post_href stringlengths 57 213 | python_solutions stringlengths 71 22.3k | slug stringlengths 3 77 | post_title stringlengths 1 100 | user stringlengths 3 29 | upvotes int64 -20 1.2k | views int64 0 60.9k | problem_title stringlengths 3 77 | number int64 1 2.48k | acceptance float64 0.14 0.91 | difficulty stringclasses 3
values | __index_level_0__ int64 0 34k |
|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/isomorphic-strings/discuss/381930/Python-multiple-solutions | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
last_position = [0]*512
for idx in range(len(s)):
if last_position[ord(s[idx])] != last_position[ord(t[idx]) + 256]:
return False
last_position[ord(s[idx])] = last_position[ord(t[idx]) + 256] = id... | isomorphic-strings | Python multiple solutions | amchoukir | 7 | 1,700 | isomorphic strings | 205 | 0.426 | Easy | 3,400 |
https://leetcode.com/problems/isomorphic-strings/discuss/1058562/Python3-simple-and-explained-line-by-line | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
#corner case: because s and t are the same length, any string of 1 character is isomorphic
if len(s) == 1:
return True
#Dictionary to map the characters from s with the characters of t
charDict = {}
... | isomorphic-strings | Python3 simple and explained line by line | luisfmiranda97 | 5 | 576 | isomorphic strings | 205 | 0.426 | Easy | 3,401 |
https://leetcode.com/problems/isomorphic-strings/discuss/656556/Simple-Python-3-solution-runtime-beats-94.7-submissions | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
i = 0
h = {}
while i < len(s):
if t[i] not in h:
if s[i] in h.values():
return False
else:
h[t[i]] = s[i]
else:
if h... | isomorphic-strings | Simple Python 3 solution; runtime beats 94.7% submissions | Swap24 | 4 | 233 | isomorphic strings | 205 | 0.426 | Easy | 3,402 |
https://leetcode.com/problems/isomorphic-strings/discuss/2377315/more-efficient-oror-using-Python-oror-very-small | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
z= zip(s,t)
if len(set(z))==len(set(s))==len(set(t)):
return True
return False | isomorphic-strings | more efficient || using Python || very small 🤞 | shane6123 | 3 | 130 | isomorphic strings | 205 | 0.426 | Easy | 3,403 |
https://leetcode.com/problems/isomorphic-strings/discuss/2350597/Python-solution-using-hash-function-to-compute-an-index-with-a-key-into-an-array-of-slots | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
if len(s)!=len(t) or len(set(s))!=len(set(t)):
return False
hashmap={}
for i in range(len(s)):
if s[i] not in hashmap:
hashmap[s[i]]=t[i]
if hashmap[s[i]]... | isomorphic-strings | Python solution using hash function to compute an index with a key into an array of slots | RohanRob | 3 | 184 | isomorphic strings | 205 | 0.426 | Easy | 3,404 |
https://leetcode.com/problems/isomorphic-strings/discuss/2439195/Simple-One-Liner-or-Python | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
return len(set(zip(s,t))) == len(set(s)) == len(set(t)) | isomorphic-strings | Simple One Liner | Python | Abhi_-_- | 2 | 199 | isomorphic strings | 205 | 0.426 | Easy | 3,405 |
https://leetcode.com/problems/isomorphic-strings/discuss/2157200/Python-3-Set-a-dict-Memory-less-than-99.41 | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
replace_dict = dict()
dict_values = replace_dict.values()
for char, rep_char in zip(s, t):
if char not in replace_dict:
if rep_char not in dict_values:
replace_dict[char] = rep_cha... | isomorphic-strings | Python 3 - Set a dict - Memory less than 99.41% | thanhinterpol | 2 | 89 | isomorphic strings | 205 | 0.426 | Easy | 3,406 |
https://leetcode.com/problems/isomorphic-strings/discuss/2709969/Easy-Python-Solution | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
m1 = [s.index(i) for i in s]
m2 = [t.index(i)for i in t]
return m1==m2 | isomorphic-strings | Easy Python Solution | mariusep | 1 | 4 | isomorphic strings | 205 | 0.426 | Easy | 3,407 |
https://leetcode.com/problems/isomorphic-strings/discuss/2179500/PYTHON-oror-Simple-and-Easy-Solution-%2B-65-faster | class Solution(object):
