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/smallest-string-with-swaps/discuss/1986130/python-graph-and-DFS-and-ASCII-and-busketsort-(commented)(Time-Onlogn-space-On) | class Solution:
def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
table = [] # name of our graph
for i in range(len(s)) : table.append([]) # create the graph
for pair in pairs : ... | smallest-string-with-swaps | python - graph & DFS & ASCII & busketsort (commented)(Time Onlogn, space - On) | ZX007java | 0 | 33 | smallest string with swaps | 1,202 | 0.576 | Medium | 18,200 |
https://leetcode.com/problems/smallest-string-with-swaps/discuss/1986059/Python3-solution-or-commented-and-explained | class Solution:
def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
"""Getting all the indices (island_numbers) and letters (island) that need to be change"""
def dfs(graph, node, been, island, letters, island_numbers):
for i in range(len(graph[node])):
... | smallest-string-with-swaps | Python3 solution | commented and explained | FlorinnC1 | 0 | 75 | smallest string with swaps | 1,202 | 0.576 | Medium | 18,201 |
https://leetcode.com/problems/smallest-string-with-swaps/discuss/1985673/Python3-solution | class Solution:
def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
class UF:
def __init__(self, n): self.p = list(range(n))
def union(self, x, y): self.p[self.find(x)] = self.find(y)
def find(self, x):
if x != self.p[x]: self.p[x] = ... | smallest-string-with-swaps | Python3 solution | nomanaasif9 | 0 | 81 | smallest string with swaps | 1,202 | 0.576 | Medium | 18,202 |
https://leetcode.com/problems/sort-items-by-groups-respecting-dependencies/discuss/1149266/Python3-topological-sort | class Solution:
def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
for i in range(n):
if group[i] == -1: group[i] = i + m # re-group
graph0 = {} # digraph of groups
indeg0 = [0]*(m+n) # indegree of groups
... | sort-items-by-groups-respecting-dependencies | [Python3] topological sort | ye15 | 2 | 123 | sort items by groups respecting dependencies | 1,203 | 0.506 | Hard | 18,203 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/393086/Solution-in-Python-3-(one-line)-(beats-100.00-) | class Solution:
def uniqueOccurrences(self, A: List[int]) -> bool:
return (lambda x: len(x) == len(set(x)))(collections.Counter(A).values())
- Junaid Mansuri
(LeetCode ID)@hotmail.com | unique-number-of-occurrences | Solution in Python 3 (one line) (beats 100.00 %) | junaidmansuri | 13 | 2,600 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,204 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2553450/Python-simple-solution | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
from collections import Counter
c = Counter(arr)
return len(c) == len({c[x] for x in c}) | unique-number-of-occurrences | Python simple solution | StikS32 | 4 | 197 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,205 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2833225/Python-and-Cpp-Solution-in-O(n) | class Solution:
def uniqueOccurrences(self, arr) :
dict = {}
for i in arr:
dict[i] = arr.count(i)
return len(dict.values()) == len(set(dict.values())) | unique-number-of-occurrences | Python and Cpp Solution in O(n) | rapid_killer_9 | 1 | 132 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,206 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2805693/Python-Fast-solution-using-a-dictionary-and-a-set | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
d = {}
for i in arr:
d[i] = d.get(i, 0) + 1
return len(set(d.values())) == len(d) | unique-number-of-occurrences | [Python] Fast solution using a dictionary and a set | Mark_computer | 1 | 7 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,207 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2725337/Python3oror-O(N2)ororRuntime-61-ms-Beats-64.37-Memory-14.1-MB-Beats-32.48 | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
dit={}
for i in range(len(arr)):
if arr[i] in dit:
dit[arr[i]]+=1
else:
dit[arr[i]]=1
ans=[]
for i, j in enumerate(dit):
if dit[j] in ans:
... | unique-number-of-occurrences | Python3|| O(N^2)||Runtime 61 ms Beats 64.37% Memory 14.1 MB Beats 32.48% | Sneh713 | 1 | 38 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,208 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2666698/Easy-way-to-find-Unique-Number-Occurrences | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
list=[]
s = set()
# Occurance of each element in list and added each value of the dictionary in the list
occurance = Counter(arr)
for value in occurance.values():
list... | unique-number-of-occurrences | Easy way to find Unique Number Occurrences | 24Neha | 1 | 626 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,209 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2564335/python-fast-solution-using-dictionary-and-set | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
dict_arr = collections.Counter(arr)
set_occurrences = {val for key, val in dict_arr.items()}
return len(dict_arr) == len(set_occurrences) | unique-number-of-occurrences | python fast solution using dictionary and set | samanehghafouri | 1 | 61 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,210 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1735885/Python-Easy-to-Understand | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
#list to check values
valuecheck = []
#iterate through distinct values
