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/maximum-number-of-groups-getting-fresh-donuts/discuss/1140716/Python3-Fastest-solution-with-explanation | class Solution:
def maxHappyGroups(self, bs: int, gs: List[int]) -> int:
c = {i: 0 for i in range(bs)}
for g in gs:
c[g % bs] += 1
ret = c[0]
c[0] = 0
def get_keys(num):
keys = []
def rec(stack):
if len(stack) == num:
if sum(stack) % bs == 0:
keys.append(Counter(stack))
else:
for i in range(stack[-1] if stack else bs - 1, - 1, - 1):
stack.append(i)
rec(stack)
stack.pop()
rec([])
return keys
def get_diff_keys(num):
keys = []
def rec(stack):
if len(stack) == num:
if sum(stack) % bs == 0:
keys.append(Counter(stack))
else:
for i in range(stack[-1] - 1 if stack else bs - 1, - 1, - 1):
stack.append(i)
rec(stack)
stack.pop()
rec([])
return keys
for tc in range(2, bs):
for keys in get_diff_keys(tc):
add = min(c[key] // keys[key] for key in keys)
if add == 0: continue
ret += add
for key in keys:
c[key] -= add * keys[key]
tc = 2
while True:
for keys in get_keys(tc):
add = min(c[key] // keys[key] for key in keys)
if add == 0: continue
ret += add
for key in keys:
c[key] -= add * keys[key]
if tc > sum(c.values()): break
tc += 1
return ret + bool(sum(c.values()))
``` | maximum-number-of-groups-getting-fresh-donuts | [Python3] Fastest solution with explanation | timetoai | 0 | 102 | maximum number of groups getting fresh donuts | 1,815 | 0.402 | Hard | 25,900 |
https://leetcode.com/problems/truncate-sentence/discuss/1142293/2-lines-of-code-with-100-less-space-used | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
words = s.split(" ")
return " ".join(words[0:k]) | truncate-sentence | 2 lines of code with 100% less space used | vashisht7 | 15 | 1,000 | truncate sentence | 1,816 | 0.821 | Easy | 25,901 |
https://leetcode.com/problems/truncate-sentence/discuss/1283308/Python-faster-than-95 | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
l = list(s.split(" "))
return (" ".join(l[:k])) | truncate-sentence | Python faster than 95% | aditi_shree | 2 | 307 | truncate sentence | 1,816 | 0.821 | Easy | 25,902 |
https://leetcode.com/problems/truncate-sentence/discuss/1141804/Python-One-Line | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split(" ")[:k]) | truncate-sentence | Python One Line | mariandanaila01 | 2 | 67 | truncate sentence | 1,816 | 0.821 | Easy | 25,903 |
https://leetcode.com/problems/truncate-sentence/discuss/1141327/Python3-1-line | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split()[:k]) | truncate-sentence | [Python3] 1-line | ye15 | 2 | 135 | truncate sentence | 1,816 | 0.821 | Easy | 25,904 |
https://leetcode.com/problems/truncate-sentence/discuss/2554328/EASY-PYTHON3-SOLUTION-FASTER | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
res = s.split()
return ' '.join(res[:k]) | truncate-sentence | ✅✔🔥 EASY PYTHON3 SOLUTION 🔥✅✔FASTER | rajukommula | 1 | 24 | truncate sentence | 1,816 | 0.821 | Easy | 25,905 |
https://leetcode.com/problems/truncate-sentence/discuss/1431216/Python-one-line | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split(' ')[:k]) | truncate-sentence | Python one line | amestri890 | 1 | 75 | truncate sentence | 1,816 | 0.821 | Easy | 25,906 |
https://leetcode.com/problems/truncate-sentence/discuss/1184331/Python3-simple-solution-one-liner | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return ' '.join(s.split(' ')[:k]) | truncate-sentence | Python3 simple solution one-liner | EklavyaJoshi | 1 | 69 | truncate sentence | 1,816 | 0.821 | Easy | 25,907 |
https://leetcode.com/problems/truncate-sentence/discuss/1157955/Python-Easy-Solution-with-Comments | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
x = s.split(' ') #make a list of given words
str1 = " " #take an empty string
return str1.join((x)[:k]) #join empty string and slice it using k | truncate-sentence | Python Easy Solution with Comments | aishwaryanathanii | 1 | 136 | truncate sentence | 1,816 | 0.821 | Easy | 25,908 |
