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 |
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https://leetcode.com/problems/grid-illumination/discuss/1497659/Beginner-Friendly-oror-96-faster-oror-Easy-to-understand | class Solution:
def gridIllumination(self, n: int, lamps: List[List[int]], queries: List[List[int]]) -> List[int]:
row,col,dig1,dig2 = defaultdict(int),defaultdict(int),defaultdict(int),defaultdict(int)
def switch(i,j,isOn):
val = 1 if isOn else -1
row[i] += val
col[j] += val
dig1[i+j]+=val
dig2[i-j]+=val
def check(x,y):
return 1 if row[x] or col[y] or dig1[x+y] or dig2[x-y] else 0
seen = set()
for x,y in set([tuple(lamp) for lamp in lamps]):
seen.add((x,y))
switch(x,y,1)
res = []
for x,y in queries:
res.append(check(x,y))
for dx,dy in [(0,0),(0,1),(0,-1),(1,0),(-1,0),(1,1),(-1,-1),(1,-1),(-1,1)]:
if (x+dx,y+dy) in seen:
seen.remove((x+dx,y+dy))
switch(x+dx,y+dy,0)
return res | grid-illumination | 📌📌 Beginner-Friendly || 96% faster || Easy-to-understand 🐍 | abhi9Rai | 0 | 58 | grid illumination | 1,001 | 0.362 | Hard | 16,300 |
https://leetcode.com/problems/find-common-characters/discuss/721548/Python-Faster-than-77.67-and-Better-Space-than-86.74 | class Solution:
def commonChars(self, A: List[str]) -> List[str]:
alphabet = string.ascii_lowercase
d = {c: 0 for c in alphabet}
for k, v in d.items():
d[k] = min([word.count(k) for word in A])
res = []
for c, n in d.items():
if n > 0:
res += [c] * n
return res | find-common-characters | Python Faster than 77.67% and Better Space than 86.74 | parkershamblin | 8 | 997 | find common characters | 1,002 | 0.683 | Easy | 16,301 |
https://leetcode.com/problems/find-common-characters/discuss/1037047/Python3-easy-solution | class Solution:
def commonChars(self, A: List[str]) -> List[str]:
ans = []
for i in set(A[0]):
x = []
for j in A:
x.append(j.count(i))
a = 0
while a < min(x):
ans.append(i)
a += 1
return ans | find-common-characters | Python3 easy solution | EklavyaJoshi | 4 | 421 | find common characters | 1,002 | 0.683 | Easy | 16,302 |
https://leetcode.com/problems/find-common-characters/discuss/1150217/Python3-Simple-And-Readable-Solution | class Solution:
def commonChars(self, A: List[str]) -> List[str]:
arr = []
for i in set(A[0]):
ans = [A[0].count(i)]
for j in A[1:]:
if(i in j):
ans.append(j.count(i))
if(len(ans) == len(A)):
arr += ([i] * min(ans))
return arr | find-common-characters | [Python3] Simple And Readable Solution | Lolopola | 3 | 279 | find common characters | 1,002 | 0.683 | Easy | 16,303 |
https://leetcode.com/problems/find-common-characters/discuss/1310060/O(n2)-Solution(python3) | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
if len(words)==1:
return list(*words)
output=[]
for i in words[0]:
temp=0
for j,k in enumerate(words[1:]):
if i in k:
words[j+1]=k.replace(i,"_",1)
temp+=1
if temp==len(words)-1:
output.append(i)
return output | find-common-characters | O(n^2) Solution(python3) | _jorjis | 2 | 276 | find common characters | 1,002 | 0.683 | Easy | 16,304 |
https://leetcode.com/problems/find-common-characters/discuss/1053340/Python-Faster-than-98-Easy-to-Understand | class Solution:
def commonChars(self, A: List[str]) -> List[str]:
A.sort(key=lambda x: len(x)) # in-place sort by shortest to longest length
letter_count = {letter: A[0].count(letter) for letter in A[0]} # dict for shortest word: key = letter, value=count of letter
for letter in letter_count.keys():
for word in A[1:]: # No need to check A[0] as that is the reference point (i.e shortest word)
tmp_count = word.count(letter)
if tmp_count == 0:
# If letter not found in word, skip this letter
letter_count[letter] = 0
break
if tmp_count < letter_count[letter]:
letter_count[letter] = word.count(letter)
return [letter for letter, count in letter_count.items() for _ in range(count)] | find-common-characters | Python Faster than 98% Easy to Understand | peatear-anthony | 2 | 628 | find common characters | 1,002 | 0.683 | Easy | 16,305 |
https://leetcode.com/problems/find-common-characters/discuss/355203/Python-Concise-Solution-which-beats-97 | class Solution:
def commonChars(self, A: List[str]) -> List[str]:
spec=A[0]
answer=[]
for each in spec:
flag=True
for each_str in A[1:]:
if each in each_str:
continue
else:
flag=False
break
if flag:
answer.append(each)
for i in range(1,len(A)):
A[i]=A[i].replace(each,"",1)
return answer | find-common-characters | Python Concise Solution which beats 97% | Sprinter1999 | 2 | 661 | find common characters | 1,002 | 0.683 | Easy | 16,306 |
https://leetcode.com/problems/find-common-characters/discuss/2619650/two-dictionaries-python | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
d1=defaultdict(int)
for l in range(len(words[0])):
z =words[0][l]
d1[z]+=1
for i in range(1,len(words)):
d2=defaultdict(int)
z=words[i]
for l in z:
d2[l]+=1
for k in d1:
d1[k]=min(d1[k],d2[k])
ret=[]
for k in sorted(d1):
while(d1[k]>0):
ret.append(k)
d1[k]-=1
return ret | find-common-characters | two dictionaries python | abhayCodes | 1 | 36 | find common characters | 1,002 | 0.683 | Easy | 16,307 |
https://leetcode.com/problems/find-common-characters/discuss/1990526/easy-python-code | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
d = []
n = len(words)
for i in words[0]:
flag = 0
for j in words:
if i not in j:
flag = 1
break
if flag == 0:
d.append(i)
for k in range(len(words)):
for l in range(len(words[k])):
if words[k][l] == i:
words[k] = words[k][:l] + words[k][l+1:]
break
return d | find-common-characters | easy python code | dakash682 | 1 | 379 | find common characters | 1,002 | 0.683 | Easy | 16,308 |
https://leetcode.com/problems/find-common-characters/discuss/1793561/85-lesser-memory-or-Simple-python-solution | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
f = words[0]
if len(words)==1:
return list(f)
words = [list(i) for i in words[1:]]
lst = []
c= 0
for i in f:
for j in range(len(words)):
if i in words[j]:
words[j].remove(i)
c += 1
if c == len(words):
lst.append(i)
c = 0
return lst | find-common-characters | 85% lesser memory | Simple python solution | Coding_Tan3 | 1 | 203 | find common characters | 1,002 | 0.683 | Easy | 16,309 |
https://leetcode.com/problems/find-common-characters/discuss/1563368/Python3-dollarolution | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
d, d1, l = {}, {}, []
for i in words[0]:
if i not in d:
d[i] = 1
else:
d[i] += 1
for i in words[1:]:
d1 = {}
for j in i:
if j not in d:
continue
elif j in d and j not in d1:
d1[j] = 1
elif j in d1 and d[j] > d1[j]:
d1[j] += 1
d = d1
for i in d1:
for j in range(d1[i]):
l.append(i)
return l | find-common-characters | Python3 $olution | AakRay | 1 | 339 | find common characters | 1,002 | 0.683 | Easy | 16,310 |
https://leetcode.com/problems/find-common-characters/discuss/2847921/Python-Solution | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
res=[]
for i in words[0]:
k=0
for j in range(1,len(words)):
words[j]=list(words[j])
if(i not in words[j]):
k+=1
else:
words[j].remove(i)
if(k==0):
res.append(i)
return res | find-common-characters | Python Solution | CEOSRICHARAN | 0 | 1 | find common characters | 1,002 | 0.683 | Easy | 16,311 |
