new_str=left_str+move_str+right_str
return new_str
print(biggerIsGreater(w))
sample input:
ab
bb
hefg
dhck
dkhc
Sample output:
ba
no answer
hegf
dhkc
hcdk
would you please have review the code and suggest more elegant way to solve this challange.
Thanks in advance
Reagrds
Ahad
python algorithm
asked 5 mins ago
ahad pashayevahad pashayev
1
New contributor
ahad pashayev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
I am new in python.I am developing my python skill by solving problems in Hackerrank. I am currently experiencing the problem with finding solution one interesting challenges. The task of the problem is the following:
I need to find a bigger string (within lexicographical order) that greater than given string.
Condition:
Given a word, create a new word by swapping some or all of its characters. This new word must meet two criteria:
It must be greater than the original word
It must be the smallest word that meets the first condition
For example, given the w="hega", the next largest word is "gahe" or w="abcd", the next largest word is "abdc".
I first try to do it through permutations methods in python. for some reason, that code did not pass all test cases.
Then, I wrote the following code.
This code works for majority cases (i believe), however, when the given string is too long, it takes a lot of time to converge (times out) and therefore it fails the test.
import string
w="dcba"
number=[n for n in range(1,27)]
letter=[x for x in string.ascii_lowercase]
dict_storage=dict(zip(string.ascii_lowercase,number))
def biggerIsGreater(w):
total_len=len(w)
count1=0
for _ in range(1,len(w)):
if count1==1:
break
else:
total_len-=1
count=total_len
for _ in range(0,len(w[:total_len])):
move_letter=dict_storage.get(w[total_len])
s1_max=dict_storage.get(max(set(w[:total_len])))
if move_letter>s1_max:
break
else:
count-=1
iter_letter=dict_storage.get(w[count])
if move_letter>iter_letter and count1==0:
move_index=total_len
loc_index=count
count1+=1
break
if count1==0:
return "no answer"
else:
left_str=w[:loc_index]
right_str=w[loc_index:move_index]
move_str=w[move_index:]
new_str=left_str+move_str+right_str
return new_str
print(biggerIsGreater(w))
sample input:
ab
bb
hefg
dhck
dkhc
Sample output:
ba
no answer
hegf
dhkc
hcdk
would you please have review the code and suggest more elegant way to solve this challange.
Thanks in advance
Reagrds
Ahad
python algorithm
asked 5 mins ago
ahad pashayevahad pashayev
1
New contributor
ahad pashayev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am new in python.I am developing my python skill by solving problems in Hackerrank. I am currently experiencing the problem with finding solution one interesting challenges. The task of the problem is the following:
I need to find a bigger string (within lexicographical order) that greater than given string.
Condition:
Given a word, create a new word by swapping some or all of its characters. This new word must meet two criteria:
It must be greater than the original word
It must be the smallest word that meets the first condition
For example, given the w="hega", the next largest word is "gahe" or w="abcd", the next largest word is "abdc".
I first try to do it through permutations methods in python. for some reason, that code did not pass all test cases.
Then, I wrote the following code.
This code works for majority cases (i believe), however, when the given string is too long, it takes a lot of time to converge (times out) and therefore it fails the test.
import string
w="dcba"
number=[n for n in range(1,27)]
letter=[x for x in string.ascii_lowercase]
dict_storage=dict(zip(string.ascii_lowercase,number))
def biggerIsGreater(w):
total_len=len(w)
count1=0
for _ in range(1,len(w)):
if count1==1:
break
else:
total_len-=1
count=total_len
for _ in range(0,len(w[:total_len])):
move_letter=dict_storage.get(w[total_len])
s1_max=dict_storage.get(max(set(w[:total_len])))
if move_letter>s1_max:
break
else:
count-=1
iter_letter=dict_storage.get(w[count])
if move_letter>iter_letter and count1==0:
move_index=total_len
loc_index=count
count1+=1
break
if count1==0:
return "no answer"
else:
left_str=w[:loc_index]
right_str=w[loc_index:move_index]
move_str=w[move_index:]
new_str=left_str+move_str+right_str
return new_str
print(biggerIsGreater(w))
sample input:
ab
bb
hefg
dhck
dkhc
Sample output:
ba
no answer
hegf
dhkc
hcdk
would you please have review the code and suggest more elegant way to solve this challange.
Thanks in advance
Reagrds
Ahad
python algorithm
asked 5 mins ago
ahad pashayevahad pashayev
1
New contributor
ahad pashayev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am new in python.I am developing my python skill by solving problems in Hackerrank. I am currently experiencing the problem with finding solution one interesting challenges. The task of the problem is the following:
I need to find a bigger string (within lexicographical order) that greater than given string.
Condition:
Given a word, create a new word by swapping some or all of its characters. This new word must meet two criteria:
It must be greater than the original word
It must be the smallest word that meets the first condition
For example, given the w="hega", the next largest word is "gahe" or w="abcd", the next largest word is "abdc".
I first try to do it through permutations methods in python. for some reason, that code did not pass all test cases.
Then, I wrote the following code.
This code works for majority cases (i believe), however, when the given string is too long, it takes a lot of time to converge (times out) and therefore it fails the test.
import string
w="dcba"
number=[n for n in range(1,27)]
letter=[x for x in string.ascii_lowercase]
dict_storage=dict(zip(string.ascii_lowercase,number))
def biggerIsGreater(w):
total_len=len(w)
count1=0
for _ in range(1,len(w)):
if count1==1:
break
else:
total_len-=1
count=total_len
for _ in range(0,len(w[:total_len])):
move_letter=dict_storage.get(w[total_len])
s1_max=dict_storage.get(max(set(w[:total_len])))
if move_letter>s1_max:
break
else:
count-=1
iter_letter=dict_storage.get(w[count])
if move_letter>iter_letter and count1==0:
move_index=total_len
loc_index=count
count1+=1
break
if count1==0:
return "no answer"
else:
left_str=w[:loc_index]
right_str=w[loc_index:move_index]
move_str=w[move_index:]
new_str=left_str+move_str+right_str
return new_str
print(biggerIsGreater(w))
sample input:
ab
bb
hefg
dhck
dkhc
Sample output:
ba
no answer
hegf
dhkc
hcdk
would you please have review the code and suggest more elegant way to solve this challange.
Thanks in advance
Reagrds
Ahad
python algorithm
python algorithm
asked 5 mins ago
ahad pashayevahad pashayev
1
New contributor
ahad pashayev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
ahad pashayevahad pashayev
1
New contributor
ahad pashayev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
ahad pashayevahad pashayev
1
New contributor
ahad pashayev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
ahad pashayevahad pashayev
1
asked 5 mins ago
ahad pashayevahad pashayev
1
1
New contributor
ahad pashayev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ahad pashayev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ahad pashayev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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