def isIsomorphic(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
# LOGIC - Map the 1st string characters in correspondace to the 2nd one, simultaneously
# check if the 2nd string characters are fit with the maping table or not.
... | isomorphic-strings | PYTHON || Simple and Easy Solution + 65% faster | navinchndr012 | 1 | 143 | isomorphic strings | 205 | 0.426 | Easy | 3,408 |
https://leetcode.com/problems/isomorphic-strings/discuss/2168026/most-easiest-python3-with-one-loop | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
st_dict, ts_dict = {}, {}
for i,j in zip(s,t):
st_dict[i]=j
ts_dict[j]=i
return list(st_dict.keys()) == list(ts_dict.values()) | isomorphic-strings | most easiest python3 with one loop | lq291464 | 1 | 130 | isomorphic strings | 205 | 0.426 | Easy | 3,409 |
https://leetcode.com/problems/isomorphic-strings/discuss/2088126/python3-O(n)-oror-O(1)O(n)-55ms-54.84 | class Solution:
# O(N) || O(26n) or O(1) if we are granteed that s and t will be lowercase english letters else its O(n)
# 55ms 54.84%
def isIsomorphic(self, s: str, t: str) -> bool:
return self.isomorphic(s) == self.isomorphic(t)
def isomorphic(self, string):
letterDict = dict()
... | isomorphic-strings | python3 O(n) || O(1)\O(n) 55ms 54.84% | arshergon | 1 | 124 | isomorphic strings | 205 | 0.426 | Easy | 3,410 |
https://leetcode.com/problems/isomorphic-strings/discuss/1335531/Python%3A-Simple-solution-using-dictionary | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
d = {}
for i, j in zip(s, t):
if i not in d.keys():
d[i] = j
for i, j in zip(s, t):
if j != d[i] or len(list(d.values())) != len(list(set(list(d.values())))):
retur... | isomorphic-strings | Python: Simple solution using dictionary | farhan_kapadia | 1 | 308 | isomorphic strings | 205 | 0.426 | Easy | 3,411 |
https://leetcode.com/problems/isomorphic-strings/discuss/746932/Python3%3A-Simple-faster-than-75 | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
mappedChars = set()
myDic = {}
for idx in range(len(s)):
if s[idx] in myDic:
if t[idx] != myDic[s[idx]]: return False
else:
if t[idx] in mappedChars: return False
... | isomorphic-strings | Python3: Simple faster than 75% | nachiketsd | 1 | 321 | isomorphic strings | 205 | 0.426 | Easy | 3,412 |
https://leetcode.com/problems/isomorphic-strings/discuss/733261/Python3-3-approaches | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
mps, mpt = {}, {}
for i, (ss, tt) in enumerate(zip(s, t)):
if mps.get(ss) != mpt.get(tt): return False
mps[ss] = mpt[tt] = i
return True | isomorphic-strings | [Python3] 3 approaches | ye15 | 1 | 163 | isomorphic strings | 205 | 0.426 | Easy | 3,413 |
https://leetcode.com/problems/isomorphic-strings/discuss/733261/Python3-3-approaches | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
return len(set(zip(s, t))) == len(set(s)) == len(set(t)) | isomorphic-strings | [Python3] 3 approaches | ye15 | 1 | 163 | isomorphic strings | 205 | 0.426 | Easy | 3,414 |
https://leetcode.com/problems/isomorphic-strings/discuss/733261/Python3-3-approaches | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
fn = lambda x: tuple(map({}.setdefault, x, range(len(s))))
return fn(s) == fn(t) | isomorphic-strings | [Python3] 3 approaches | ye15 | 1 | 163 | isomorphic strings | 205 | 0.426 | Easy | 3,415 |
https://leetcode.com/problems/isomorphic-strings/discuss/551292/Python-simple-solution-20-ms-faster-than-96.70-O(n2) | class Solution(object):
def isIsomorphic(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
char_dict = dict()
for i in range(len(s)):
if s[i] not in char_dict:
if t[i] not in char_dict.values():
char_di... | isomorphic-strings | Python simple solution 20 ms, faster than 96.70% , O(n^2) | hemina | 1 | 391 | isomorphic strings | 205 | 0.426 | Easy | 3,416 |
https://leetcode.com/problems/isomorphic-strings/discuss/511504/Python-find-and-save-the-position-of-each-character | class Solution(object):
def isIsomorphic(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
s_index = [ s.find(i) for i in s ]
t_index = [ t.find(i) for i in t ]
return s_index == t_index | isomorphic-strings | Python find and save the position of each character | kobewang | 1 | 285 | isomorphic strings | 205 | 0.426 | Easy | 3,417 |
https://leetcode.com/problems/isomorphic-strings/discuss/359128/Solution-in-Python-3 | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
S, T, j, k = {} ,{}, 0, 0
for a,b in zip(s,t):
if a not in S: S[a], j = j, j + 1
if b not in T: T[b], k = k, k + 1
return all([S[s[i]] == T[t[i]] for i in range(len(s))])
- Junaid Mansuri
(LeetCode ID)@hotmail.com | isomorphic-strings | Solution in Python 3 | junaidmansuri | 1 | 493 | isomorphic strings | 205 | 0.426 | Easy | 3,418 |
https://leetcode.com/problems/isomorphic-strings/discuss/2847272/Python-solution-T-C-%3A-0(str1-%2B-str2) | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
if len(s)!=len(t):
return False
dic1 = {}
dic2 = {}
for i, j in zip(s ,t):
if i not in dic1 and j not in dic2:
dic1[i] = j
dic2[j] = i
elif i in dic1 an... | isomorphic-strings | Python solution T-C :- 0(str1 + str2) | kartik_5051 | 0 | 2 | isomorphic strings | 205 | 0.426 | Easy | 3,419 |
https://leetcode.com/problems/isomorphic-strings/discuss/2845992/Mapping-s-and-t | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
# character map the char from string s to string t for replacement
# eg "egg" -> "add"; # e->a, g->d
# if this mapping doesn't match then its not an isomorphic string
mapS, mapT = {}, {}
for cs, ct in zip(s, t):... | isomorphic-strings | Mapping s and t | keshan-spec | 0 | 1 | isomorphic strings | 205 | 0.426 | Easy | 3,420 |
https://leetcode.com/problems/isomorphic-strings/discuss/2837937/Isomorphic-Strings | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
return len(set(zip(s, t))) == len(set(s)) == len(set(t)) | isomorphic-strings | Isomorphic Strings | arthur54342 | 0 | 1 | isomorphic strings | 205 | 0.426 | Easy | 3,421 |
https://leetcode.com/problems/isomorphic-strings/discuss/2832077/python-oror-simple-solution-oror-hashmapdictionary | class Solution:
def isIsomorphic(self, s1: str, s2: str) -> bool:
def f(s: str, d: dict) -> list[int]:
ans = [] # answer
cnt = 0 # count of diff chars in s