https://leetcode.com/problems/find-common-characters/discuss/2840096/Python3-Beats-98-or-Optimized-lee215's-code | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
if len(words) == 1: return list(words)
r = collections.Counter(words[0])
for word in words[1:]:
r &= collections.Counter(word)
return list(r.elements()) | find-common-characters | [Python3] Beats 98% | Optimized lee215's code | m0nxt3r | 0 | 2 | find common characters | 1,002 | 0.683 | Easy | 16,312 |
https://leetcode.com/problems/find-common-characters/discuss/2831873/Python-Solution-Hashmap-oror-Explained | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
l=[]
for i in range(len(words[0])): #select words from words[0]
r=words[0][i]
c=1 #count the same character as r in every other remaining strings in words
for j in range(1,len(words)):
if r in words[j]:
c+=1
words[j]=words[j].replace(r,"",1) #delete that r from the string
#if count of the particular character is equal to len(words)
#that means every string containg it
if c==len(words):
l.append(r)
return l | find-common-characters | Python Solution - Hashmap || Explained✔ | T1n1_B0x1 | 0 | 2 | find common characters | 1,002 | 0.683 | Easy | 16,313 |
https://leetcode.com/problems/find-common-characters/discuss/2724687/Easy-Python-Solution | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
op = []
check = words[0]
new_lst = words
for i in check:
cnt=0
nxt_lst = []
for j in new_lst:
if i in j:
nxt_lst.append(j.replace(i,'',1))
cnt+=1
else:
nxt_lst.append(j)
new_lst = nxt_lst
if cnt==len(words):
op.append(i)
return op | find-common-characters | Easy Python Solution | BAparna97 | 0 | 13 | find common characters | 1,002 | 0.683 | Easy | 16,314 |
https://leetcode.com/problems/find-common-characters/discuss/2603929/Python-solution-O(n) | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
common_chs = set(words[0])
for i in range(1, len(words)):
common_chs = set(words[i]).intersection(common_chs)
dict_count = {}
for word in words:
mini_dict = {}
for ch in word:
if ch not in common_chs:
continue
if ch in mini_dict:
mini_dict[ch] += 1
else:
mini_dict[ch] = 1
for ch, cnt in mini_dict.items():
if ch in dict_count:
dict_count[ch].add(cnt)
else:
dict_count[ch] = {cnt}
result = []
for char, set_cnts in dict_count.items():
for _ in range(min(set_cnts)):
result.append(char)
return result | find-common-characters | Python solution O(n) | samanehghafouri | 0 | 75 | find common characters | 1,002 | 0.683 | Easy | 16,315 |
https://leetcode.com/problems/find-common-characters/discuss/2473907/Python-or-Counter-or-2-versions | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
res = reduce(lambda x, y: Counter(x) & Counter(y), words)
return res.elements() | find-common-characters | Python | Counter | 2 versions | Wartem | 0 | 152 | find common characters | 1,002 | 0.683 | Easy | 16,316 |
https://leetcode.com/problems/find-common-characters/discuss/2439776/Python-Beginner-Friendly | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
x=words.pop()
res=[]
for i in x:
for j in range(len(words)):
if i not in words[j]:break
words[j]=words[j].replace(i,"",1)
else:res.append(i)
return res | find-common-characters | Python Beginner Friendly | imamnayyar86 | 0 | 186 | find common characters | 1,002 | 0.683 | Easy | 16,317 |
https://leetcode.com/problems/find-common-characters/discuss/2420209/Simple-and-fast-python3-solution-88-faster | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
res = []
for _ in words[0]:
if all(_ in __ for __ in words[1:]):
# okay, all strings have the same char, but
# some of them have more than one
res.append(_)
# so we need to update other strings and remove
# the char we found
words[1:] = [__.replace(_,'',1) for __ in words[1:]]
return res | find-common-characters | Simple and fast python3 solution 🚀🚀🚀 88% faster 👌 | TheCodeBeer | 0 | 117 | find common characters | 1,002 | 0.683 | Easy | 16,318 |
https://leetcode.com/problems/find-common-characters/discuss/1982521/Pythonor-Dictionary-or-min-or-hash-table | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
word0 = Counter(words[0])
print(word0)
for key in word0.keys():
for i in range(0,len(words)):
if key in words[i]:
word0[key] = min(word0[key],words[i].count(key))
else:
word0[key] = 0
output = ''
for k , v in word0.items():
if v >= 1:
output += k * v
return list(output) | find-common-characters | Python| Dictionary | min | hash table | user4774i | 0 | 139 | find common characters | 1,002 | 0.683 | Easy | 16,319 |
https://leetcode.com/problems/find-common-characters/discuss/1848277/Python-Simple-and-Concise! | class Solution(object):
def commonChars(self, words):
counters = map(Counter, words)
counter = reduce(and_, counters)
return counter.elements() | find-common-characters | Python - Simple and Concise! | domthedeveloper | 0 | 278 | find common characters | 1,002 | 0.683 | Easy | 16,320 |
https://leetcode.com/problems/find-common-characters/discuss/1848277/Python-Simple-and-Concise! | class Solution(object):
def commonChars(self, words):
return reduce(and_, map(Counter, words)).elements() | find-common-characters | Python - Simple and Concise! | domthedeveloper | 0 | 278 | find common characters | 1,002 | 0.683 | Easy | 16,321 |
https://leetcode.com/problems/find-common-characters/discuss/1833423/Python3-solution | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
result = []
first = words.pop(0)
words_len = len(words)
for char in first:
if len(list(filter(lambda x: char in x, words))) == words_len:
result.append(char)
words = list(map(lambda x: x.replace(char, '', 1), words))
return result | find-common-characters | Python3 solution | hgalytoby | 0 | 192 | find common characters | 1,002 | 0.683 | Easy | 16,322 |
https://leetcode.com/problems/find-common-characters/discuss/1833423/Python3-solution | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
result = []
first = words.pop(0)
for char in first:
add = False
for index, word in enumerate(words):
if char in word:
words[index] = word.replace(char, '', 1)
add = True
else:
add = False
break
if add:
result.append(char)
return result | find-common-characters | Python3 solution | hgalytoby | 0 | 192 | find common characters | 1,002 | 0.683 | Easy | 16,323 |
https://leetcode.com/problems/find-common-characters/discuss/1819917/Python3-Solution-with-using-hashamp | class Solution:
def _get(self, sets, alpha):
cnt = float('inf')
for _set in sets:
if alpha not in _set:
return 0
cnt = min(cnt, _set[alpha])
return cnt
def commonChars(self, words: List[str]) -> List[str]:
sets = []
for word in words:
d = collections.defaultdict(int)
for c in word:
d[c] += 1
sets.append(d)
res = []
alphas = 'abcdefghijklmnopqrstuvwxyz'
for alpha in alphas:
present_cnt = self._get(sets, alpha)
res += [alpha] * present_cnt
return res | find-common-characters | [Python3] Solution with using hashamp | maosipov11 | 0 | 164 | find common characters | 1,002 | 0.683 | Easy | 16,324 |
https://leetcode.com/problems/find-common-characters/discuss/1757073/Python3-solution | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