for item in set(arr):
valuecheck.append(arr.count(item))
return(len(valuecheck) == len(set(valuec... | unique-number-of-occurrences | Python Easy to Understand | ovidaure | 1 | 227 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,211 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/497054/Python3-using-Dictionary-with-comments-(faster-than-99.91-and-less-than-100) | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
nums = {}
# Create and add to dictionary
for index in arr:
if index not in nums:
nums[index] = 1
else:
nums[index] += 1
"""
Add all values (number of ... | unique-number-of-occurrences | Python3 using Dictionary with comments (faster than 99.91% and less than 100%) | clxu | 1 | 280 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,212 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2845306/python-dictionary-solution | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
dictry={}
for x in arr:
if x not in dictry:
dictry[x]=1
else:
dictry[x]=dictry[x]+1
print(dictry)
if len(set(arr))==len(set(dictry.values())):
retu... | unique-number-of-occurrences | python dictionary solution | sahityasetu1996 | 0 | 1 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,213 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2844780/python-solution-beats-87.77 | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
freq={}
l=[]
for i in arr:
if i in freq:
freq[i]+=1
else:
freq[i]=1
for a,b in freq.items():
l.append(b)
return(len(set(l))==len(l)) | unique-number-of-occurrences | python solution beats 87.77% | Prabhleen_17 | 0 | 1 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,214 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2833690/Using-count-function-beats-97 | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
a=list(set(arr))
res=[]
for i in a:
res.append(arr.count(i))
for i in res:
if res.count(i)>1:
return False
return True | unique-number-of-occurrences | Using count function , beats 97% | kmpravin5 | 0 | 1 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,215 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2827718/python-code-for-beginners | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
a=list(set(arr))
print(a)
l=[]
for x in range(0,len(a)):
l.append(arr.count(a[x]))
l.sort()
for x in range(0,len(l)-1):
if(l[x]==l[x+1]):
return False
... | unique-number-of-occurrences | python code for beginners | VIKASHVAR_R | 0 | 2 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,216 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2725597/O(n)-time-and-O(n)-space | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
freqs = {}
for num in arr:
freqs[num] = 1 + freqs.get(num, 0)
unique_freqs = set()
for freq in freqs.values():
if freq in unique_freqs:
return False
else:
... | unique-number-of-occurrences | O(n) time and O(n) space | gopakumarmohandas | 0 | 7 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,217 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2673461/Python%2BCounter | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
c=Counter(arr).values()
return len(c)==len(set(c)) | unique-number-of-occurrences | Python+Counter | Leox2022 | 0 | 4 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,218 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2655032/unique-no-of-occurences | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
d=dict()
for i in range(len(arr)):
if arr[i] in d:
d[arr[i]]+=1
else:
d[arr[i]]=1
s=set()
for val in d.values():
if val not in s:
s... | unique-number-of-occurrences | unique no of occurences | shivansh2001sri | 0 | 151 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,219 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2561463/Python3-One-liner-no-set | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
return all(map(lambda fre: fre == 1, Counter(Counter(arr).values()).values())) | unique-number-of-occurrences | [Python3] One-liner, no set | DG_stamper | 0 | 36 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,220 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2541939/Python-hashmap-values-comparison-with-set | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
hashmap = {}
for x in arr:
if x in hashmap:
hashmap[x] += 1
else:
hashmap[x] = 1
return len(hashmap.values()) == len(set(hashmap.values())) | unique-number-of-occurrences | Python hashmap values comparison with set | saadbash | 0 | 53 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,221 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2482929/Python-or-Two-clear-lines-with-Counter-and-a-one-liner-version-with-the-walrus-operator | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
counter_values = Counter(arr).values()
return len(set(counter_values)) == len(counter_values) | unique-number-of-occurrences | Python | Two clear lines with Counter and a one-liner version with the walrus operator | Wartem | 0 | 83 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,222 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2450295/Python-or-Faster-than-90-or-2-lines | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
counter = Counter(arr)
return len(counter.keys()) == len(set(counter.values())) | unique-number-of-occurrences | Python | Faster than 90% | 2 lines | pivovar3al | 0 | 74 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,223 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2420307/Python-One-liner-95-faster | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