https://leetcode.com/problems/truncate-sentence/discuss/2846098/Python3-Solution-with-using-space-counting | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
idx = 0
while k and idx < len(s):
if s[idx] == ' ':
k -= 1
idx += 1
return s[:idx - 1] if s[idx - 1] == ' ' else s[:idx] | truncate-sentence | [Python3] Solution with using space counting | maosipov11 | 0 | 3 | truncate sentence | 1,816 | 0.821 | Easy | 25,909 |
https://leetcode.com/problems/truncate-sentence/discuss/2839917/Python-Easy-1-liner-Runtime-34-ms | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return ' '.join(s.split(" ")[:k]) | truncate-sentence | [Python] Easy 1 liner, Runtime 34 ms | m0nxt3r | 0 | 3 | truncate sentence | 1,816 | 0.821 | Easy | 25,910 |
https://leetcode.com/problems/truncate-sentence/discuss/2829874/python | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
n = len(s)
for i in range(n):
if s[i] == ' ':
k -= 1
if k == 0:
return s[:i]
return s | truncate-sentence | python | xy01 | 0 | 1 | truncate sentence | 1,816 | 0.821 | Easy | 25,911 |
https://leetcode.com/problems/truncate-sentence/discuss/2829578/1-LINE-OF-CODE-oror-Python-3oror-Easy-to-understand | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split()[:k]) | truncate-sentence | 1 LINE OF CODE✅ || Python 3🔥|| Easy to understand✌🏼 | jhadevansh0809 | 0 | 2 | truncate sentence | 1,816 | 0.821 | Easy | 25,912 |
https://leetcode.com/problems/truncate-sentence/discuss/2826387/Easy-and-Fast-Python-One-Liner-(Beats-99) | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split()[0: k])
''' | truncate-sentence | Easy and Fast Python One-Liner (Beats 99%) | PranavBhatt | 0 | 4 | truncate sentence | 1,816 | 0.821 | Easy | 25,913 |
https://leetcode.com/problems/truncate-sentence/discuss/2821204/Simple-Python-Solution-in-just-3-line | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
arr = s.split()
arr[::] = arr[0:k]
return ' '.join(arr) | truncate-sentence | Simple Python Solution in just 3 line | ameenusyed09 | 0 | 1 | truncate sentence | 1,816 | 0.821 | Easy | 25,914 |
https://leetcode.com/problems/truncate-sentence/discuss/2787840/Python-one-liner | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return ' '.join([word for word in s.split()[:k]]) | truncate-sentence | Python one liner | ibozkurt79 | 0 | 3 | truncate sentence | 1,816 | 0.821 | Easy | 25,915 |
https://leetcode.com/problems/truncate-sentence/discuss/2777607/OneLiner-easy-Python-solution. | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split()[:k]) | truncate-sentence | OneLiner easy Python solution. | suyog_097 | 0 | 1 | truncate sentence | 1,816 | 0.821 | Easy | 25,916 |
https://leetcode.com/problems/truncate-sentence/discuss/2752248/Easy-Fast-3-Line-Solution-Python3 | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
a=""
s_list =s.split(" ")
for i in range(k):
a += s_list[i] + " "
return a.rstrip() | truncate-sentence | Easy Fast 3 Line Solution Python3 | avs-abhishek123 | 0 | 2 | truncate sentence | 1,816 | 0.821 | Easy | 25,917 |
https://leetcode.com/problems/truncate-sentence/discuss/2743997/Python-1-line-code | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return' '.join((s.split(' ')[:k])) | truncate-sentence | Python 1 line code | kumar_anand05 | 0 | 3 | truncate sentence | 1,816 | 0.821 | Easy | 25,918 |
https://leetcode.com/problems/truncate-sentence/discuss/2739309/one-line-python-solutionor97-space-complexity | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return (' ').join(s.split(' ')[0:k]) | truncate-sentence | one line python solution|97% space complexity | vishwas1451 | 0 | 3 | truncate sentence | 1,816 | 0.821 | Easy | 25,919 |
https://leetcode.com/problems/truncate-sentence/discuss/2731764/python-simpler-way | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
sentence_list = s.split(" ")