words_len = len(words)
if words_len == 1:
return list(words[0])
letters = list(words[0])
check_words = words[1:]
words_len -= 1
result = ""
for l in letters:
letter_in_all_word = False
for word_index in range(words_len):
if l in check_words[word_index]:
letter_in_all_word = True
check_words[word_index] = check_words[word_index].replace(l, "", 1)
else:
letter_in_all_word = False
break
if letter_in_all_word:
result += l
return list(result) | find-common-characters | Python3 solution | khalidhassan3011 | 0 | 225 | find common characters | 1,002 | 0.683 | Easy | 16,325 |
https://leetcode.com/problems/find-common-characters/discuss/1738745/Python3-99.6-or-Detailed-Commented-or-Beginenr-Friendly | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
# build a filter based on the first word
w = words[0]
d = Counter(w)
for s in words:
ds = Counter(s)
del_list = []
for k in d.keys():
if not k in ds.keys():
del_list.append(k)
# 1. remove all the difference (d.keys - d intersect ds)
for dl in del_list:
del d[dl]
# 2. update the existing letter's frequency
for k, v in ds.items():
if k in d.keys() and ds[k] <= d[k]:
d[k] = ds[k]
ans = []
# populate ans from the original filter
for k, v in d.items():
for _ in range(v):
ans.append(k)
return ans | find-common-characters | Python3 99.6% | Detailed Commented | Beginenr Friendly | doneowth | 0 | 314 | find common characters | 1,002 | 0.683 | Easy | 16,326 |
https://leetcode.com/problems/find-common-characters/discuss/1592045/Python-Solution-without-Collections | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
cchars={}
res = []
first = words[0]
inter = set(first)
#Finds the characters common to all words
for i in range(1,len(words)):
new = set(first) & set(words[i])
if inter != new:
inter &= new
#Create a dict with occurence for each common character
for i in range(0,len(words)):
for j in inter:
val = words[i].count(j)
if j not in cchars:
cchars[j] = val
else:
if val < cchars[j]:
cchars[j] = val
#Returns a list
for k in cchars:
cnt = cchars[k]
while cnt !=0:
res.append(k)
cnt-=1
return res | find-common-characters | Python Solution without Collections | kravisha | 0 | 271 | find common characters | 1,002 | 0.683 | Easy | 16,327 |
https://leetcode.com/problems/find-common-characters/discuss/1581531/described-python-solution-runtime-(36-ms-greater97.94)-memory-(14.1-MB-less95.88) | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
freq = {}
for c in words[0]:
freq[c] = freq.get(c, 0) + 1
for w in words[1:]:
for k, v in freq.items():
if k in w:
freq[k] = min(v, w.count(k))
else:
freq[k] = 0
ans = []
for k,v in freq.items():
ans.extend([k] * v)
return(ans) | find-common-characters | described python solution, runtime (36 ms, >97.94%), memory (14.1 MB, <95.88%) | motyl | 0 | 192 | find common characters | 1,002 | 0.683 | Easy | 16,328 |
https://leetcode.com/problems/find-common-characters/discuss/1578505/ez-python-on-solution | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
cache = []
for i in range(len(words)):
cache.append([0] * 26)
for i in range(len(words)):
word = words[i]
for char in word:
cache[i][ord(char)-ord('a')] += 1
output = []
for i in range(26):
curr_comm = len(words)
for j in range(len(words)):
curr_comm = min(curr_comm,cache[j][i])
for j in range(curr_comm):
output.append(chr(i+ord('a')))
return output | find-common-characters | ez python on solution | yingziqing123 | 0 | 115 | find common characters | 1,002 | 0.683 | Easy | 16,329 |
https://leetcode.com/problems/find-common-characters/discuss/1506730/python-3-oror-beginner-oror-easy-oror-9884 | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
if len(words)==1:
return list(words[0])
output=list(words[0])
for word in words[1:]:
temp=[]
for ch in output:
if ch in word:
temp.append(ch)
word=word.replace(ch,"*",1)
else:
continue
output=temp
return output | find-common-characters | python 3 || beginner || easy || 98%,84% | minato_namikaze | 0 | 360 | find common characters | 1,002 | 0.683 | Easy | 16,330 |
https://leetcode.com/problems/find-common-characters/discuss/1386001/Python3-Faster-Than-98.81-Memory-Less-Than-85.47 | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
s = set(words[0])
for i in words[1:]:
s = s & set(i)
v = []
s = list(s)
for i in s:
mn = 101
for z in words:
mn = min(mn, z.count(i))
for j in range(mn):
v.append(i)
return v | find-common-characters | Python3 Faster Than 98.81%, Memory Less Than 85.47% | Hejita | 0 | 225 | find common characters | 1,002 | 0.683 | Easy | 16,331 |
https://leetcode.com/problems/find-common-characters/discuss/1379555/Python-Using-Dict-faster-than-88.00-of-Python3 | class Solution:
def commonChars(self, n: List[str]) -> List[str]:
l =len(n)
d = {}
for x in n:
for y in set(x):
if y not in d:
d[y] = [x.count(y)]
else:
d[y].append(x.count(y))
res = []
for x, y in d.items():
if len(y) >= l:
z = min(y)
if z > 1:
for _ in range(z):
res.append(x)
else:
res.append(x)
return(res) | find-common-characters | [Python] Using Dict, faster than 88.00% of Python3 | sushil021996 | 0 | 148 | find common characters | 1,002 | 0.683 | Easy | 16,332 |
https://leetcode.com/problems/find-common-characters/discuss/1275165/Python-Faster-than-86-or-Use-ord-function | class Solution:
def commonChars(self, words: List[str]) -> List[str]:
temp = []
for word in words:
les = [0] * 26
for char in word:
les[ord(char) - ord('a')] += 1
temp.append(les)
# zip fuction => list(zip([1,2,3],[4,5,6])) = [(1, 4), (2, 5), (3, 6)]
# zip => It is to count the number of occurrences of the same letter of each word and group them into the same array
# *temp => * unpack array
record_value = list(zip(*temp))
ans = []
for i,v in enumerate(record_value):
# use the min value to express that this word appears at least several times in each word
for _ in range(min(v)):
ans.append(chr(ord('a') + i))
return ans | find-common-characters | Python Faster than 86 % | Use ord function | doimustz | 0 | 226 | find common characters | 1,002 | 0.683 | Easy | 16,333 |
https://leetcode.com/problems/find-common-characters/discuss/1204027/python-easy-solution-or-No-extra-DS-used | class Solution:
def commonChars(self, A: List[str]) -> List[str]:
letters = list(A[0])
for i in range(1,len(A)):
j=0
while j < len(letters):
if letters[j] not in A[i]:
letters.remove(letters[j])
else:
A[i] = A[i].replace(letters[j],"",1)
j = j+1
return letters | find-common-characters | [python] easy-solution | No extra DS used | arkumari2000 | 0 | 357 | find common characters | 1,002 | 0.683 | Easy | 16,334 |
https://leetcode.com/problems/find-common-characters/discuss/1139150/Python-pythonic-solution | class Solution:
def commonChars(self, A: List[str]) -> List[str]:
result = []
# iterrate set of shortest word
for ch in set(min(A, key=len)):
result.extend(ch for _ in range(min([word.count(ch) for word in A], default=0)))
return result | find-common-characters | [Python], pythonic solution | cruim | 0 | 201 | find common characters | 1,002 | 0.683 | Easy | 16,335 |
https://leetcode.com/problems/find-common-characters/discuss/786642/Python3-solution-easy-to-understand-92.42-faster! | class Solution:
def commonChars(self, A: List[str]) -> List[str]:
c = None
for i in range(len(A)):
if c != None:
x = c
c = Counter(A[i])
# Finding intersection of dictionaries
newc = x & c
c = newc
continue
c = Counter(A[i])
news = ""
for x in c:
news += str(x)*newc[x]
return list(news) | find-common-characters | Python3 solution, easy-to-understand; 92.42% faster! | rocketrikky | 0 | 184 | find common characters | 1,002 | 0.683 | Easy | 16,336 |
https://leetcode.com/problems/find-common-characters/discuss/367553/Python-solution-beats-98 | class Solution:
def commonChars(self, A: List[str]) -> List[str]:
alpha = list(A[0])
A = list(map(list,A))
n = len(A)
result = []
for c in alpha:
for i in range(1,n):
s = A[i]
if c not in s:
break
else:
s.remove(c)
else:
result.append(c)
return result | find-common-characters | Python solution beats 98% | oumoussmehdi | 0 | 201 | find common characters | 1,002 | 0.683 | Easy | 16,337 |
https://leetcode.com/problems/find-common-characters/discuss/305744/Python-solution-using-frequency-maps | class Solution:
def commonChars(self, A: List[str]) -> List[str]:
n = len(A)
if n == 0:
return []
master = None
for word in A:
freq = get_freq(word)
if master is None:
master = freq
else:
master = get_new_master(master, freq)
output = []
for char in master:
output = output + [char]*master[char]
return output
def get_new_master(m, s):
dict = {}
for char in m:
if char in s:
if s[char] >= m[char]:
dict[char] = m[char]
else:
dict[char] = s[char]
return dict
def get_freq(word):
dict = {}
for char in word:
if char in dict:
dict[char]+=1
else:
dict[char] = 1
return dict | find-common-characters | Python solution using frequency maps | nzelei | 0 | 179 | find common characters | 1,002 | 0.683 | Easy | 16,338 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/1291378/python-solution-using-stack | class Solution:
def isValid(self, s: str) -> bool:
stack=[]
for i in s:
if i == 'a':stack.append(i)
elif i=='b':
if not stack:return False
else:
if stack[-1]=='a':stack.pop()
else:return False
stack.append(i)
else:
if not stack:return False
else:
if stack[-1]=='b':stack.pop()
else:return False
return len(stack)==0 | check-if-word-is-valid-after-substitutions | python solution using stack | chikushen99 | 2 | 94 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,339 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/2238315/EasySimple-Stack-And-String-Replace-Approaches | class Solution:
def isValid(self, s: str) -> bool:
incomplete = True
while incomplete:
if 'abc' in s:
s= s.replace('abc','')
else:
incomplete = False
return s == '' | check-if-word-is-valid-after-substitutions | [Easy]Simple Stack And String Replace Approaches | ankitshetty | 1 | 54 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,340 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/2238315/EasySimple-Stack-And-String-Replace-Approaches | class Solution:
def isValid(self, s: str) -> bool:
stack = []
for i in s:
if i == 'c' and len(stack) >= 2 and stack[-1] == 'b' and stack[-2] == 'a':
stack.pop()
stack.pop()
else:
stack.append(i)
if ''.join(stack) == 'abc': stack = []
return stack == [] | check-if-word-is-valid-after-substitutions | [Easy]Simple Stack And String Replace Approaches | ankitshetty | 1 | 54 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,341 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/1793602/98.67-Lesser-Memory-or-only-3-lines-of-code-or-Very-simple-python-Solution | class Solution:
def isValid(self, s: str) -> bool:
while("abc" in s):
s = s.replace("abc","") # continuously replace "abc" by "". By the end if we end up with "", then the word is valid.
return s == "" | check-if-word-is-valid-after-substitutions | ✔98.67% Lesser Memory | only 3 lines of code | Very simple python Solution | Coding_Tan3 | 1 | 58 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,342 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/2835036/easy-python-solution | class Solution:
def isValid(self, s: str) -> bool:
while len(s) > 0 :
if 'abc' in s :
s = s.replace('abc', '')
else :
return False
return True | check-if-word-is-valid-after-substitutions | easy python solution | sghorai | 0 | 3 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,343 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/2794355/python-super-easy-using-stack | class Solution:
def isValid(self, s: str) -> bool:
if s[0] != "a":
return False
stack = []
for i in s:
if i == "c":
if not stack:
return False
prev = i
pop_count = 0
while stack and ord(prev) - ord(stack[-1]) == 1:
prev = stack[-1]
stack.pop()
pop_count +=1
if pop_count != 2:
return False
else:
stack.append(i)
return not stack | check-if-word-is-valid-after-substitutions | python super easy using stack | harrychen1995 | 0 | 2 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,344 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/2673561/Python-clean-and-concise | class Solution:
def isValid(self, s: str) -> bool:
while s:
temp = s.replace("abc", "")
if temp == s:
return False
else:
s = temp
return True | check-if-word-is-valid-after-substitutions | Python - clean and concise | phantran197 | 0 | 2 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,345 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/2661248/Short-and-Clean-Python-Solution | class Solution:
def isValid(self, s: str) -> bool:
while "abc" in s:
s = s.replace("abc", "")
return s == "" | check-if-word-is-valid-after-substitutions | Short and Clean Python Solution | emrecoltu | 0 | 3 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,346 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/2334578/Easy-python-solution-Noob-must-watch-and-upvote | class Solution:
def isValid(self, s: str) -> bool:
ans = ''
for i in s:
ans+=i
while len(ans)>=3:
if ans[-3:]=="abc":
ans=ans[0:-3]
else:
break
if ans=='':
return True
else:
return False | check-if-word-is-valid-after-substitutions | Easy python solution Noob must watch and upvote | Brillianttyagi | 0 | 36 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,347 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/2133703/Easy-Python3 | class Solution:
def isValid(self, s: str) -> bool:
while True:
if 'abc' in s:
s = s.replace('abc','')
elif len(s) == 0:
return True
else:
return False | check-if-word-is-valid-after-substitutions | Easy Python3 | jayeshmaheshwari555 | 0 | 41 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,348 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/1998113/ororPYTHON-SOL-oror-STACK-oror-WELL-EXPLAINED-oror-EASY-oror-FAST-oror | class Solution:
def isValid(self, s: str) -> bool:
stack = []
for i in s:
if i == "c" and len(stack) > 1 and stack[-1] == 'b' and stack[-2] == 'a':
stack.pop()
stack.pop()
else:
stack.append(i)
return stack == [] | check-if-word-is-valid-after-substitutions | ||PYTHON SOL || STACK || WELL EXPLAINED || EASY || FAST || | reaper_27 | 0 | 55 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,349 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/1934252/python-3-oror-stack-solution-oror-O(n)O(n) | class Solution:
def isValid(self, s: str) -> bool:
stack = []
for c in s:
if c == 'a':
stack.append(c)
elif c == 'b':
if not stack or stack[-1] != 'a':
return False
stack[-1] = 'ab'
elif not stack or stack.pop() != 'ab':
return False
return not stack | check-if-word-is-valid-after-substitutions | python 3 || stack solution || O(n)/O(n) | dereky4 | 0 | 53 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,350 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/1498519/Python-oror-Easy-Solution-oror-Beat-~-99.6 | class Solution:
def isValid(self, s: str) -> bool:
while "abc" in s:
s = s.replace("abc", "")
if s == "":
return (True)
return (False) | check-if-word-is-valid-after-substitutions | Python || Easy Solution || Beat ~ 99.6% | naveenrathore | 0 | 58 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,351 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/1480073/Python3-solution-using-stack | class Solution:
def isValid(self, s: str) -> bool:
st = []
for i in s:
st.append(i)
while len(st) > 2:
flag = False
i = 2
while i < len(st):
if st[i] == 'c' and st[i-1] == 'b' and st[i-2] == 'a':
flag = True
st.pop(i)
st.pop(i-1)
st.pop(i-2)
i += 1
if not flag:
return False
return True if len(st) == 0 else False | check-if-word-is-valid-after-substitutions | Python3 solution using stack | EklavyaJoshi | 0 | 36 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,352 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/1326366/2-liner-Faster-than-~80 | class Solution:
def isValid(self, s: str) -> bool:
while "abc" in s : s = s.replace("abc","")
return not s | check-if-word-is-valid-after-substitutions | 2 liner , Faster than ~80% | abhijeetgupto | 0 | 47 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,353 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/1048410/Python3-Solution | class Solution:
def isValid(self, s: str) -> bool:
def abc(s):
mil=0
for i in range(len(s)-2):
if s[i:i+3]=="abc":
mil=1
s=s[:i]+s[i+3:]
if s=="":
return True
else:
return abc(s)
if mil==0:
return False
return abc(s) | check-if-word-is-valid-after-substitutions | Python3 Solution | samarthnehe | 0 | 62 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,354 |
https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/985317/Python3-stack-O(N) | class Solution:
def isValid(self, s: str) -> bool:
stack = []
for c in s:
if c == "c" and stack[-2:] == ["a", "b"]:
stack.pop()
stack.pop()
else: stack.append(c)
return not stack | check-if-word-is-valid-after-substitutions | [Python3] stack O(N) | ye15 | 0 | 63 | check if word is valid after substitutions | 1,003 | 0.582 | Medium | 16,355 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1793625/Python-3-Very-typical-sliding-window-%2B-hashmap-problem. | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
left = 0
answer = 0
counts = {0: 0, 1: 0}
for right, num in enumerate(nums):
counts[num] += 1
while counts[0] > k:
counts[nums[left]] -= 1
left += 1
curr_window_size = right - left + 1
answer = max(answer, curr_window_size)
return answer | max-consecutive-ones-iii | [Python 3] Very typical sliding window + hashmap problem. | seankala | 5 | 152 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,356 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/247678/Python-3-Solution%3A-sliding-window-for-zeros'-indexes.-Detailed-explanation-included. | class Solution:
def longestOnes(self, A: List[int], K: int) -> int:
zero_index = [i for i, v in enumerate(A) if v == 0]
if K >= len(zero_index):
return len(A)
res = 0
for i in range(0, len(zero_index) - K + 1):
one_start = zero_index[i-1] + 1 if i > 0 else 0
one_end = zero_index[i+K] - 1 if i+K < len(zero_index) else len(A) - 1
res = max(res, one_end - one_start + 1)
return res | max-consecutive-ones-iii | Python 3 Solution: sliding window for zeros' indexes. Detailed explanation included. | jinjiren | 4 | 732 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,357 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2794365/Beats-98.68-!-Simple-moving-window-without-shrinking-window-size | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
l=r=0
for r in range(len(nums)):
if nums[r] == 0:
k-=1
if k<0:
if nums[l] == 0:
k+=1
l+=1
return r-l+1 | max-consecutive-ones-iii | Beats 98.68% ! Simple moving window without shrinking window size | avinash_konduri | 1 | 91 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,358 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2194773/Python3-SlidingWindow-O(n)-oror-O(1) | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
firstStart, secondStart = 0, 0
maxWindow = float('-inf')
while firstStart < len(nums):
if nums[firstStart] == 0:
k -= 1
if k < 0:
if nums[secondStart] == 0:
k += 1
secondStart += 1
firstStart += 1
maxWindow = max(maxWindow, (firstStart - secondStart))
return maxWindow | max-consecutive-ones-iii | Python3 SlidingWindow O(n) || O(1) | arshergon | 1 | 59 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,359 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1602308/Easy-to-understand-Python-sliding-window-solution | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
p1,p2,res,seenZeros=0,0,0,0
while p2<len(nums):
if nums[p2]==0:
seenZeros+=1
while seenZeros>k:
seenZeros=seenZeros-1 if nums[p1]==0 else seenZeros
p1+=1
res=max(res,p2-p1+1)
p2+=1
return res | max-consecutive-ones-iii | Easy to understand Python 🐍 sliding window solution | InjySarhan | 1 | 143 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,360 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1468182/Python3-Sliding-Window | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
begin = 0
for end in range(len(nums)):
if nums[end] == 0:
k -= 1
if k < 0:
if nums[begin] == 0:
k += 1
begin += 1
return end - begin + 1 | max-consecutive-ones-iii | [Python3] Sliding Window | maosipov11 | 1 | 151 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,361 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/884066/Python3-concise-sliding-window-solution | class Solution:
def longestOnes(self, A: List[int], K: int) -> int:
res = 0
i = 0
for j in range(len(A)):
if A[j] == 0:
K -= 1
res = max(res, j-i)
while K < 0:
if A[i] == 0:
K += 1
i += 1
return max(res, len(A)-i) | max-consecutive-ones-iii | [Python3] concise sliding window solution | hwsbjts | 1 | 138 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,362 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2842264/O(K)-Memory-O(N)-Time-oror-Python | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
# the go-to sliding window approach is to use:
# 1. use a queue to store the current sequence.
# 2. use a count variable to keep track of 0 one can take.
# Then:
# a. whenever the next value is 1, add to queue.
# b. otherwise it's a 0, and then:
# b.1 if 0-counter is at K: pop queue until a zero is popped and decrease it.
# b.2 if 0-counter is less than k: add the 0 to the queue.
# This is O(N) memory (sliding window, one iteration) and O(N) worst-case memory.
# we can further opt memory to O(K), which is not bounded in this question - but it could be less than the array length.
ans=0
l=0
zero_queue=[] # O(K) memory at most.