# go through every char in s
for c in s:
# if we haven't seen c yet
... | isomorphic-strings | python || simple solution || hashmap/dictionary | wduf | 0 | 8 | isomorphic strings | 205 | 0.426 | Easy | 3,422 |
https://leetcode.com/problems/isomorphic-strings/discuss/2830649/Isomorphic-Strings-in-Python | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
f = ''
d = {}
if len(set(s)) == len(set(t)):
for i, j in enumerate(s):
d[j] = t[i]
for k in s:
if k in d:
f+=d[k]
if f == t:
... | isomorphic-strings | Isomorphic Strings in Python | prink879 | 0 | 4 | isomorphic strings | 205 | 0.426 | Easy | 3,423 |
https://leetcode.com/problems/isomorphic-strings/discuss/2828049/Python-Get-and-compare-2-signatures-based-on-new-char-ordering | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
return self.signature(s) == self.signature(t)
def signature(self, s):
'''
The idea is we can isomorphise 2 strings by taking their new char oder
As in, for "egg" and "add", it is 101111 and 101111,
... | isomorphic-strings | [Python] Get and compare 2 signatures based on new char ordering | graceiscoding | 0 | 5 | isomorphic strings | 205 | 0.426 | Easy | 3,424 |
https://leetcode.com/problems/isomorphic-strings/discuss/2827582/Python-Solution | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
m1=dict()
m2=dict()
for i in range(len(s)):
if(s[i] in m1.keys() and t[i] not in m2.keys()):
return False
if(s[i] not in m1.keys() and t[i] in m2.keys()):
return False
... | isomorphic-strings | Python Solution | CEOSRICHARAN | 0 | 3 | isomorphic strings | 205 | 0.426 | Easy | 3,425 |
https://leetcode.com/problems/isomorphic-strings/discuss/2819042/Python3-Solution | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
char_map = {}
new_chars = set()
for i, char in enumerate(s):
if char not in char_map:
if t[i] in new_chars:
return False
new_chars.add(t[i])
char_... | isomorphic-strings | Python3 Solution | patvera1 | 0 | 6 | isomorphic strings | 205 | 0.426 | Easy | 3,426 |
https://leetcode.com/problems/isomorphic-strings/discuss/2819032/Python-My-O(n)-Solution | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
char_mapping = {}
t_chars_used = set()
for i, (s_char, t_char) in enumerate(zip(s,t)):
if s_char in char_mapping:
# fail here if previous mapping does not hold
... | isomorphic-strings | [Python] My O(n) Solution | manytenks | 0 | 4 | isomorphic strings | 205 | 0.426 | Easy | 3,427 |
https://leetcode.com/problems/isomorphic-strings/discuss/2814161/Very-simple-python-O(n) | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
_map={}
for i in range(len(s)):
if s[i] in _map.keys():
if (_map[s[i]] != t[i]):
return False
_map[s[i]]=t[i]
if len(set(list(_map.values()))) != len(set(list(_map.ke... | isomorphic-strings | Very simple python O(n) | ATHBuys | 0 | 4 | isomorphic strings | 205 | 0.426 | Easy | 3,428 |
https://leetcode.com/problems/isomorphic-strings/discuss/2812080/python-easy-solution | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
map_dict = {}
for i,j in zip(s, t):
if i not in map_dict:
map_dict[i] = j
elif map_dict[i] != j:
return False
if len(set(map_dict.keys())) != len(set(map_dict.values()))... | isomorphic-strings | python easy solution | cool_rabbit | 0 | 5 | isomorphic strings | 205 | 0.426 | Easy | 3,429 |
https://leetcode.com/problems/isomorphic-strings/discuss/2811844/using-set-and-zip | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
return len(set(s)) == len(set(t)) ==len(set(zip(s,t))) | isomorphic-strings | using set and zip | roger880327 | 0 | 2 | isomorphic strings | 205 | 0.426 | Easy | 3,430 |
https://leetcode.com/problems/isomorphic-strings/discuss/2807240/Solved-with-Dictionary | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
x = {}
for i in range(0,len(s)):
if s[i] in x:
# it is in x so check if the respecting t[i] is the same value as before
if x[s[i]]== t[i]:
# if yesit is good means the val... | isomorphic-strings | Solved with Dictionary | user6407Ay | 0 | 3 | isomorphic strings | 205 | 0.426 | Easy | 3,431 |
https://leetcode.com/problems/isomorphic-strings/discuss/2804329/hashmap-%2B-hash-set-0(n)-time%2Bspace-99-faster-Python3 | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
if len(t)!=len(s): return False
map = {}
invalid = set()
for i,c in enumerate(s):
if c not in map:
if t[i] in invalid: return False
map[c] = t[i]
invalid.add(t[... | isomorphic-strings | hashmap + hash set 0(n) time+space 99% faster Python3 | iSyqozz512 | 0 | 5 | isomorphic strings | 205 | 0.426 | Easy | 3,432 |
https://leetcode.com/problems/isomorphic-strings/discuss/2801213/Python-O(n)-solution-90 | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
if len(s) != len(t):
return true
values = set()
mappy={}
for i in range(len(s)):
a,b = s[i],t[i]
if a in mappy:
if mappy[a] != b:
return ... | isomorphic-strings | Python O(n) solution 90% | shreeshail | 0 | 3 | isomorphic strings | 205 | 0.426 | Easy | 3,433 |
https://leetcode.com/problems/isomorphic-strings/discuss/2800894/Python-or-single-line | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
return len(set(s))==len(set(t))==len(set(zip(s,t))) | isomorphic-strings | Python | single line | SAI_KRISHNA_PRATHAPANENI | 0 | 1 | isomorphic strings | 205 | 0.426 | Easy | 3,434 |
https://leetcode.com/problems/isomorphic-strings/discuss/2791890/Solution%3A-Python3-98-faster | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
comDict = {}
if len(s) != len(t):
return False
for i in range(len(s)):
if s[i] in comDict.keys():
if comDict[s[i]] != t[i]:
return False
else:
... | isomorphic-strings | Solution: Python3 98% faster | wanatabeyuu | 0 | 4 | isomorphic strings | 205 | 0.426 | Easy | 3,435 |
https://leetcode.com/problems/isomorphic-strings/discuss/2791849/Python-Easy-solution-using-Hashmap-with-explaination. | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
if len(s) != len(t):
return False
if len(set(s)) != len(set(t)):
return False
conn = {}
for i in range(len(s)):
if s[i] not in conn:
conn[s[i]] = t[i]
elif... | isomorphic-strings | ✅ Python Easy solution using Hashmap with explaination. ✅ | raghupalash | 0 | 2 | isomorphic strings | 205 | 0.426 | Easy | 3,436 |
https://leetcode.com/problems/isomorphic-strings/discuss/2786767/Python-easy-solution | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