result = []
for i in range(0, k):
result.append(sentence_list[i])
return " ".join(result) | truncate-sentence | python simpler way | arshadali7860 | 0 | 3 | truncate sentence | 1,816 | 0.821 | Easy | 25,920 |
https://leetcode.com/problems/truncate-sentence/discuss/2730453/Python3-Solution-with-explanation | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return ' '.join(s.split(' ')[:k]) | truncate-sentence | Python3 Solution - with explanation | sipi09 | 0 | 4 | truncate sentence | 1,816 | 0.821 | Easy | 25,921 |
https://leetcode.com/problems/truncate-sentence/discuss/2719449/Python-solutionoror | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return(" ".join(s.split(" ")[:k])) | truncate-sentence | Python solution|| | sinjan_singh | 0 | 3 | truncate sentence | 1,816 | 0.821 | Easy | 25,922 |
https://leetcode.com/problems/truncate-sentence/discuss/2700066/Python-3-One-Line-Solution-and-Explanation | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split(" ")[x] for x in range(k)) | truncate-sentence | [Python 3] One-Line Solution and Explanation | keioon | 0 | 4 | truncate sentence | 1,816 | 0.821 | Easy | 25,923 |
https://leetcode.com/problems/truncate-sentence/discuss/2681776/simple-python-3-solution-oror-83ms | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
l = []
for i in range(len(s.split())):
if i == k:
break
l.append(s.split()[i])
return ' '.join(l) | truncate-sentence | simple python 3 solution || 83ms | Sumeet07 | 0 | 5 | truncate sentence | 1,816 | 0.821 | Easy | 25,924 |
https://leetcode.com/problems/truncate-sentence/discuss/2613997/O(n)-time-O(1)-space-approach-using-a-loop | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
for i in range(len(s)):
if s[i] == ' ':
k -= 1
if k == 0:
return s[0:i]
return s | truncate-sentence | O(n) time, O(1) space approach using a loop | kcstar | 0 | 5 | truncate sentence | 1,816 | 0.821 | Easy | 25,925 |
https://leetcode.com/problems/truncate-sentence/discuss/2478740/Very-easy-Python-solution | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
s_list = s.split(" ")
res = []
for i in range(k):
res.append(s_list[i])
return " ".join(res) | truncate-sentence | Very easy Python solution | aruj900 | 0 | 26 | truncate sentence | 1,816 | 0.821 | Easy | 25,926 |
https://leetcode.com/problems/truncate-sentence/discuss/2388901/One-line-solution | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split()[:k]) | truncate-sentence | One line solution | samanehghafouri | 0 | 14 | truncate sentence | 1,816 | 0.821 | Easy | 25,927 |
https://leetcode.com/problems/truncate-sentence/discuss/2224609/1-line-Python-solution | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split(" ")[:k]) | truncate-sentence | 1 line Python solution | lyubol | 0 | 43 | truncate sentence | 1,816 | 0.821 | Easy | 25,928 |
https://leetcode.com/problems/truncate-sentence/discuss/2155191/one-line-solution-Easy-to-understand | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return ' '.join(s.split(maxsplit = k)[:k]) | truncate-sentence | one line solution, Easy to understand | writemeom | 0 | 26 | truncate sentence | 1,816 | 0.821 | Easy | 25,929 |
https://leetcode.com/problems/truncate-sentence/discuss/2122612/one-liner-efficient-python-solution-or-96.28-efficient | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split(" ")[:k]) | truncate-sentence | one liner efficient python solution | 96.28% efficient | nikhitamore | 0 | 27 | truncate sentence | 1,816 | 0.821 | Easy | 25,930 |
https://leetcode.com/problems/truncate-sentence/discuss/2046236/Python3-simple-with-python | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split(' ')[:k]) | truncate-sentence | [Python3] simple with python | Shiyinq | 0 | 42 | truncate sentence | 1,816 | 0.821 | Easy | 25,931 |