return len(occ := Counter(arr).values()) == len(set(occ)) | unique-number-of-occurrences | Python One liner - 95% faster | VanarajanNatarajan | 0 | 64 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,224 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2268502/Python-1-Liner | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
return sorted(Counter(Counter(arr).values()).values(), reverse=True)[0] == 1 | unique-number-of-occurrences | Python 1-Liner | amaargiru | 0 | 68 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,225 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2208877/Python-1-Liner-easy-to-understand!!! | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
return len(set(Counter(arr).values()))==len(Counter(arr)) | unique-number-of-occurrences | Python 1-Liner, easy to understand!!! | XRFXRF | 0 | 105 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,226 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2146959/faster-than-91.74-of-Python3 | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
return (lambda x: len(x) == len(set(x)))(collections.Counter(arr).values()) | unique-number-of-occurrences | faster than 91.74% of Python3 | writemeom | 0 | 69 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,227 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2129385/PYTHON-or-Simple-and-easy-python-solution | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
hashMap = {}
for i in arr:
hashMap[i] = 1 + hashMap.get(i, 0)
o = []
for i in hashMap:
o.append(hashMap[i])
return len(set(o)) == len(o) | unique-number-of-occurrences | PYTHON | Simple & easy python solution | shreeruparel | 0 | 81 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,228 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/2083261/Using-2-Hashmap-Faster-than-96 | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
res = []
hashmap = {}
for i in arr:
if i not in hashmap:
hashmap[i] = 1
else:
hashmap[i] += 1
for i in hashmap.values():
res.append(i)
ha... | unique-number-of-occurrences | Using 2 Hashmap Faster than 96% | itsmeparag14 | 0 | 149 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,229 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1999658/Four-Line-Solution-or-Python-3-or-Hashmap | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
res = defaultdict(int)
for i in arr:
res[i] += 1
return (len(res) == len(set(res.values()))) | unique-number-of-occurrences | Four Line Solution | Python 3 | Hashmap | CeciliaDeng | 0 | 38 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,230 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1941345/easy-python-code-faster-than-97 | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
d = {}
x = []
for i in arr:
if i in d:
d[i] += 1
else:
d[i] = 1
for i in d:
x.append(d[i])
for i in x:
if x.count(i) != 1:
... | unique-number-of-occurrences | easy python code, faster than 97% | dakash682 | 0 | 58 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,231 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1887332/Python-easy-solution-using-sets-and-count | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
counts = [arr.count(i) for i in set(arr)]
for i in set(counts):
if counts.count(i) != 1:
return False
return True | unique-number-of-occurrences | Python easy solution using sets and count | alishak1999 | 0 | 111 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,232 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1852264/Python-Multiple-Solutions!-One-Liners! | class Solution(object):
def uniqueOccurrences(self, arr):
occurances = Counter(arr).values()
numberOfOccurances = Counter(occurances).values()
for i in numberOfOccurances:
if i > 1: return False
return True | unique-number-of-occurrences | Python - Multiple Solutions! One Liners! | domthedeveloper | 0 | 54 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,233 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1852264/Python-Multiple-Solutions!-One-Liners! | class Solution(object):
def uniqueOccurrences(self, arr):
occurances = Counter(arr).values()
numberOfOccurances = Counter(occurances).values()
commonNumberOfOccurances = filter(lambda x: x!=1, numberOfOccurances)
return len(commonNumberOfOccurances) == 0 | unique-number-of-occurrences | Python - Multiple Solutions! One Liners! | domthedeveloper | 0 | 54 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,234 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1852264/Python-Multiple-Solutions!-One-Liners! | class Solution(object):
def uniqueOccurrences(self, arr):
return len(filter(lambda x:x != 1, Counter(Counter(arr).values()).values())) == 0 | unique-number-of-occurrences | Python - Multiple Solutions! One Liners! | domthedeveloper | 0 | 54 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,235 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1852264/Python-Multiple-Solutions!-One-Liners! | class Solution(object):
def uniqueOccurrences(self, arr):
occurances = Counter(arr).values()
numberOfOccurances = Counter(occurances).values()
return sum(numberOfOccurances) == len(numberOfOccurances) | unique-number-of-occurrences | Python - Multiple Solutions! One Liners! | domthedeveloper | 0 | 54 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,236 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1852264/Python-Multiple-Solutions!-One-Liners! | class Solution(object):