intS = 0
listS = []
dictS = {}
for char in s:
if char not in dictS:
dictS[char] = intS
intS += 1
listS.append(dictS[char])
intT = 0
listT... | isomorphic-strings | Python easy solution | kamman626 | 0 | 3 | isomorphic strings | 205 | 0.426 | Easy | 3,437 |
https://leetcode.com/problems/isomorphic-strings/discuss/2785865/Isomorphic-Strings-or-Easy-Python-solution | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
mapST,mapTS = {}, {}
if len(s) != len(t):
return False
else:
for i in range(len(s)):
if (s[i] in mapST) and mapST[s[i]] != t[i]:
return False
elif (t[i]... | isomorphic-strings | Isomorphic Strings | Easy Python solution | nishanrahman1994 | 0 | 5 | isomorphic strings | 205 | 0.426 | Easy | 3,438 |
https://leetcode.com/problems/isomorphic-strings/discuss/2769178/Simple-Python-Solution | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
n = len(s)
mapping = dict()
seen = set()
for i in range(n):
if s[i] not in mapping:
if t[i] in seen:
return False
else:
mapping... | isomorphic-strings | Simple Python Solution | mansoorafzal | 0 | 7 | isomorphic strings | 205 | 0.426 | Easy | 3,439 |
https://leetcode.com/problems/isomorphic-strings/discuss/2766221/Using-maps-and-array-easy-to-understand. | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
if len(s) != len(t):
return False
map = {}
set = []
for i in range(len(s)):
if s[i] not in map:
if t[i] not in set:
map[s[i]] = t[i]
... | isomorphic-strings | Using maps and array, easy to understand. | karanvirsagar98 | 0 | 3 | isomorphic strings | 205 | 0.426 | Easy | 3,440 |
https://leetcode.com/problems/isomorphic-strings/discuss/2751659/Solution-using-ONE-HashTable | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
map = {}
if len(s) == len(t):
for i in range(len(s)):
if t[i] in map.values():
if s[i] not in map.keys():
return False
... | isomorphic-strings | Solution using ONE HashTable | zaberraiyan | 0 | 5 | isomorphic strings | 205 | 0.426 | Easy | 3,441 |
https://leetcode.com/problems/isomorphic-strings/discuss/2749030/Simple-Python-Solution-using-length-and-zip | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
z = zip(s, t)
return len(set(s)) == len(set(z)) == len(set(t)) | isomorphic-strings | Simple Python Solution using length and zip | vivekrajyaguru | 0 | 3 | isomorphic strings | 205 | 0.426 | Easy | 3,442 |
https://leetcode.com/problems/isomorphic-strings/discuss/2747260/python-Hashmap-solution | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
# O(n), O(n)
st, ts = {}, {}
for c1, c2 in zip(s, t):
# for i in range(len(s)):
# c1, c2 = s[i], t[i]
if (c1 in st and st[c1] != c2 or
c2 in ts and ts[c2] != c1):
... | isomorphic-strings | python Hashmap solution | sahilkumar158 | 0 | 4 | isomorphic strings | 205 | 0.426 | Easy | 3,443 |
https://leetcode.com/problems/isomorphic-strings/discuss/2740162/simple-Python-solution-using-Dictionary | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
s_len = len(s)
t_len = len(t)
if s_len != t_len:
return False
else:
diff_dict = {}
for i in range(s_len):
if s[i] in diff_dict and diff_dict[s[i]] != t[i]:
... | isomorphic-strings | simple Python solution, using Dictionary | don_masih | 0 | 4 | isomorphic strings | 205 | 0.426 | Easy | 3,444 |
https://leetcode.com/problems/isomorphic-strings/discuss/2737891/easy-way-in-python | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
d={}
for i in range(len(s)):
if s[i] not in d:
if t[i] in d.values():
return False
else:
d[s[i]]=t[i]
else:
if d[s[i]]!=t[i]... | isomorphic-strings | easy way in python | sindhu_300 | 0 | 7 | isomorphic strings | 205 | 0.426 | Easy | 3,445 |
https://leetcode.com/problems/isomorphic-strings/discuss/2716910/Short-and-easy-Python-solution-with-dictionaries | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
#check that strings are of equal length
if len(s) != len(t): return False
n = len(s)
if n == 0: return True
D_s_t, D_t_s = dict(), dict()
for i in range(n):
if s[i] in D_s_t.keys() and D_s_t[s[i]] != ... | isomorphic-strings | Short and easy Python solution with dictionaries | tatiana_ospv | 0 | 3 | isomorphic strings | 205 | 0.426 | Easy | 3,446 |
https://leetcode.com/problems/isomorphic-strings/discuss/2712155/Python-simple-solution-using-Dictionary | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
mapST,mapTS={},{}
for c1,c2 in zip(s,t):
if( (c1 in mapST and mapST[c1]!=c2) or (c2 in mapTS and mapTS[c2]!=c1) ):
return False
mapST[c1]=c2
mapTS[c2]=c1
return True | isomorphic-strings | Python simple solution using Dictionary | Raghunath_Reddy | 0 | 10 | isomorphic strings | 205 | 0.426 | Easy | 3,447 |
https://leetcode.com/problems/isomorphic-strings/discuss/2710776/iterative-python-method-(most-unique-solution) | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
if not len(s)==len(t):
return False
s = list(s)
t = list(t)
hash = {}
a=0
for i in s:
if i in hash:
hash[i].append(t[a])
if i not in hash:
... | isomorphic-strings | iterative python method (most unique solution) | sahityasetu1996 | 0 | 2 | isomorphic strings | 205 | 0.426 | Easy | 3,448 |
https://leetcode.com/problems/isomorphic-strings/discuss/2685418/Pythondictionary-Easy-understanding-by-record-each-position | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
dicS = {}
dicT = {}
for i in range(len(s)):
if s[i] not in dicS:
dicS[s[i]] = [i]
else:
dicS[s[i]].append(i)
if t[i] not in dicT:
dicT[... | isomorphic-strings | [Python/dictionary] Easy understanding by record each position | Allen_Huang | 0 | 48 | isomorphic strings | 205 | 0.426 | Easy | 3,449 |
https://leetcode.com/problems/isomorphic-strings/discuss/2675105/Isomorphic-Strings | class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
MAX_CHARS = 256
n = len(s)
m = len(t)
if n != m:
return False
marked = [False] * MAX_CHARS
map = [-1] * MAX_CHARS
for i in range(n):
if map[ord(s[i])] == -1:
... | isomorphic-strings | Isomorphic Strings | jashii96 | 0 | 6 | isomorphic strings | 205 | 0.426 | Easy | 3,450 |
https://leetcode.com/problems/reverse-linked-list/discuss/2458632/Easy-oror-0-ms-oror-100-oror-Fully-Explained-oror-Java-C%2B%2B-Python-JS-C-Python3-(Recursive-and-Iterative) | class Solution(object):
def reverseList(self, head):