https://leetcode.com/problems/truncate-sentence/discuss/2013859/Python-oneliner | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return ' '.join(s.split()[:k]) | truncate-sentence | Python oneliner | StikS32 | 0 | 30 | truncate sentence | 1,816 | 0.821 | Easy | 25,932 |
https://leetcode.com/problems/truncate-sentence/discuss/1942304/Python-One-Liner-%2B-Optimization | class Solution:
def truncateSentence(self, s, k):
return " ".join(s.split()[:k]) | truncate-sentence | Python - One-Liner + Optimization | domthedeveloper | 0 | 40 | truncate sentence | 1,816 | 0.821 | Easy | 25,933 |
https://leetcode.com/problems/truncate-sentence/discuss/1942304/Python-One-Liner-%2B-Optimization | class Solution:
def truncateSentence(self, s, k):
return " ".join(s.split(maxsplit=k)[:k]) | truncate-sentence | Python - One-Liner + Optimization | domthedeveloper | 0 | 40 | truncate sentence | 1,816 | 0.821 | Easy | 25,934 |
https://leetcode.com/problems/truncate-sentence/discuss/1868092/Python-Easy-and-Fast-Python3 | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
sentence = ''
l = s.split()
i = 0
while i < k:
sentence += l[i]
sentence += ' '
i += 1
return sentence[:-1] | truncate-sentence | [Python] Easy and Fast Python3 | natscripts | 0 | 48 | truncate sentence | 1,816 | 0.821 | Easy | 25,935 |
https://leetcode.com/problems/truncate-sentence/discuss/1867153/Python-one-line-solution | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return ' '.join(s.split()[0:k]) | truncate-sentence | Python one line solution | alishak1999 | 0 | 16 | truncate sentence | 1,816 | 0.821 | Easy | 25,936 |
https://leetcode.com/problems/truncate-sentence/discuss/1859170/Python-dollarolution-(one-line) | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return (' '.join(s.split()[:k])) | truncate-sentence | Python $olution (one line) | AakRay | 0 | 17 | truncate sentence | 1,816 | 0.821 | Easy | 25,937 |
https://leetcode.com/problems/truncate-sentence/discuss/1831370/Python-easy-to-read-and-understand | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
l = s.split(" ")
if k >= len(l):
return s
ans = ''
i = 0
while i < k:
ans = ans + l[i] + ' '
i = i+1
return ans[:-1] | truncate-sentence | Python easy to read and understand | sanial2001 | 0 | 28 | truncate sentence | 1,816 | 0.821 | Easy | 25,938 |
https://leetcode.com/problems/truncate-sentence/discuss/1780823/Python-oneline | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split()[:k]) | truncate-sentence | Python oneline | kakchaudhary | 0 | 23 | truncate sentence | 1,816 | 0.821 | Easy | 25,939 |
https://leetcode.com/problems/truncate-sentence/discuss/1755694/Python-One-Liner%3A-Beats-98 | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return ' '.join(s.split(' ')[:k]) | truncate-sentence | Python One-Liner: Beats 98% | dos_77 | 0 | 44 | truncate sentence | 1,816 | 0.821 | Easy | 25,940 |
https://leetcode.com/problems/truncate-sentence/discuss/1739151/1816.-Truncate-Sentence-solution | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
sent=s.split()
str=""
for x in range(0,k):
if x>=0 and x<(k-1):
str+=(sent[x]+" ")
elif x==(k-1):
str+=sent[x]
return str | truncate-sentence | 1816. Truncate Sentence solution | seabreeze | 0 | 42 | truncate sentence | 1,816 | 0.821 | Easy | 25,941 |
https://leetcode.com/problems/truncate-sentence/discuss/1727006/Python3-or-One-line-solution | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split()[:k]) | truncate-sentence | Python3 | One line solution | khalidhassan3011 | 0 | 33 | truncate sentence | 1,816 | 0.821 | Easy | 25,942 |
https://leetcode.com/problems/truncate-sentence/discuss/1685184/Python-one-line-solution-using-split-and-join-methods | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return " ".join(s.split()[:k]) #s.split() method will split each word after space and append it to the empty string using the .join() method where [:k] is same as [0:k] which defines the range of the words which we want to output. | truncate-sentence | Python one line solution using split and join methods | pawelborkar | 0 | 57 | truncate sentence | 1,816 | 0.821 | Easy | 25,943 |