# O(N) runtime.
for r,num in enumerate(nums):
if num!=1:
if zero_queue and len(zero_queue)==k:
zero_idx=zero_queue.pop(0)
l=zero_idx+1
if len(zero_queue)<k:
zero_queue.append(r)
else:
l=r+1
ans=max(ans, r-l+1)
return ans | max-consecutive-ones-iii | O(K) Memory, O(N) Time || Python | d4mahu | 0 | 2 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,363 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2835663/Simple-and-Easy-Solution-or-Python | class Solution(object):
def longestOnes(self, nums, k):
max_, zero, ptr = 0, 0, 0
for index, value in enumerate(nums):
if value == 0: zero += 1
while zero > k:
if nums[ptr] == 0: zero -= 1
ptr += 1
max_ = max(max_, (index - ptr) + 1)
return max_ | max-consecutive-ones-iii | Simple and Easy Solution | Python | its_krish_here | 0 | 3 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,364 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2830016/Python-Solution-Sliding-Window-Approach-oror-Explained | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
m=0 #max length of window
c=0 #count of 0 found
i=0 #initial window index
j=0 #window's end's index
while j<len(nums):
if nums[j]==0: #count if 0 found
c+=1
if c==k: #if 0's equal to k in window then store the max
m=max(m,j-i+1) #j-i+1 <- valid window length
elif c>k: #if somehow 0's > k in window then move the window one step forward
if nums[i]==0: #on moving the index i, its necessaray to check if that's 0 or not
c-=1 #decrement the count if it is
i+=1 #and move the index
j+=1 #increment everytime, its making the window
if c<k: #if 0's count < k in window, return whole nums length -> forms valid substring
return len(nums)
#else return max window length
return m | max-consecutive-ones-iii | Python Solution - Sliding Window Approach || Explained✔ | T1n1_B0x1 | 0 | 6 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,365 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2826469/Sliding-window%3A-runtime-greater-90.62-memory-less-61.79 | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
repeated_one, win_start, max_length = 0, 0, 0
sub_arr = []
for win_end in range(len(nums)):
right_char = nums[win_end]
if right_char == 1:
repeated_one += 1
if (win_end - win_start + 1 - repeated_one) > k:
if nums[win_start] == 1:
repeated_one -= 1
win_start += 1
max_length = max(max_length, win_end - win_start + 1)
return max_length | max-consecutive-ones-iii | Sliding window: runtime > 90.62%, memory < 61.79% | samratM | 0 | 4 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,366 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2788245/Python3-solution | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
left = 0
nFlips =0
for right in range(len(nums)):
if nums[right] == 0:
nFlips +=1
if nFlips > k:
# if the left index points to zero then decrease nFlips
# Why ? because we are going to create a new sub array by moving the leftIndex by one.
if nums[left] == 0:
nFlips -=1
left+=1
return right - left +1 | max-consecutive-ones-iii | Python3 solution | user5580aS | 0 | 5 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,367 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2786521/sliding-window-approach-python | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
# O(n), O(1)
# sliding window approach
left = max_cons = 0
for right, num in enumerate(nums):
k -= 1 - num
if k < 0:
k += 1 - nums[left]
left += 1
else:
max_cons = max(max_cons, right - left + 1)
return max_cons | max-consecutive-ones-iii | sliding window approach python | sahilkumar158 | 0 | 5 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,368 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2755162/max-consecutive-ones-iii | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
count=0
ans=0
start=0
end=0
while(end<len(nums)):
if nums[end]==0:
if k==0:
while(start<end and nums[start]!=0):
count=count-1
start=start+1
start=start+1
end=end+1
ans=max(ans,count)
else:
count=count+1
k=k-1
ans=max(ans,count)
end=end+1
else:
count=count+1
end=end+1
ans=max(ans,count)
return ans | max-consecutive-ones-iii | max-consecutive-ones-iii | Aviraj_Battan | 0 | 2 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,369 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2746983/Python-Sliding-Window-short | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
mx, l = 0, 0
zeros = 0
for r in range(len(nums)):
zeros += nums[r] == 0
while zeros > k:
zeros -= nums[l] == 0
l += 1
mx = max(mx, r - l + 1)
return mx | max-consecutive-ones-iii | Python Sliding Window short | JSTM2022 | 0 | 4 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,370 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2718672/Python-Sliding-Window | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
beg, end, zeros, ans = 0,0,0,0
while end < len(nums):
if zeros + (nums[end] == 0) <= k:
zeros += (nums[end] == 0)
end += 1
ans = max(ans, end - beg)
else:
zeros -= (nums[beg] == 0)
beg +=1
return ans | max-consecutive-ones-iii | [Python] Sliding Window | yzhao05 | 0 | 5 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,371 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2681685/Python-easy-solution | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
start = 0
zero_count = 0
max_ones = 0
for end in range(len(nums)):
if nums[end] == 0:
zero_count = zero_count + 1
while(zero_count > k):
if nums[start] == 0:
zero_count = zero_count - 1
start = start + 1
max_ones = max(max_ones, end-start + 1)
return max_ones | max-consecutive-ones-iii | Python easy solution | rupalimishra_v2 | 0 | 37 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,372 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2663849/Sliding-window-plus-occurrences | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
occ = [0, 0]
ans = l = r = 0
while r < len(nums):
occ[nums[r]] += 1
while occ[0] > k:
occ[nums[l]] -= 1
l += 1
ans = max(ans, r-l+1)
r += 1
return ans | max-consecutive-ones-iii | Sliding window plus occurrences | tomfran | 0 | 3 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,373 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2648233/FAST-and-Simple-Approach | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
l,r,longest = 0,0,0
while r<len(nums):
if nums[r] == 1:
r+=1
elif nums[r] == 0:
if k>0:
#0 shift to 1
k-=1
r+=1
else:
#exhausted flips so shift pointer from left and increase k if nums[l] was zero
if nums[l] == 0:
k+=1
l+=1
longest= max(longest,r-l)
return longest | max-consecutive-ones-iii | FAST and Simple Approach | Avtansh_Sharma | 0 | 4 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,374 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2645769/Python3-Solution-oror-O(N)-Time-and-O(1)-Space-Complexity | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
count_0=0
n=len(nums)
maxSize=0
currSize=0
left=0
for i in range(n):
currSize+=1
if nums[i]==0:
count_0+=1
if count_0>k:
while count_0>k and left<=i:
if nums[left]==0:
count_0-=1
currSize-=1
left+=1
if currSize>maxSize and count_0<=k:
maxSize=currSize
return maxSize | max-consecutive-ones-iii | Python3 Solution || O(N) Time & O(1) Space Complexity | akshatkhanna37 | 0 | 5 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,375 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2613786/Python-Sliding-Window-Technique-(Faster-than-90) | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