# Initialize prev pointer as NULL...
prev = None
# Initialize the curr pointer as the head...
curr = head
# Run a loop till curr points to NULL...
while curr:
# Initialize next pointer as the next pointe... | reverse-linked-list | Easy || 0 ms || 100% || Fully Explained || Java, C++, Python, JS, C, Python3 (Recursive & Iterative) | PratikSen07 | 92 | 6,800 | reverse linked list | 206 | 0.726 | Easy | 3,451 |
https://leetcode.com/problems/reverse-linked-list/discuss/2458632/Easy-oror-0-ms-oror-100-oror-Fully-Explained-oror-Java-C%2B%2B-Python-JS-C-Python3-(Recursive-and-Iterative) | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
# Initialize prev pointer as NULL...
prev = None
# Initialize the curr pointer as the head...
curr = head
# Run a loop till curr points to NULL...
while curr:
# Initiali... | reverse-linked-list | Easy || 0 ms || 100% || Fully Explained || Java, C++, Python, JS, C, Python3 (Recursive & Iterative) | PratikSen07 | 92 | 6,800 | reverse linked list | 206 | 0.726 | Easy | 3,452 |
https://leetcode.com/problems/reverse-linked-list/discuss/469204/PythonJSJavaC%2B%2B-O(n)-by-recursion-w-Comment | class Solution:
def helper(self, prev, cur):
if cur:
# locate next hopping node
next_hop = cur.next
# reverse direction
cur.next = prev
return self.helper( cur, next_hop)
else:
# ne... | reverse-linked-list | Python/JS/Java/C++ O(n) by recursion [w/ Comment] | brianchiang_tw | 7 | 1,400 | reverse linked list | 206 | 0.726 | Easy | 3,453 |
https://leetcode.com/problems/reverse-linked-list/discuss/469204/PythonJSJavaC%2B%2B-O(n)-by-recursion-w-Comment | class Solution:
def reverseList(self, head: ListNode) -> ListNode:
prev, cur = None, head
while cur:
# locate next hoppoing node
next_hop = cur.next
# reverse direction
cur.next = prev
... | reverse-linked-list | Python/JS/Java/C++ O(n) by recursion [w/ Comment] | brianchiang_tw | 7 | 1,400 | reverse linked list | 206 | 0.726 | Easy | 3,454 |
https://leetcode.com/problems/reverse-linked-list/discuss/642637/faster-than-95.78-in-python-or-Iterative-Solution | class Solution:
def reverseList(self, head: ListNode) -> ListNode:
#I will use the following head, prev, temp
prev = None
while head: #while head is present then this loop executes
#first I will assign the value of head to the temp variable
temp = he... | reverse-linked-list | faster than 95.78% in python | Iterative Solution | pritomlily | 6 | 633 | reverse linked list | 206 | 0.726 | Easy | 3,455 |
https://leetcode.com/problems/reverse-linked-list/discuss/2684169/Python-2-Easy-Way-To-Reverse-Linked-List-or-99-Faster-or-Fast-and-Simple-Solution | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
curr = head
prev = None
while curr:
next = curr.next
curr.next = prev
prev = curr
curr = next
return prev | reverse-linked-list | ✔️ Python 2 Easy Way To Reverse Linked List | 99% Faster | Fast and Simple Solution | pniraj657 | 4 | 706 | reverse linked list | 206 | 0.726 | Easy | 3,456 |
https://leetcode.com/problems/reverse-linked-list/discuss/2684169/Python-2-Easy-Way-To-Reverse-Linked-List-or-99-Faster-or-Fast-and-Simple-Solution | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
if not head:
return head
stack = []
temp = head
while temp:
stack.append(temp)
temp = temp.next
head = temp = stack.pop()
... | reverse-linked-list | ✔️ Python 2 Easy Way To Reverse Linked List | 99% Faster | Fast and Simple Solution | pniraj657 | 4 | 706 | reverse linked list | 206 | 0.726 | Easy | 3,457 |
https://leetcode.com/problems/reverse-linked-list/discuss/2375440/Python-Accurate-Solution-Two-Pointers-oror-Documented | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
resultNode = ListNode() # reversed list container
curNode = head
while curNode:
keyNode = curNode # before swapping links, take the backup
curNode = curNod... | reverse-linked-list | [Python] Accurate Solution - Two Pointers || Documented | Buntynara | 4 | 190 | reverse linked list | 206 | 0.726 | Easy | 3,458 |
https://leetcode.com/problems/reverse-linked-list/discuss/1630780/Python3-Short-and-Simple-or-In-place-O(1)-space-or-Iterative | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
prev = None
while head:
nextHead = head.next
head.next, prev = prev, head
head = nextHead
return prev | reverse-linked-list | [Python3] Short & Simple | In-place O(1) space | Iterative | PatrickOweijane | 4 | 431 | reverse linked list | 206 | 0.726 | Easy | 3,459 |
https://leetcode.com/problems/reverse-linked-list/discuss/1049857/Python3 | class Solution:
def reverseList(self, head: ListNode) -> ListNode:
curr=head
prev=None
while curr:
next=curr.next
curr.next=prev
prev=curr
curr=next
return prev | reverse-linked-list | Python3 | samarthnehe | 4 | 511 | reverse linked list | 206 | 0.726 | Easy | 3,460 |
https://leetcode.com/problems/reverse-linked-list/discuss/2415236/Python-Iterative%3A-Beats-99-oror-Recursive%3A-Beats-87-with-full-working-explanation | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: # Time: O(n) and Space: O(1)
prev, cur = None, head
while cur: # let cur 3
temp = cur.next # nxt = 4
cur.next = prev # 3 -> 2
prev = cur # pr... | reverse-linked-list | Python [Iterative: Beats 99% || Recursive: Beats 87%] with full working explanation | DanishKhanbx | 3 | 311 | reverse linked list | 206 | 0.726 | Easy | 3,461 |
https://leetcode.com/problems/reverse-linked-list/discuss/2415236/Python-Iterative%3A-Beats-99-oror-Recursive%3A-Beats-87-with-full-working-explanation | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: # Time: O(n) and Space: O(n)
if head == None or head.next == None: # head = 2 & 2 -> 3
return head
newHead = self.reverseList(head.next) # newHead = Func(head=3) returns head = 3
