Length of longest non-repeating substring challenge using sliding window
$begingroup$
I am working on Leetcode challenge #3 (https://leetcode.com/problems/longest-substring-without-repeating-characters/)
Here is my solution using sliding window and a dictionary. I specifically added start = seen[s[i]]+1 to skip ahead. I am still told I am far slower than most people (for example, given abcdefgdabc, I am skipping abc when I see the second d. I thought this would save ton of time, but apparently this algorithm has a poor run time.
class Solution(object):
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
seen = {}
start = 0
max_size = 0
# check whether left pointer has reached the end yet
while start < len(s):
size = 0
for i in range(start, len(s)):
if s[i] in seen:
start = seen[s[i]]+1
seen = {}
break
else:
seen[s[i]] = i
size += 1
max_size = max(size, max_size)
return max_size
python algorithm
asked 16 mins ago
CppLearnerCppLearner
247210
$endgroup$
add a comment |
$begingroup$
I am working on Leetcode challenge #3 (https://leetcode.com/problems/longest-substring-without-repeating-characters/)
Here is my solution using sliding window and a dictionary. I specifically added start = seen[s[i]]+1 to skip ahead. I am still told I am far slower than most people (for example, given abcdefgdabc, I am skipping abc when I see the second d. I thought this would save ton of time, but apparently this algorithm has a poor run time.
class Solution(object):
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
seen = {}
start = 0
max_size = 0
# check whether left pointer has reached the end yet
while start < len(s):
size = 0
for i in range(start, len(s)):
if s[i] in seen:
start = seen[s[i]]+1
seen = {}
break
else:
seen[s[i]] = i
size += 1
max_size = max(size, max_size)
return max_size
python algorithm
asked 16 mins ago
CppLearnerCppLearner
247210
$endgroup$
add a comment |
$begingroup$
I am working on Leetcode challenge #3 (https://leetcode.com/problems/longest-substring-without-repeating-characters/)
Here is my solution using sliding window and a dictionary. I specifically added start = seen[s[i]]+1 to skip ahead. I am still told I am far slower than most people (for example, given abcdefgdabc, I am skipping abc when I see the second d. I thought this would save ton of time, but apparently this algorithm has a poor run time.
class Solution(object):
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
seen = {}
start = 0
max_size = 0
# check whether left pointer has reached the end yet
while start < len(s):
size = 0
for i in range(start, len(s)):
if s[i] in seen:
start = seen[s[i]]+1
seen = {}
break
else:
seen[s[i]] = i
size += 1
max_size = max(size, max_size)
return max_size
python algorithm
asked 16 mins ago
CppLearnerCppLearner
247210
$endgroup$
I am working on Leetcode challenge #3 (https://leetcode.com/problems/longest-substring-without-repeating-characters/)
Here is my solution using sliding window and a dictionary. I specifically added start = seen[s[i]]+1 to skip ahead. I am still told I am far slower than most people (for example, given abcdefgdabc, I am skipping abc when I see the second d. I thought this would save ton of time, but apparently this algorithm has a poor run time.
class Solution(object):
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
seen = {}
start = 0
max_size = 0
# check whether left pointer has reached the end yet
while start < len(s):
size = 0
for i in range(start, len(s)):
if s[i] in seen:
start = seen[s[i]]+1
seen = {}
break
else:
seen[s[i]] = i
size += 1
max_size = max(size, max_size)
return max_size
python algorithm
python algorithm
asked 16 mins ago
CppLearnerCppLearner
247210
asked 16 mins ago
CppLearnerCppLearner
247210
edited 9 mins ago
CppLearner
asked 16 mins ago
CppLearnerCppLearner
247210
asked 16 mins ago
CppLearnerCppLearner
247210
asked 16 mins ago
CppLearnerCppLearner
247210
247210
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
});
}
});
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%2f214211%2flength-of-longest-non-repeating-substring-challenge-using-sliding-window%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
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%2f214211%2flength-of-longest-non-repeating-substring-challenge-using-sliding-window%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