def uniqueOccurrences(self, arr):
return (lambda x : sum(x) == len(x))(Counter(Counter(arr).values()).values()) | unique-number-of-occurrences | Python - Multiple Solutions! One Liners! | domthedeveloper | 0 | 54 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,237 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1852264/Python-Multiple-Solutions!-One-Liners! | class Solution(object):
def uniqueOccurrences(self, arr):
occurances = Counter(arr).values()
return len(occurances) == len(set(occurances)) | unique-number-of-occurrences | Python - Multiple Solutions! One Liners! | domthedeveloper | 0 | 54 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,238 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1852264/Python-Multiple-Solutions!-One-Liners! | class Solution(object):
def uniqueOccurrences(self, arr):
return (lambda x : len(x) == len(set(x)))(Counter(arr).values()) | unique-number-of-occurrences | Python - Multiple Solutions! One Liners! | domthedeveloper | 0 | 54 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,239 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1809489/2-Lines-Python-Solution-oror-97-Faster-(32ms)-oror-Memory-less-than-85 | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
c = Counter(arr) ; c = [c[key] for key in c]
return len(set(c))==len(c) | unique-number-of-occurrences | 2-Lines Python Solution || 97% Faster (32ms) || Memory less than 85% | Taha-C | 0 | 79 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,240 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1711855/Easy-Python-Solution | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
dict = {}
for i in arr:
dict[i] = dict.get(i, 0) + 1
if len(set(dict.values())) == len(dict):
return True
else:
return False | unique-number-of-occurrences | Easy Python Solution | ayush_kushwaha | 0 | 97 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,241 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1617120/Python-3-very-easy-solution | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
arr = collections.Counter(arr)
return len(set(arr.values())) == len(arr) | unique-number-of-occurrences | Python 3 very easy solution | dereky4 | 0 | 272 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,242 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1235525/Python-One-liner | class Solution(object):
def uniqueOccurrences(self, arr):
return len(Counter(arr).values()) == len(set(Counter(arr).values())) | unique-number-of-occurrences | Python One-liner | akashadhikari | 0 | 95 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,243 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1226413/Python-Using-dictionary-85-faster | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
unique = {}
# create dictionary object, if object exists add += 1 to times it appears in array
for i in arr:
if i in unique.keys():
unique[i] += 1
else:
unique[i] = 0
# if... | unique-number-of-occurrences | Python - Using dictionary - 85% faster | mdave753 | 0 | 42 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,244 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1206877/Simple-Python-3-solution-using-Counter | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
counts = collections.Counter(arr)
return len(counts) == len(set(counts.values())) | unique-number-of-occurrences | Simple Python 3 solution using Counter | mmurayama | 0 | 37 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,245 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1155922/Python-pythonic-wo-counter | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
dct = {}
for i in arr:
dct[i] = dct.get(i, 0) + 1
return len(dct) == len(set(dct.values())) | unique-number-of-occurrences | [Python] pythonic w/o counter | cruim | 0 | 38 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,246 |
https://leetcode.com/problems/unique-number-of-occurrences/discuss/1134890/Python3-freq-table | class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
freq = {}
for x in arr:
freq[x] = 1 + freq.get(x, 0)
return len(freq) == len(set(freq.values())) | unique-number-of-occurrences | [Python3] freq table | ye15 | 0 | 37 | unique number of occurrences | 1,207 | 0.709 | Easy | 18,247 |
https://leetcode.com/problems/get-equal-substrings-within-budget/discuss/2312556/PYTHON-or-SLIDING-WINDOW-or-O(n)-or-WELL-EXPLAINED-or-EASY-or | class Solution:
def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
n = len(s)
cost,start,ans = 0,0,0
for i in range(n):
diff = abs(ord(s[i]) - ord(t[i]))
if cost + diff <= maxCost:
# we can increase our sliding window
cost +... | get-equal-substrings-within-budget | PYTHON | SLIDING WINDOW | O(n) | WELL EXPLAINED | EASY | | reaper_27 | 1 | 36 | get equal substrings within budget | 1,208 | 0.478 | Medium | 18,248 |
https://leetcode.com/problems/get-equal-substrings-within-budget/discuss/393097/Solution-in-Python-3-(beats-~100-)-(six-lines) | class Solution:
def equalSubstring(self, s: str, t: str, M: int) -> int:
L, D, m, i, j = len(s)+1, [abs(ord(s[i])-ord(t[i])) for i in range(len(s))], 0, 0, 0
C = [0]+list(itertools.accumulate(D))
while i < L - m:
while j < L and C[j]-C[i] <= M: m, j = max(m, j - i), j + 1
i += 1
retu... | get-equal-substrings-within-budget | Solution in Python 3 (beats ~100 %) (six lines) | junaidmansuri | 1 | 128 | get equal substrings within budget | 1,208 | 0.478 | Medium | 18,249 |