head.n... | reverse-linked-list | Python [Iterative: Beats 99% || Recursive: Beats 87%] with full working explanation | DanishKhanbx | 3 | 311 | reverse linked list | 206 | 0.726 | Easy | 3,462 |
https://leetcode.com/problems/reverse-linked-list/discuss/1767848/Python-Simple-Python-Solution-Using-Iterative-Approach-With-While-Loop | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
current_node = head
previous_node = None
while current_node != None:
next_node = current_node.next
current_node.next = previous_node
previous_node = current_node
current_node = next_node
return previous_node | reverse-linked-list | [ Python ] ✔✔ Simple Python Solution Using Iterative Approach With While Loop 🔥✌ | ASHOK_KUMAR_MEGHVANSHI | 3 | 228 | reverse linked list | 206 | 0.726 | Easy | 3,463 |
https://leetcode.com/problems/reverse-linked-list/discuss/2299710/Very-simple-python-solution | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
ans = None
while head:
ans = ListNode(head.val, ans)
head = head.next
return ans | reverse-linked-list | Very simple python solution | IshanKute | 2 | 321 | reverse linked list | 206 | 0.726 | Easy | 3,464 |
https://leetcode.com/problems/reverse-linked-list/discuss/2249495/206.-My-Python-Solution | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
rev = None
while head:
temp = rev
rev = head
head = head.next
rev.next = temp
return rev | reverse-linked-list | 206. My Python Solution | JunyiLin | 2 | 130 | reverse linked list | 206 | 0.726 | Easy | 3,465 |
https://leetcode.com/problems/reverse-linked-list/discuss/2249495/206.-My-Python-Solution | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
rev = None
return self.__reverse(head, rev)
def __reverse(self, head, rev):
if not head:
return rev
temp = rev
rev = head
head = head.next
rev.next = t... | reverse-linked-list | 206. My Python Solution | JunyiLin | 2 | 130 | reverse linked list | 206 | 0.726 | Easy | 3,466 |
https://leetcode.com/problems/reverse-linked-list/discuss/1987977/Python-easiest-recursion-O(n) | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
def helper(head):
if not head.next:
self.final = head
return head
# At the penultimate note, below will return the last node
next_ = helper(head.next)
# revers... | reverse-linked-list | Python easiest recursion O(n) | diet_pepsi | 2 | 293 | reverse linked list | 206 | 0.726 | Easy | 3,467 |
https://leetcode.com/problems/reverse-linked-list/discuss/1946364/Python-Solution-or-Iterative-or-Recursion-or-O(n) | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
# Two pointer solution itertaively where T O(n) and M O(1)
prev, curr = None, head
while curr:
temp = curr.next
curr.next = prev
prev = curr
curr = temp
return prev | reverse-linked-list | Python Solution | Iterative | Recursion | O(n) | nikhitamore | 2 | 229 | reverse linked list | 206 | 0.726 | Easy | 3,468 |
https://leetcode.com/problems/reverse-linked-list/discuss/1946364/Python-Solution-or-Iterative-or-Recursion-or-O(n) | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
#recursive T O(n) and M O(n)
if not head:
return None
newHead = head
if head.next:
newHead = self.reverseList(head.next)
head.next.next = head
head.next = None
return newHead | reverse-linked-list | Python Solution | Iterative | Recursion | O(n) | nikhitamore | 2 | 229 | reverse linked list | 206 | 0.726 | Easy | 3,469 |
https://leetcode.com/problems/reverse-linked-list/discuss/1380420/Python3-faster-than-99.95 | class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if head is None or head.next is None:
return head
cur = head
prev = None
while cur is not None:
p1 = cur
cur = cur.next
p1.next = prev
prev = p1
retu... | reverse-linked-list | Python3 - faster than 99.95% | CC_CheeseCake | 2 | 203 | reverse linked list | 206 | 0.726 | Easy | 3,470 |
https://leetcode.com/problems/reverse-linked-list/discuss/2314942/Python-solution-using-two-pointer-or-Reverse-Linked-List | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
prevNode, currNode = None, head
while currNode:
nextNode = currNode.next
currNode.next = prevNode
prevNode = currNode
currNode = nextNode
return prevNode | reverse-linked-list | Python solution using two pointer | Reverse Linked List | nishanrahman1994 | 1 | 192 | reverse linked list | 206 | 0.726 | Easy | 3,471 |
https://leetcode.com/problems/reverse-linked-list/discuss/2128833/Python3-Solution | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
if not head or not head.next:#判断为空或长度为1时的情况
return head
last = self.reverseList(head.next) #迭代的含义就是使head不停的前进，做reverse
head.next.next = head #
... | reverse-linked-list | Python3 Solution | qywang | 1 | 80 | reverse linked list | 206 | 0.726 | Easy | 3,472 |
https://leetcode.com/problems/reverse-linked-list/discuss/2128833/Python3-Solution | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
rever = None
while head:
nextnode = head.next
head.next = rever
rever = head
head = nextnode
... | reverse-linked-list | Python3 Solution | qywang | 1 | 80 | reverse linked list | 206 | 0.726 | Easy | 3,473 |
https://leetcode.com/problems/reverse-linked-list/discuss/1977436/Python3-Runtime%3A-43ms-66.97-memory%3A-15.5mb-57.44 | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
if not head: return head
prev = None
current = head
while current is not None:
next = current.next
current.next = prev
prev = current
current = next
... | reverse-linked-list | Python3 Runtime: 43ms 66.97% memory: 15.5mb 57.44% | arshergon | 1 | 100 | reverse linked list | 206 | 0.726 | Easy | 3,474 |
https://leetcode.com/problems/reverse-linked-list/discuss/1977436/Python3-Runtime%3A-43ms-66.97-memory%3A-15.5mb-57.44 | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
if not head: return head
stack = []
current = head
while current is not None:
stack.append(current.val)
current = current.next
newCurrentNode = head