https://leetcode.com/problems/truncate-sentence/discuss/1678149/easy-simple-solution | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
answer=""
s=s.split()
for i in range(k-1):
answer+=s[i]
answer += " "
answer+=s[k-1]
return answer | truncate-sentence | easy simple solution | Buyanjargal | 0 | 60 | truncate sentence | 1,816 | 0.821 | Easy | 25,944 |
https://leetcode.com/problems/truncate-sentence/discuss/1627395/Python3%3A-simple-two-solution%3A-with-split-without-split | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
#return " ".join(s.split()[0:k])
counter = 0
new_s = ''
for c in s:
if c == ' ':
counter += 1
if counter == k:
break
new_s += c
return new_s | truncate-sentence | Python3: simple two solution: with split, without split | CleverUzbek | 0 | 47 | truncate sentence | 1,816 | 0.821 | Easy | 25,945 |
https://leetcode.com/problems/truncate-sentence/discuss/1610490/c%2B%2Bpython3-easy-approach | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return ' '.join(s.split(' ')[:k]) | truncate-sentence | c++/python3 easy approach | qikang1994 | 0 | 38 | truncate sentence | 1,816 | 0.821 | Easy | 25,946 |
https://leetcode.com/problems/truncate-sentence/discuss/1589639/soln-without-using-any-built-in-methods-like-split()-or-python-or-faster-than-95 | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
count = idx = 0 # count refers to the no.of white spaces
ans = ""
while count != k:
# if no.of words > no.of white spaces then count the last index as a white space
if idx == len(s)-1 or s[idx] == " ": count += 1
ans += s[idx]
idx += 1
return ans[:-1] if idx < len(s) else ans | truncate-sentence | soln without using any built in methods like split() | python | faster than 95% | anandanshul001 | 0 | 57 | truncate sentence | 1,816 | 0.821 | Easy | 25,947 |
https://leetcode.com/problems/truncate-sentence/discuss/1537221/Being-Pythonic%3A-Exploiting-py-features | class Solution:
def truncateSentence(self, s: str, k: int) -> str:
return ' '.join(s.split()[0:k]) | truncate-sentence | Being Pythonic: Exploiting py features | _rust | 0 | 26 | truncate sentence | 1,816 | 0.821 | Easy | 25,948 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1141356/Python3-hash-map | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
mp = {}
for i, t in logs:
mp.setdefault(i, set()).add(t)
ans = [0]*k
for v in mp.values():
if len(v) <= k:
ans[len(v)-1] += 1
return ans | finding-the-users-active-minutes | [Python3] hash map | ye15 | 9 | 597 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,949 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1143301/Simple-Python3-solution-with-2-layer-dictionary | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
# use two-layer dict
# first layer record user; second layer record the active minutes
# 1. Put info. in 2-layer dict O(n)
d = {}
for (user, active_min) in logs:
if not (user in d):
d[user] = {}
user_log = d[user]
user_log[active_min] = True
# 2. Extract info. from 2-layer dict (at most O(n))
result = [0] * k
for user in d:
result[len(d[user])-1] += 1 # len(d[user]) must >= 1 to exist
return result | finding-the-users-active-minutes | Simple Python3 solution with 2-layer dictionary | tkuo-tkuo | 2 | 188 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,950 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1277381/Python3-solution-using-single-dictionary | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
d = {}
for i in logs:
if i[0] in d:
if i[1] not in d[i[0]]:
d[i[0]] = d[i[0]] + [i[1]]
else:
d[i[0]] = [i[1]]
res = [0]*k
for i in d.values():
res[len(i)-1] += 1
return res | finding-the-users-active-minutes | Python3 solution using single dictionary | EklavyaJoshi | 1 | 142 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,951 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/2837107/golang-python-rust-solution | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