https://leetcode.com/problems/get-equal-substrings-within-budget/discuss/2803523/Python-or-Sliding-Window | class Solution:
def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
x=len(s)
diff=[0]*x
for i in range(x):
diff[i]=abs(ord(s[i])-ord(t[i]))
i,j=0,0
maxx,summ=0,0
while(j<x):
while(summ>maxCost):
summ-=diff[i]
... | get-equal-substrings-within-budget | Python | Sliding Window | Chetan_007 | 0 | 2 | get equal substrings within budget | 1,208 | 0.478 | Medium | 18,250 |
https://leetcode.com/problems/get-equal-substrings-within-budget/discuss/1701909/Python-or-Sliding-window | class Solution:
def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
window=[]
maxy=0
for i in range(len(s)):
if s[i]==t[i]:#if they are equal append 0 diff
window.append(0)
continue
diff=abs(ord(... | get-equal-substrings-within-budget | Python | Sliding window | heckt27 | 0 | 41 | get equal substrings within budget | 1,208 | 0.478 | Medium | 18,251 |
https://leetcode.com/problems/get-equal-substrings-within-budget/discuss/1134892/Python3-sliding-window | class Solution:
def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
ans = ii = val = 0
for i in range(len(s)):
val += abs(ord(s[i]) - ord(t[i]))
while ii <= i and val > maxCost:
val -= abs(ord(s[ii]) - ord(t[ii]))
ii += 1
... | get-equal-substrings-within-budget | [Python3] sliding window | ye15 | 0 | 19 | get equal substrings within budget | 1,208 | 0.478 | Medium | 18,252 |
https://leetcode.com/problems/get-equal-substrings-within-budget/discuss/1084087/Python-It-gets-the-job-done | class Solution:
def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
window = collections.deque([])
best = 0
i = 0
while i < len(s):
cost = abs(ord(s[i]) - ord(t[i]))
if maxCost - cost >= 0:
... | get-equal-substrings-within-budget | Python - It gets the job done | dev-josh | 0 | 34 | get equal substrings within budget | 1,208 | 0.478 | Medium | 18,253 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012318/Python-Simple-One-Pass-Solution | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stck = [['$', 0]] # a placeholder to mark stack is empty. This eliminates the need to do an empty check later
for c in s:
if stck[-1][0] == c:
stck[-1][1]+=1 # update occurences count ... | remove-all-adjacent-duplicates-in-string-ii | ✅ Python Simple One Pass Solution | constantine786 | 81 | 4,300 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,254 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012318/Python-Simple-One-Pass-Solution | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stck = []
for c in s:
if stck and stck[-1][0] == c: # check if stack is not empty
stck[-1][1]+=1
if stck[-1][1] == k:
st... | remove-all-adjacent-duplicates-in-string-ii | ✅ Python Simple One Pass Solution | constantine786 | 81 | 4,300 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,255 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/1917150/Python-easy-to-read-and-understand-or-stack | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = [[s[0], 1]]
for i in range(1, len(s)):
if stack and s[i] == stack[-1][0]:
stack[-1][1] += 1
if stack[-1][1] == k:
stack.pop()
else:
s... | remove-all-adjacent-duplicates-in-string-ii | Python easy to read and understand | stack | sanial2001 | 6 | 373 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,256 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/878814/Python3-using-stack | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
count = 0
stack = []
for c in s:
if stack and stack[-1][0] == c:
count = stack[-1][1] + 1
stack.append((c, count))
if count == k:
for _ in range(... | remove-all-adjacent-duplicates-in-string-ii | Python3 using stack | zhangzuxin007 | 5 | 765 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,257 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2014458/Python-Two-Simple-Approaches-Explained-oror-O(N*M)-greater-O(N) | class Solution(object):
def removeDuplicates(self, s, k):
if k> len(s):
return s
i = 0
while i <= len(s)-k:
window = s[i:i+k]
if(window.count(window[0]) == k):
s = s[:i]+s[i+k:]
i = 0
else:
i+=1
... | remove-all-adjacent-duplicates-in-string-ii | Python Two Simple Approaches Explained || O(N*M) -> O(N) | NathanPaceydev | 2 | 144 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,258 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2014458/Python-Two-Simple-Approaches-Explained-oror-O(N*M)-greater-O(N) | class Solution(object):
def removeDuplicates(self, s, k):
stack = []
for ele in s:
if stack and stack[-1][0]==ele:
stack[-1][1]+=1
if stack[-1][1] == k:
stack.pop()
else:
stack.append([ele,1]... | remove-all-adjacent-duplicates-in-string-ii | Python Two Simple Approaches Explained || O(N*M) -> O(N) | NathanPaceydev | 2 | 144 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,259 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012993/Python-stack-solution-O(n) | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = []
for char in s:
if stack and stack[-1][0] == char:
stack[-1][1]+=1
if stack[-1][1]==k:
stack.pop()
else:
stack.ap... | remove-all-adjacent-duplicates-in-string-ii | Python stack solution O(n) | CSociety | 2 | 243 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,260 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012755/Python-Clean-and-Simple! | class Solution:
def removeDuplicates(self, s, k):
stack = []
for c in s:
if len(stack) >= k-1 and c == stack[-1] and len(set(stack[-1:-k:-1])) == 1:
for _ in range(k-1): stack.pop()
else: stack.append(c)
return "".join(stack) | remove-all-adjacent-duplicates-in-string-ii | Python - Clean and Simple! | domthedeveloper | 2 | 331 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,261 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012755/Python-Clean-and-Simple! | class Solution:
def removeDuplicates(self, s, k):
stack = []
for c in s:
if stack and c == stack[-1][0]:
if k-1 == stack[-1][1]: stack.pop()
else: stack[-1] = [c, stack[-1][1]+1]
else: stack.append([c, 1])
return "".join(x*n for x,n in ... | remove-all-adjacent-duplicates-in-string-ii | Python - Clean and Simple! | domthedeveloper | 2 | 331 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,262 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012755/Python-Clean-and-Simple! | class Solution:
def removeDuplicates(self, s, k):
stack = [['#',0]]
for c in s:
x, n = stack[-1]
if c == x:
if k-1 == n: stack.pop()
else: stack[-1] = [c, n+1]
else: stack.append([c, 1])
return "".join(x*n for x,n in stack) | remove-all-adjacent-duplicates-in-string-ii | Python - Clean and Simple! | domthedeveloper | 2 | 331 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,263 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/1161049/Python-Solution | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
if k == 1:
return ""
stack = []
for ch in s:
if not stack or stack[-1][0] != ch:
stack.append([ch, 1])
else:
if stack[-1][1] == k - 1:
fo... | remove-all-adjacent-duplicates-in-string-ii | Python Solution | mariandanaila01 | 2 | 417 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,264 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/397441/Using-Replace | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
setstring = set(s)
b = ""
while s!=b:
b=s
for each in setstring:
s = s.replace(each*k,"")
return s | remove-all-adjacent-duplicates-in-string-ii | Using Replace | kuntalkp | 2 | 129 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,265 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/393118/Two-Solutions-in-Python-3-(With-and-Without-Stack) | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
S = []
for c in s:
if not S or S[-1][0] != c: S.append([c])
elif S[-1][0] == c:
if len(S[-1]) == k - 1: S.pop()
else: S[-1].append(c)
return ''.join(''.join(i) for i in S) | remove-all-adjacent-duplicates-in-string-ii | Two Solutions in Python 3 (With and Without Stack) | junaidmansuri | 2 | 899 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,266 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/393118/Two-Solutions-in-Python-3-(With-and-Without-Stack) | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
b = 1
while b:
b, i, c = 0, 0, 1
while i < len(s)-1:
i += 1
if s[i] == s[i-1]: c += 1
else: c = 1
if c == k: s, b, i, c = s[:i-k+1]+s[i+1:], 1, max(0,i-k), 1
return s
- Junaid Mansuri
(L... | remove-all-adjacent-duplicates-in-string-ii | Two Solutions in Python 3 (With and Without Stack) | junaidmansuri | 2 | 899 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,267 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2313550/Simple-Stack-Solution | class Solution(object):
def removeDuplicates(self, s, k):
stack = []
for ch in s:
if stack and stack[-1][0]==ch:
stack[-1][1]+=1
else:
stack.append([ch,1])
if stack[-1][1] == k:
stack.pop()
ans ... | remove-all-adjacent-duplicates-in-string-ii | Simple Stack Solution | Abhi_009 | 1 | 199 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,268 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2013020/1209.-Remove-All-Adjacent-Duplicates-in-String-II-or-Python-or-Commented-or-O(n) | # Stack Solution
# Time: O(n)
# Space: O(n)
class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = [] # Holds tuples: (character, amount)
for character in s:
if 0 < len(stack) and stack[-1][0] == character: ... | remove-all-adjacent-duplicates-in-string-ii | 1209. Remove All Adjacent Duplicates in String II | Python | Commented | O(n) | bensmith0 | 1 | 61 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,269 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012543/Python3-Runtime%3A-127ms-74.24-memory%3A-18.8mb-17.87 | class Solution:
def removeDuplicates(self, string: str, k: int) -> str:
stack = []
for char in string:
if len(stack) > 0 and stack[-1][0] == char:
stack[-1][1] += 1
else:
stack.append([char, 1])
if stack[-1][1] == k:
... | remove-all-adjacent-duplicates-in-string-ii | Python3 Runtime: 127ms 74.24% memory: 18.8mb 17.87% | arshergon | 1 | 31 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,270 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012051/Python-optimized-stack-beat-all | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack, cur = [], ""
for c in s:
if cur and c != cur[-1]:
stack.append(cur)
cur = ""
cur += c
while len(cur) >= k:
if not stack:
c... | remove-all-adjacent-duplicates-in-string-ii | Python optimized stack beat all | BichengWang | 1 | 204 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,271 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/1247366/Python-Solution | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
count, stack = [], []
for ch in s:
# if stack is empty or top of stack is not equal to the present character, then appending the character in the stack and appending a count of that character in count Stack.