... | reverse-linked-list | Python3 Runtime: 43ms 66.97% memory: 15.5mb 57.44% | arshergon | 1 | 100 | reverse linked list | 206 | 0.726 | Easy | 3,475 |
https://leetcode.com/problems/reverse-linked-list/discuss/1926119/Python-Simple-Elegant-Iterative-Solution-Memory-Less-Than-94.93-With-Comments | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
# prev will be the new head
# curr used so we don't manipulate head
prev, curr = None, head
while curr:
# To place current here after each proccesing
temp ... | reverse-linked-list | Python Simple Elegant Iterative Solution Memory Less Than 94.93%, With Comments, | Hejita | 1 | 87 | reverse linked list | 206 | 0.726 | Easy | 3,476 |
https://leetcode.com/problems/reverse-linked-list/discuss/1510577/Python-or-Recursive-24-ms-or-Iterative-32-ms-or-Extra-List-28-ms | class Solution:
def reverseList(self, head):
if not head or not head.next:
return head
def rec(node):
global new_head
if not node.next:
new_head = node
return new_head
rec(node.next).next = nod... | reverse-linked-list | Python | Recursive 24 ms | Iterative 32 ms | Extra List 28 ms | ocancemal1996 | 1 | 327 | reverse linked list | 206 | 0.726 | Easy | 3,477 |
https://leetcode.com/problems/reverse-linked-list/discuss/1510577/Python-or-Recursive-24-ms-or-Iterative-32-ms-or-Extra-List-28-ms | class Solution:
def reverseList(self, head):
if not head or not head.next:
return head
left = head
right = head.next
while left and right:
right.next, left = left, right.next
if left and right:
left.next, right = right, lef... | reverse-linked-list | Python | Recursive 24 ms | Iterative 32 ms | Extra List 28 ms | ocancemal1996 | 1 | 327 | reverse linked list | 206 | 0.726 | Easy | 3,478 |
https://leetcode.com/problems/reverse-linked-list/discuss/1510577/Python-or-Recursive-24-ms-or-Iterative-32-ms-or-Extra-List-28-ms | class Solution:
def reverseList(self, head):
if not head or not head.next:
return head
temp = head
node_list = []
while temp:
node_list.append(temp)
temp = temp.next
node_list[0].next = None
for idx, node in en... | reverse-linked-list | Python | Recursive 24 ms | Iterative 32 ms | Extra List 28 ms | ocancemal1996 | 1 | 327 | reverse linked list | 206 | 0.726 | Easy | 3,479 |
https://leetcode.com/problems/reverse-linked-list/discuss/407408/Python-Easy-solution | class Solution:
def reverseList(self, head: ListNode) -> ListNode:
a = []
temp = head
if not temp:
return None
while(temp):
a.append(temp.val)
temp = temp.next
a = a[::-1]
head = ListNode(a[0])
temp = head
for i in range(1,len(a)):
temp.next = ListNode(a[i])
temp = temp.next
return hea... | reverse-linked-list | Python Easy solution | saffi | 1 | 605 | reverse linked list | 206 | 0.726 | Easy | 3,480 |
https://leetcode.com/problems/reverse-linked-list/discuss/2834118/python-oror-simple-solution-oror-iterative | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
# if linkedlist empty or only one node
if (not head) or (not head.next):
return head
# previous node, next node
prev = next = None
# iterate through list
while head:
... | reverse-linked-list | python || simple solution || iterative | wduf | 0 | 6 | reverse linked list | 206 | 0.726 | Easy | 3,481 |
https://leetcode.com/problems/reverse-linked-list/discuss/2813071/python3-solution-with-simple-explanation | class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
prev = None
while head:
head.next,head,prev = prev,head.next,head
return prev | reverse-linked-list | python3 solution with simple explanation | pardunmeplz | 0 | 4 | reverse linked list | 206 | 0.726 | Easy | 3,482 |
https://leetcode.com/problems/course-schedule/discuss/1627381/Simple-and-Easy-Topological-Sorting-code-beats-97.63-python-submissions | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
graph=defaultdict(list)
indegree={}
#initialising dictionary
for i in range(numCourses):
indegree[i]=0
#filling graph and indegree dictionaries
for child,parent in prerequisites:
graph[parent].append(chil... | course-schedule | Simple and Easy Topological Sorting code, beats 97.63% python submissions | RaghavGupta22 | 11 | 1,400 | course schedule | 207 | 0.454 | Medium | 3,483 |
https://leetcode.com/problems/course-schedule/discuss/1802039/Python-Iterative-DFS | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
graph = { courseNum : [] for courseNum in range(numCourses) }
for course, prerequisite in prerequisites:
graph[course].append(prerequisite)
Course = namedtuple('Course... | course-schedule | Python Iterative DFS | Rush_P | 2 | 318 | course schedule | 207 | 0.454 | Medium | 3,484 |
https://leetcode.com/problems/course-schedule/discuss/1546562/Python3-Commented-Khans-Algorithm-code-85-Speed | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
#Data Structures and Variables
in_degree = [0] * numCourses
adj_list = [[] for x in range(numCourses)]
queue = []
counter = 0
#building in_degree list and adj_list
... | course-schedule | Python3- Commented Khans Algorithm code - 85% Speed | 17pchaloori | 2 | 206 | course schedule | 207 | 0.454 | Medium | 3,485 |
https://leetcode.com/problems/course-schedule/discuss/2642520/Python-DFS-a-stack-dict-and-visited-array | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
courseGraph = [[] for _ in range(numCourses)]
for ai, bi in prerequisites:
courseGraph[ai].append(bi)
visited = [False] * numCourses
stack = defaultdict(bool)
for c in r... | course-schedule | Python DFS, a stack dict and visited array | hellboy11 | 1 | 263 | course schedule | 207 | 0.454 | Medium | 3,486 |
https://leetcode.com/problems/course-schedule/discuss/2505782/Python-Solution-using-DFS | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
#map each course with preq list
preMap = {i : [] for i in range(numCourses)}
for crs, preq in prerequisites:
preMap[crs].append(preq)
#visitSet
