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 8 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 8 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 8 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 8 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 8 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 8 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 8 mins ago
ahad pashayevahad pashayev
1
asked 8 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.
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "196"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
ahad pashayev is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f212513%2fgenerate-next-bigger-combination-in-lexicographic-order-in-python%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
ahad pashayev is a new contributor. Be nice, and check out our Code of Conduct.
ahad pashayev is a new contributor. Be nice, and check out our Code of Conduct.
ahad pashayev is a new contributor. Be nice, and check out our Code of Conduct.
ahad pashayev is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Code Review Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f212513%2fgenerate-next-bigger-combination-in-lexicographic-order-in-python%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