count1=0
ws=0
maxlen=0
for we in range(len(nums)):
if nums[we]==1:
count1+=1
if we-ws+1-count1>k:
if nums[ws]==1:
count1-=1
ws+=1
maxlen=max(maxlen,we-ws+1)
return maxlen | max-consecutive-ones-iii | Python - Sliding Window Technique (Faster than 90%) | utsa_gupta | 0 | 18 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,376 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2611335/Sliding-window-Solution | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
ws=0
c1=0
c0=0
maxlen=0
for we in range(len(nums)):
c=nums[we]
if c==0:
c0+=1
else:
c1+=1
if c0>k:
leftchar=nums[ws]
if leftchar==0:
c0-=1
ws+=1
maxlen=max(maxlen,we-ws+1)
return maxlen | max-consecutive-ones-iii | Sliding window Solution | shagun_pandey | 0 | 19 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,377 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2566023/Python-easy-solution-using-sliding-windows-TC%3A(N)-SC(1) | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
start = 0
mini = 0
maxFreq = 0
for end in range(len(nums)):
if(nums[end] == 1):
maxFreq += 1
if((end - start + 1) - maxFreq ) > k:
if(nums[start] == 1):
maxFreq -= 1
start += 1
mini = max(mini , end - start + 1)
return mini | max-consecutive-ones-iii | Python easy solution using sliding windows TC:(N) SC(1) | rajitkumarchauhan99 | 0 | 28 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,378 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2401243/Python-Solution-or-Classic-Two-Pointer-Sliding-Window-or-Count-Valid-Zeroes | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
ans = 0
start = 0
end = 0
# to count zeroes encountered
zero = 0
while end < len(nums):
# count valid zeroes
if nums[end] == 0:
zero += 1
# if zeroes amount more than valid, remove them while sliding the window
while zero > k:
if nums[start] == 0:
zero -= 1
start += 1
# check for max length
ans = max(ans,end-start+1)
end += 1
return ans | max-consecutive-ones-iii | Python Solution | Classic Two Pointer - Sliding Window | Count Valid Zeroes | Gautam_ProMax | 0 | 37 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,379 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2338974/Python-easy-solution | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
j = 0
ans = 0
result = -999999999
for i in range(len(nums)):
if nums[i]==0:
ans+=1
elif ans>k:
result = max(result,j-i)
while ans>k:
if nums[j]==0:
ans-=1
j+=1
return max(result,i-j+1) | max-consecutive-ones-iii | Python easy solution | Brillianttyagi | 0 | 60 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,380 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2175930/Sliding-Window-Approach-oror-Simplest-Solution-with-Dry-Run | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
l = 0
for r, val in enumerate(nums):
k -= (1 - val)
if k < 0:
k += (1 - nums[l])
l += 1
return r - l + 1 | max-consecutive-ones-iii | Sliding Window Approach || Simplest Solution with Dry Run | Vaibhav7860 | 0 | 72 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,381 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2124836/Python-or-SImple-Sliding-window-solution | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
zeroes_in_window = 0
max_len=0
slow=0
for fast,f in enumerate(nums) :
if f == 0:
zeroes_in_window+=1
# Maintain Valid Window.
# I.E If Number of zeroes has exceeded k shrink from left till zeroes_in_window<=k
while zeroes_in_window > k :
#Decrement count of zeroes in window
if nums[slow]==0 :
zeroes_in_window -= 1
slow+=1
#Update max window if current window is larger
max_len=max(fast-slow+1,max_len)
return max_len | max-consecutive-ones-iii | Python | SImple Sliding window solution | pbprasad_99 | 0 | 28 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,382 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/2061486/Python-two-pointers | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
start = flip = result = 0
for end in range(len(nums)):
flip += nums[end] == 0
while flip > k:
flip -= nums[start] == 0
start += 1
result = max(result, end - start + 1)
return result | max-consecutive-ones-iii | Python, two pointers | blue_sky5 | 0 | 44 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,383 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1998136/PYTHON-SOL-oror-SLIDING-WINDOW-oror-FASTER-THAN-99.46-oror-COMMENTED-oror-EXPLAINED-oror | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
#sliding window
ans = 0
n = len(nums)
left = k
prev = 0
for i in range(n):
if nums[i] == 1:
# no need to flip
continue
else:
# check if we can flip
if left > 0:
left -= 1
continue
else:
# we cannot flip so sliding window is stopped
if i - prev > ans : ans = i - prev
# now sliding window will shrink
while prev < i and nums[prev] != 0:
prev += 1
prev += 1
if n - prev > ans : ans = n - prev
return ans | max-consecutive-ones-iii | PYTHON SOL || SLIDING WINDOW || FASTER THAN 99.46% || COMMENTED || EXPLAINED || | reaper_27 | 0 | 126 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,384 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1949596/5-Lines-Python-Solution-oror-99-Faster-oror-Memory-less-than-95 | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
l=0
for r in range(len(nums)):
k-=1-nums[r]
if k<0: k+=1-nums[l] ; l+=1
return r-l+1 | max-consecutive-ones-iii | 5-Lines Python Solution || 99% Faster || Memory less than 95% | Taha-C | 0 | 101 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,385 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1949596/5-Lines-Python-Solution-oror-99-Faster-oror-Memory-less-than-95 | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
l=0 ; ans=0 ; C={0:0, 1:0}
for r,num in enumerate(nums):
C[num]+=1
while C[0]>k: C[nums[l]]-=1 ; l+=1
ans=max(ans,r-l+1)
return ans | max-consecutive-ones-iii | 5-Lines Python Solution || 99% Faster || Memory less than 95% | Taha-C | 0 | 101 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,386 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1827489/Python-easy-to-read-and-understand-or-sliding-window | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
cnt, ans, i = 0, 0, 0
for j in range(len(nums)):
if nums[j] == 0:
cnt += 1
while cnt > k:
if nums[i] == 0:
cnt -= 1
i = i+1
ans = max(ans, j-i+1)
return ans | max-consecutive-ones-iii | Python easy to read and understand | sliding window | sanial2001 | 0 | 102 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,387 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1784602/Python-Sliding-window-approach-Simple-to-understand-explained-with-comments | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
zeros = 0 # Keep track of all the zeros in the window
left = 0 # Left pointer
res = 0 # result to return
# Just like most window problems, iterate through the array and imagine the index is your right pointer
for i in range(len(nums)):
# If the number is zero we need to increment our zeros counter, we know
# that k zeros can be flipped, that means our window can have at max k 0s
#
# If we see that the window has more than k zeros we need to start moving the left pointer
# once we work passed a zero on the left we can continue iterating through the rest of the numbers
if nums[i] == 0:
zeros += 1 # increment our zeros counter every time we spot one
while zeros > k:
if nums[left] == 0:
zeros -= 1 # Subtract our counter because our window is moving passed a zero
left += 1 # Move the left pointer
# In most sliding window problems you'll use a similar approach to keep track of the result,
# (i + 1 - left) is basically calculating the window size, you add 1 to the right pointer because arrays
# start at 0, an array of 5 numbers has a max index of 4.
res = max(res, i + 1 - left)
return res | max-consecutive-ones-iii | Python - Sliding window approach - Simple to understand, explained with comments | iamricks | 0 | 65 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,388 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1740528/Unique-EASY-Solution-or-python3 | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
lo, hi, N, ones, zeros, max_sum = 0, 0, len(nums), 0, 0, -math.inf
while hi < N:
if nums[hi] == 1: ones+=1
if nums[hi] == 0: zeros+=1
while lo <= hi and zeros > k:
if nums[lo] == 1: ones-=1
if nums[lo] == 0: zeros-=1
lo+=1
max_sum = max(max_sum, ones+zeros)
hi+=1
return max_sum | max-consecutive-ones-iii | Unique EASY Solution | python3 | SN009006 | 0 | 50 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,389 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1720185/Python-Simple-sliding-window-explained | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
l, r = 0, 0
while r < len(nums):
# if the right pointer sees a 0
# decrement the available number
# of zeroes we have
if nums[r] == 0: k -= 1
# if at any point we don't
# have num(zeroes) to flip,
# we'll increment the left pointer
# BUT, while incrementing the left
# pointer if we come across a 0
# it means we're going to remove
# that from our window [l..r] so
# we increment the available zeroes too
if k < 0:
if nums[l] == 0: k += 1
l += 1
# right pointer is going to
# march ahead irrespectively
r += 1
# the length of the required
# window can be determined by
# the distance bw left and righ
return r - l | max-consecutive-ones-iii | [Python] Simple sliding window explained | buccatini | 0 | 110 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,390 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1625066/python-or-96.68 | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
zeros = [-1]
for i in range(len(nums)):
if nums[i]==0:
zeros.append(i)
zeros.append(len(nums))
if len(zeros)-2<=k:
return len(nums)
ret = 0
for i in range(k+1, len(zeros)):
ret = max(ret, zeros[i] - zeros[i-k-1] - 1)
return ret | max-consecutive-ones-iii | python | 96.68% | 1579901970cg | 0 | 120 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,391 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1616582/python3-sliding-window-solutionvery-easy-to-understand | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
countZeros=0
res=0
left=0
for right in range(len(nums)):
if nums[right]==0:
countZeros+=1
while countZeros>k:
if nums[left]==0:
countZeros-=1
left+=1
res=max(res,right-left+1)
return res | max-consecutive-ones-iii | python3 sliding window solution,very easy to understand | Karna61814 | 0 | 56 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,392 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1572697/Easiest-Solution-Python-or-Faster-than-80 | class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
start=0
end=0
ans=0
d={}
while end<=len(nums)-1:
d[nums[end]]=d.get(nums[end],0)+1
if 0 not in d:
ans=max(ans,end-start+1)
end+=1
else:
if d[0]<=k:
ans=max(ans,end-start+1)
end+=1
else:
if d[nums[start]]==1:
del d[nums[start]]
start+=1
end+=1
else:
d[nums[start]]-=1
start+=1
end+=1
return ans | max-consecutive-ones-iii | Easiest Solution Python | Faster than 80% | AdityaTrivedi88 | 0 | 88 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,393 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/1087613/Python-or-Sliding-Window | class Solution:
def longestOnes(self, A: List[int], K: int) -> int:
max_ones, window_start, max_size = 0, 0, 0
for window_end in range(len(A)):
if A[window_end] == 1:
max_ones += 1
if (window_end - window_start + 1 - max_ones) > K:
if A[window_start] == 1:
max_ones -= 1
window_start += 1
max_size = max(max_size, window_end-window_start+1)
return max_size | max-consecutive-ones-iii | Python | Sliding Window | Rakesh301 | 0 | 134 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,394 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/985341/Python3-sliding-window-O(N) | class Solution:
def longestOnes(self, A: List[int], K: int) -> int:
ans = cnt = 0
seen = {0: -1}
for i in range(len(A)):
if A[i] == 0: cnt += 1 # count of 0s
seen.setdefault(cnt, i)
ans = max(ans, i - seen.get(cnt-K, -1))
return ans | max-consecutive-ones-iii | [Python3] sliding window O(N) | ye15 | 0 | 78 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,395 |
https://leetcode.com/problems/max-consecutive-ones-iii/discuss/985341/Python3-sliding-window-O(N) | class Solution:
def longestOnes(self, A: List[int], K: int) -> int:
ans = ii = cnt = 0
for i, x in enumerate(A):
if not x: cnt += 1
while ii <= i and cnt > K:
if not A[ii]: cnt -= 1
ii += 1
ans = max(ans, i - ii + 1)
return ans | max-consecutive-ones-iii | [Python3] sliding window O(N) | ye15 | 0 | 78 | max consecutive ones iii | 1,004 | 0.634 | Medium | 16,396 |
https://leetcode.com/problems/maximize-sum-of-array-after-k-negations/discuss/1120243/WEEB-EXPLAINS-PYTHON-SOLUTION | class Solution:
def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
A.sort()
i = 0
while i < len(A) and K>0:
if A[i] < 0: # negative value
A[i] = A[i] * -1 # update the list, change negative to positive
K-=1
elif A[i] > 0: # positive value
if K % 2 == 0: # let K==2(must be even value), this means -1*-1==1 so it has no effect on sum
return sum(A)
else: return sum(A) - 2 * min(A) # let A==[1,2,3],K=1, so equation is 6-2(1)==4, same as -1+2+3=4 after taking the minimum in the list to give the largest possible sum required in the question
else: return sum(A) # if A[i]==0,just sum cuz 0 is neutral: 1-0==1 or 1+0==1 thus no change just sum
i+=1
if K > len(A): # that means we have changed all values to positive
A.sort() # cuz now its the opposite let A = [-4,-2,-3], K = 8, now flipping all negatives to positives, we have a new minimum which is 2
if K % 2 == 0: # Here onwards is basically the same thing from before
return sum(A)
else: return sum(A) - 2 * min(A)
return sum(A) | maximize-sum-of-array-after-k-negations | WEEB EXPLAINS PYTHON SOLUTION | Skywalker5423 | 6 | 361 | maximize sum of array after k negations | 1,005 | 0.51 | Easy | 16,397 |
https://leetcode.com/problems/maximize-sum-of-array-after-k-negations/discuss/350855/Solution-in-Python-3-(beats-~100) | class Solution:
def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
S, a = sum(A), sorted([i for i in A if i < 0])
L, b = len(a), min([i for i in A if i >= 0])
if L == 0: return S if K % 2 == 0 else S - 2*b
if K <= L or (K - L) % 2 == 0: return S - 2*sum(a[:min(K,L)])
return S - 2*sum(a[:-1]) if -a[-1] < b else S - 2*sum(a) - 2*b
- Junaid Mansuri | maximize-sum-of-array-after-k-negations | Solution in Python 3 (beats ~100%) | junaidmansuri | 3 | 477 | maximize sum of array after k negations | 1,005 | 0.51 | Easy | 16,398 |
https://leetcode.com/problems/maximize-sum-of-array-after-k-negations/discuss/1809612/Greedy-approach-python | class Solution:
def largestSumAfterKNegations(self, nums: List[int], k: int) -> int:
nums.sort(key=abs, reverse=True)
for i in range(len(nums)):
if nums[i] < 0 and k > 0:
nums[i] *= -1
k -= 1
if k == 0:
break
if k % 2 != 0:
nums[-1] *= -1
return sum(nums)
``` | maximize-sum-of-array-after-k-negations | Greedy approach - python | lister777 | 2 | 93 | maximize sum of array after k negations | 1,005 | 0.51 | Easy | 16,399 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.