Time complexity regarding O(LogN)
Let's say we have the following code.
def problem(n):
list =
for i in range(n):
list.append(i)
length = len(list)
return list
The program has time complexity of O(n) if we don't calculate len(list). But if we do, will the time complexity be O(n * log(n)) or O(n^2)? .
python time-complexity
edited Nov 15 '18 at 5:55
Clock Slave
2,29352461
asked Nov 15 '18 at 5:51
Nick SunNick Sun
1
add a comment |
Let's say we have the following code.
def problem(n):
list =
for i in range(n):
list.append(i)
length = len(list)
return list
The program has time complexity of O(n) if we don't calculate len(list). But if we do, will the time complexity be O(n * log(n)) or O(n^2)? .
python time-complexity
edited Nov 15 '18 at 5:55
Clock Slave
2,29352461
asked Nov 15 '18 at 5:51
Nick SunNick Sun
1
len(list)is O(1).
– Warren Weckesser
Nov 15 '18 at 5:55
1
Why on earth would it be O(n log n) or O(n^2)? Even iflenwas not O(1), the "worst implementation possible" would still be at most O(n)...
– Julien
Nov 15 '18 at 6:07
add a comment |
Let's say we have the following code.
def problem(n):
list =
for i in range(n):
list.append(i)
length = len(list)
return list
The program has time complexity of O(n) if we don't calculate len(list). But if we do, will the time complexity be O(n * log(n)) or O(n^2)? .
python time-complexity
edited Nov 15 '18 at 5:55
Clock Slave
2,29352461
asked Nov 15 '18 at 5:51
Nick SunNick Sun
1
Let's say we have the following code.
def problem(n):
list =
for i in range(n):
list.append(i)
length = len(list)
return list
The program has time complexity of O(n) if we don't calculate len(list). But if we do, will the time complexity be O(n * log(n)) or O(n^2)? .
python time-complexity
python time-complexity
edited Nov 15 '18 at 5:55
Clock Slave
2,29352461
asked Nov 15 '18 at 5:51
Nick SunNick Sun
1
edited Nov 15 '18 at 5:55
Clock Slave
2,29352461
asked Nov 15 '18 at 5:51
Nick SunNick Sun
1
edited Nov 15 '18 at 5:55
Clock Slave
2,29352461
edited Nov 15 '18 at 5:55
Clock Slave
2,29352461
edited Nov 15 '18 at 5:55
Clock Slave
2,29352461
2,29352461
asked Nov 15 '18 at 5:51
Nick SunNick Sun
1
asked Nov 15 '18 at 5:51
Nick SunNick Sun
1
asked Nov 15 '18 at 5:51
Nick SunNick Sun
1
1
len(list)is O(1).
– Warren Weckesser
Nov 15 '18 at 5:55
1
Why on earth would it be O(n log n) or O(n^2)? Even iflenwas not O(1), the "worst implementation possible" would still be at most O(n)...
– Julien
Nov 15 '18 at 6:07
add a comment |
len(list)is O(1).
– Warren Weckesser
Nov 15 '18 at 5:55
1
Why on earth would it be O(n log n) or O(n^2)? Even iflenwas not O(1), the "worst implementation possible" would still be at most O(n)...
– Julien
Nov 15 '18 at 6:07
len(list) is O(1).– Warren Weckesser
Nov 15 '18 at 5:55
len(list) is O(1).– Warren Weckesser
Nov 15 '18 at 5:55
1
1
Why on earth would it be O(n log n) or O(n^2)? Even if
len was not O(1), the "worst implementation possible" would still be at most O(n)...– Julien
Nov 15 '18 at 6:07
Why on earth would it be O(n log n) or O(n^2)? Even if
len was not O(1), the "worst implementation possible" would still be at most O(n)...– Julien
Nov 15 '18 at 6:07
add a comment |
1 Answer
1
active
oldest
votes
No, the len() function has constant time in python and it is not dependent on the length of the element, your time complexity for the above code would remain O(N) governed by your for i in range(n) loop. Here is the time complexity for many CPython functions, like len()! (Get Length in table)
answered Nov 15 '18 at 5:56
d_kennetzd_kennetz
1
add a comment |
Your Answer
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: "1"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
});
}
});
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%2fstackoverflow.com%2fquestions%2f53313215%2ftime-complexity-regarding-ologn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
No, the len() function has constant time in python and it is not dependent on the length of the element, your time complexity for the above code would remain O(N) governed by your for i in range(n) loop. Here is the time complexity for many CPython functions, like len()! (Get Length in table)
answered Nov 15 '18 at 5:56
d_kennetzd_kennetz
1
add a comment |
No, the len() function has constant time in python and it is not dependent on the length of the element, your time complexity for the above code would remain O(N) governed by your for i in range(n) loop. Here is the time complexity for many CPython functions, like len()! (Get Length in table)
answered Nov 15 '18 at 5:56
d_kennetzd_kennetz
1
add a comment |
No, the len() function has constant time in python and it is not dependent on the length of the element, your time complexity for the above code would remain O(N) governed by your for i in range(n) loop. Here is the time complexity for many CPython functions, like len()! (Get Length in table)
answered Nov 15 '18 at 5:56
d_kennetzd_kennetz
1
No, the len() function has constant time in python and it is not dependent on the length of the element, your time complexity for the above code would remain O(N) governed by your for i in range(n) loop. Here is the time complexity for many CPython functions, like len()! (Get Length in table)
answered Nov 15 '18 at 5:56
d_kennetzd_kennetz
1
answered Nov 15 '18 at 5:56
d_kennetzd_kennetz
1
answered Nov 15 '18 at 5:56
d_kennetzd_kennetz
1
answered Nov 15 '18 at 5:56
d_kennetzd_kennetz
1
1
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- 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.
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%2fstackoverflow.com%2fquestions%2f53313215%2ftime-complexity-regarding-ologn%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
len(list)is O(1).– Warren Weckesser
Nov 15 '18 at 5:55
1
Why on earth would it be O(n log n) or O(n^2)? Even if
lenwas not O(1), the "worst implementation possible" would still be at most O(n)...– Julien
Nov 15 '18 at 6:07