visitSet = set()
def dfs(crs):
#found loop
if c... | course-schedule | Python Solution using DFS | nikhitamore | 1 | 88 | course schedule | 207 | 0.454 | Medium | 3,487 |
https://leetcode.com/problems/course-schedule/discuss/2318527/Python-DFS-Beats-92-with-full-working-explanation | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool: # Time: O(Nodes(V) + Edges(prerequisites)) and Space: O(V + E)
preMap = {i: [] for i in range(numCourses)} # init hashmap for storing course as a key for prerequisite as values
for crs, pre in prerequi... | course-schedule | Python [DFS / Beats 92%] with full working explanation | DanishKhanbx | 1 | 141 | course schedule | 207 | 0.454 | Medium | 3,488 |
https://leetcode.com/problems/course-schedule/discuss/2117700/simple-iterative-dfs | class Solution:
def canFinish(self, numCourses: int, req: List[List[int]]) -> List[int]:
courses = [[] for _ in range(numCourses)]
for c, pre in req:
courses[c].append(pre)
visiting = {}
for i in range(numCourses):
if visiting.get(i... | course-schedule | simple iterative dfs | gabhinav001 | 1 | 60 | course schedule | 207 | 0.454 | Medium | 3,489 |
https://leetcode.com/problems/course-schedule/discuss/2000011/Python3-Runtime%3A-136ms-49.50-Memory%3A-17mb-34.96 | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
reqMap = {i:[] for i in range(numCourses)}
for crs, pre in prerequisites:
reqMap[crs].append(pre)
visited = set()
def dfs(crs):
if crs in visited:
... | course-schedule | Python3 Runtime: 136ms 49.50% Memory: 17mb 34.96% | arshergon | 1 | 64 | course schedule | 207 | 0.454 | Medium | 3,490 |
https://leetcode.com/problems/course-schedule/discuss/1768311/Python3-solution-or-DFS-or-commmented | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
def dfs(graph, node, been):
if not graph[node]:
return False
if node in been:
return True # cycle detected
been.add(node)
for n in gra... | course-schedule | Python3 solution | DFS | commmented | FlorinnC1 | 1 | 224 | course schedule | 207 | 0.454 | Medium | 3,491 |
https://leetcode.com/problems/course-schedule/discuss/1672351/Python3-BFS-97-or-Detailed-Explained-or-Beginner-Friendly | class Solution:
def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
# init DAG: Course pre -> Course
d = defaultdict(list)
# indegree list: there are # courses as pre-requisites for takign Course A
indeg = [0] * numCourses
# c: course; p: pre ... | course-schedule | Python3 BFS 97% | Detailed Explained | Beginner Friendly | doneowth | 1 | 188 | course schedule | 207 | 0.454 | Medium | 3,492 |
https://leetcode.com/problems/course-schedule/discuss/1394342/Clear-Python3-DFS-Solution-using-DFS-template-98-time | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
"""
>>> sol = Solution()
>>> numCourses = 2
>>> prerequisites = [[1,0]]
>>> sol.canFinish(numCourses, prerequisites)
True
>>> numCourses = 2
>>> prerequisites... | course-schedule | Clear Python3 DFS Solution using DFS template 98% time | alfrednerd | 1 | 144 | course schedule | 207 | 0.454 | Medium | 3,493 |
https://leetcode.com/problems/course-schedule/discuss/1364906/Straightforward-%2B-Clean-Python | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
preqs = collections.defaultdict(set)
graph = collections.defaultdict(set)
for c, p in prerequisites:
preqs[c].add(p)
graph[c].add(p)
graph[p].ad... | course-schedule | Straightforward + Clean Python | Pythagoras_the_3rd | 1 | 166 | course schedule | 207 | 0.454 | Medium | 3,494 |
https://leetcode.com/problems/course-schedule/discuss/658915/Python3-topological-sort | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
graph = {} # digraph
for u, v in prerequisites:
graph.setdefault(v, []).append(u)
def fn(x):
"""Return True if cycle is detected."""
if visited[x]: re... | course-schedule | [Python3] topological sort | ye15 | 1 | 216 | course schedule | 207 | 0.454 | Medium | 3,495 |
https://leetcode.com/problems/course-schedule/discuss/658915/Python3-topological-sort | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
indeg = [0]*numCourses
graph = {}
for u, v in prerequisites:
indeg[u] += 1
graph.setdefault(v, []).append(u)
stack = [i for i, x in enumerate(indeg) if not ... | course-schedule | [Python3] topological sort | ye15 | 1 | 216 | course schedule | 207 | 0.454 | Medium | 3,496 |
https://leetcode.com/problems/course-schedule/discuss/2825170/Course-Schedule-DFS | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
#create a graph based on prerequisites
graph = self.buildGraph(numCourses,prerequisites)
#record visited nodes, record nodes in path
self.visited = [False]*numCourses
self.path = [F... | course-schedule | Course Schedule DFS | Romantic_taxi_driver | 0 | 5 | course schedule | 207 | 0.454 | Medium | 3,497 |
https://leetcode.com/problems/course-schedule/discuss/2815458/Python-Error-(Dictionary-changes-size-during-iteration)-Requires-Assistance | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
adjList = defaultdict(list)
courses = {}
for course, prereq in prerequisites:
adjList[course].append(prereq)
def dfs(course):
if course in courses:
... | course-schedule | Python Error (Dictionary changes size during iteration) - Requires Assistance | BENJI_GAO | 0 | 4 | course schedule | 207 | 0.454 | Medium | 3,498 |
https://leetcode.com/problems/course-schedule/discuss/2815409/Python-solution | class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
# 0: Unchecked, 1:Checking, 2:Completed
Status = [0] * numCourses
Prerequisite = defaultdict(list)
for cur, pre in prerequisites:
Prerequisite[cur].append(pre)
... | course-schedule | Python solution | maomao1010 | 0 | 9 | course schedule | 207 | 0.454 | Medium | 3,499 |
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Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.