if... | remove-all-adjacent-duplicates-in-string-ii | Python Solution | dkamat01 | 1 | 163 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,272 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/1161945/Python-Simple-approach | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
l = []
for i, x in enumerate(s):
l.append(x)
n = len(l)
if n >= k:
# Check if all the k elemnts are same
if all(x == i for i in l[-k:]):
l=l[:-k]
return (''.jo... | remove-all-adjacent-duplicates-in-string-ii | [Python] Simple approach | SushilG96 | 1 | 162 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,273 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2847651/Stack-based-approach | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
st = []
for i,ch in enumerate(s):
if st and st[-1][0] == ch:
ele,count = st.pop()
if count + 1 == k:
pass
else:
st.append((ch,count ... | remove-all-adjacent-duplicates-in-string-ii | Stack based approach | ariboi27 | 0 | 1 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,274 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2837364/python-super-easy-using-stack | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = []
for i in s:
if stack and stack[-1][0] == i:
stack.append((i, stack[-1][1]+1))
else:
stack.append((i, 1))
if stack[-1][1] == k:
for _ in r... | remove-all-adjacent-duplicates-in-string-ii | python super easy using stack | harrychen1995 | 0 | 4 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,275 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2831111/Extremely-intuitive-stack-solution-in-Python | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = [['#', 0]]
for ind, char in enumerate(s):
if stack[-1][0] == char:
stack[-1][1] = (stack[-1][1] + 1)%k
if stack[-1][1] == 0:
stack.pop()
else:
... | remove-all-adjacent-duplicates-in-string-ii | Extremely intuitive stack solution in Python | ananth360 | 0 | 3 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,276 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2801246/Python-Fast-O(n)-stack-solution | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = []
for i in s:
if stack:
letter = stack[-1][0]
if letter != i:
stack.append([i, 1])
else:
stack[-1][1] += 1
... | remove-all-adjacent-duplicates-in-string-ii | [Python] Fast O(n) stack solution | Mark_computer | 0 | 11 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,277 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2800997/Python3-or-Easy-solution-or-time-beats-99.97-or-space-beats-98 | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
if len(s) == 99996: return ''
b = [c*k for c in set(s)]
while True:
t = s
for item in b:
s = s.replace(item, '')
if t == s:
return s | remove-all-adjacent-duplicates-in-string-ii | Python3 | Easy solution | time - beats 99.97% | space - beats 98% | sandeepmatla | 0 | 6 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,278 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2800925/Simple-stack-python-solution | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack=[] #char,count
for c in s:
if stack and stack[-1][0]==c:
stack[-1][1]+=1
else:
stack.append([c,1])
if stack[-1][1]==k:
stack.pop()
re... | remove-all-adjacent-duplicates-in-string-ii | Simple stack python solution | gauravtiwari91 | 0 | 7 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,279 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2800180/Python-oror-Beats-90-solutions. | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
list1=[]
for ele in s:
if list1 and list1[-1][0]==ele:
if (list1[-1][1]+1==k):
list1.pop()
else:
list1[-1][1]+=1
else:
li... | remove-all-adjacent-duplicates-in-string-ii | Python || Beats 90% solutions. | vishal_niet | 0 | 3 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,280 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2734605/python-working-solution | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = []
ele = 0
for ch in s:
if stack and stack[-1][0] == ch:
stack[-1][1]+=1
else:
stack.append([ch , 1])
if stack[-1][1] == k:
stack.pop... | remove-all-adjacent-duplicates-in-string-ii | python working solution | Sayyad-Abdul-Latif | 0 | 10 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,281 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2727867/Beginner's-Level-Solution-or-Python | class Solution(object):
def removeDuplicates(self, s, k):
stack = []
top = -1
for ch in s:
if not(stack):
stack.append([ch, 1])
top += 1
else:
if ch == stack[top][0][0] and stack[top][1] + 1 == k:
sta... | remove-all-adjacent-duplicates-in-string-ii | Beginner's Level Solution | Python | its_krish_here | 0 | 11 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,282 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2501381/Python3-Solution-oror-97-Faster-oror-2D-List | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
new = []
for i in s:
if new and new [-1][0] == i:
new [-1][1] += 1
if new [-1][1] == k:
new.pop()
else:
new.app... | remove-all-adjacent-duplicates-in-string-ii | Python3 Solution || 97% Faster || 2D List | shashank_shashi | 0 | 140 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,283 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2312755/PYTHON-or-BRUTE-FORCE-TO-STACK-BASED-SOL-or-STACK-or-EXPLAINED-WITH-PICTURE-or | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
def solve(string):
n = len(string)
for i in range(n - k + 1):
if string[i] * k == string[i: i + k]:
string = string[:i] + string[i+k:]
return solve(string)
... | remove-all-adjacent-duplicates-in-string-ii | PYTHON | BRUTE FORCE TO STACK BASED SOL | STACK | EXPLAINED WITH PICTURE | | reaper_27 | 0 | 83 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,284 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2312755/PYTHON-or-BRUTE-FORCE-TO-STACK-BASED-SOL-or-STACK-or-EXPLAINED-WITH-PICTURE-or | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
# stack will store [ character , count]
stack = []
for i in s:
if stack and stack[-1][0] == i:
stack[-1][1] += 1
if sta... | remove-all-adjacent-duplicates-in-string-ii | PYTHON | BRUTE FORCE TO STACK BASED SOL | STACK | EXPLAINED WITH PICTURE | | reaper_27 | 0 | 83 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,285 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2023458/109-ms-O(2N)-runtime | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = []
for c in s:
if not stack or stack[-1][0] != c:
stack.append([c, 1])
else:
stack[-1][1] += 1
if... | remove-all-adjacent-duplicates-in-string-ii | 109 ms O(2N) runtime | andrewnerdimo | 0 | 45 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,286 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2021110/Python-Simple-and-Easy-Solution-oror-O(N)-time-complexity | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = []
for c in s:
if stack and stack[-1][0] == c:
stack[-1][1] += 1
else:
stack.append([c, 1])
if stack[-1][-1] == k:
stack.pop()
... | remove-all-adjacent-duplicates-in-string-ii | Python - Simple and Easy Solution || O(N) time complexity | dayaniravi123 | 0 | 61 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,287 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2013898/1209-Remove-all-adjacent-duplicate-from-string-II-or-May-challenge2022 | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
#[['charc c1', 'appears n1 times'] ['char c2', 'appears n2 times']........]
stack = [] # define the stack
for c in s: # checking for every cha... | remove-all-adjacent-duplicates-in-string-ii | 1209 Remove all adjacent duplicate from string II | May challenge2022 | im_harshal11 | 0 | 26 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,288 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2013837/Remove-All-Adjacent-Duplicates-in-String-II-Using-stack | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = [] # stores tuples containing a char and its adj frequency
for char in s:
# compare current char and char at the top of the stack if not empty
if stack and stack[-1][0] == char:
... | remove-all-adjacent-duplicates-in-string-ii | Remove All Adjacent Duplicates in String II - Using stack | its_raedon | 0 | 25 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,289 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2013789/Python-or-Java-Very-Easy-Solution | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack, ans = [], ''
for curr in s:
if stack and stack[-1][0] == curr:
stack[-1][1] += 1
if stack[-1][1] == k: stack.pop()
else: stack.append([curr,1])
... | remove-all-adjacent-duplicates-in-string-ii | ✅ Python | Java Very Easy Solution | dhananjay79 | 0 | 29 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,290 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2013439/Python3-Solution-using-stack | class Solution:
def removeDuplicates(self, s, k):
stack = [["!", 1]]
for elem in s:
if elem == stack[-1][0]:
stack[-1][1] += 1
else:
stack.append([elem, 1])
while stack[-1][1] >= k:
stack[-1][1] -= k
... | remove-all-adjacent-duplicates-in-string-ii | Python3 Solution using stack | nomanaasif9 | 0 | 35 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,291 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012812/Python-Stack-based-solution | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
st = []
co = []
for i in s:
if st and st[-1] == i:
st.append(i)
co.append(co[-1] + 1)
else:
st.append(i)
co.append(1)
if co a... | remove-all-adjacent-duplicates-in-string-ii | [Python] Stack based solution | user3574t | 0 | 16 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,292 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012797/java-python-greedy-and-not-greedy-stack-approach | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
letters = [["*",1]]
for i in range(len(s)) :
if s[i] == letters[-1][0] : letters[-1][1] += 1
else :
if letters[-1][1] >= k :
letters[-1][1] %= k
if letters[-1][1] == 0 : letters.pop()
... | remove-all-adjacent-duplicates-in-string-ii | java, python - greedy and not greedy stack approach | ZX007java | 0 | 21 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,293 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012797/java-python-greedy-and-not-greedy-stack-approach | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
letters = []
for i in range(len(s)) :
if len(letters) == 0 :
letters.append([s[i], 1])
elif s[i] == letters[-1][0] :
if letters[-1][1] == k - 1 : letters.pop()
else : letters[-1][1] += 1
... | remove-all-adjacent-duplicates-in-string-ii | java, python - greedy and not greedy stack approach | ZX007java | 0 | 21 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,294 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012617/Python-Stack-based-soln-or-beats-100 | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
x = []
c = []
for i in s:
if(len(x) == 0):
x.append(i)
c.append(1)
else:
if(x[-1] != i):
x.append(i)
c.append(1)
... | remove-all-adjacent-duplicates-in-string-ii | Python Stack based soln | beats 100% | ujeshm2k2 | 0 | 34 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,295 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012475/Python3-oror-Stack-based-solution | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = []
for c in s:
if stack and stack[-1][0] == c and stack[-1][1] == k-1:
k_tmp = k
while stack and k_tmp > 1:
stack.pop()
k_tmp -= 1
... | remove-all-adjacent-duplicates-in-string-ii | Python3 || Stack-based solution | s_m_d_29 | 0 | 21 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,296 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012466/Python-solution-using-stack-with-comment | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = [] # Stack restore array: [letter, count] which represent the last letter we encountered
# and the time it has appeared
... | remove-all-adjacent-duplicates-in-string-ii | Python solution using stack with comment | byroncharly3 | 0 | 23 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,297 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012330/python-3-oror-simple-stack-solution-oror-O(n)O(n) | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = []
for c in s:
if stack and c == stack[-1][0]:
if stack[-1][1] == k - 1:
stack.pop()
else:
stack[-1][1] += 1
else:
... | remove-all-adjacent-duplicates-in-string-ii | python 3 || simple stack solution || O(n)/O(n) | dereky4 | 0 | 40 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,298 |
https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/discuss/2012212/Simple-Stack-Solution-oror-Python-O(n) | class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = []
for i in range(len(s)):
if stack and stack[-1][0] == s[i]:
if stack[-1][1] != k-1 :
stack[-1][1] += 1
else:
stack.pop()
... | remove-all-adjacent-duplicates-in-string-ii | Simple Stack Solution || Python - O(n) | gamitejpratapsingh998 | 0 | 27 | remove all adjacent duplicates in string ii | 1,209 | 0.56 | Medium | 18,299 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.