res = [0] * k
users = {}
for id,time in logs:
if id not in users:
users[id] = set()
users[id].add(time)
else:
users[id].add(time)
for user in users:
res[len(users[user])-1]+=1
return res | finding-the-users-active-minutes | golang python rust solution | anshsharma17 | 0 | 2 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,952 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/2775142/Python3-Solution-beats-81 | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
dct = {}
ans = [0] * k
for log in logs:
if log[0] in dct:
dct[log[0]].add(log[1])
else:
dct[log[0]] = {log[1]}
for i in dct.values():
ans[len(i)-1] = ans[len(i)-1] + 1
return ans | finding-the-users-active-minutes | Python3 Solution, beats 81% | sipi09 | 0 | 2 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,953 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/2719369/Python-3-Set-Counter-Counter-w-List-Comprehension | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
c = Counter(Counter([i for i, j in set([(id, tim) for [id, tim] in logs])]).values())
return [c[i+1] for i in range(k)] | finding-the-users-active-minutes | Python 3 Set-Counter-Counter w/ List Comprehension | godshiva | 0 | 4 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,954 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/2700068/MAX-TIME-PYTHON | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
h={}
for (i,j) in logs:
if i not in h:
h[i]=[j]
else:
if j not in h[i]:
h[i].append(j)
a=[0]*k
for keys in h.values():
a[len(keys)-1]+=1
return a | finding-the-users-active-minutes | MAX TIME PYTHON | 2001640100048_2C | 0 | 2 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,955 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/2666049/python3-Faster-than-97 | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
users = {}
for log in logs:
user = log[0]
time = log[1]
if user in users:
users[user].add(time)
else:
users[user] = set()
users[user].add(time)
sol = {}
for i in users.values():
length = len(i)
if length in sol:
sol[length] += 1
else:
sol[length] = 1
ans = [0]*k
for i , j in sol.items():
ans[i-1] = j
return ans | finding-the-users-active-minutes | python3 Faster than 97% | Noisy47 | 0 | 6 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,956 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/2647969/Can-you-tell-me-why-this-is-slow-Python3-soon | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
res = []
for i in range (0,k):
res.append(0)
d = {} # id : [minute1, minute2]
for pair in logs:
id, min = pair[0], pair[1]
if id not in d:
d[id] = [min]
res[0] += 1
else:
if min not in d[id]:
d[id].append(min)
newUam = len(d[id])
res[newUam-2] -= 1
res[newUam-1] += 1
return res | finding-the-users-active-minutes | Can you tell me why this is slow? Python3 soon | egeergull | 0 | 3 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,957 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/2204539/Hash-map-by-Python-easy-to-understand | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
'''
We can use a dict to record the action minutes of every user,
the key is the ID, the value are the minutes this user performs actions.
Then, use set to remove the same minutes, and count the UAMs.
Time complexity: O(n), space comlexity: O(n)
'''
counter=collections.defaultdict(list)
for i in logs:
counter[i[0]].append(i[1])
ans=[0]*k
for i in counter.keys():
ans[len(set(counter[i]))-1]+=1
return ans | finding-the-users-active-minutes | Hash map by Python, easy to understand | XRFXRF | 0 | 49 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,958 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/2198998/Faster-than-88-python | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
users = {}
ans = [0] * k
for el in logs:
if el[0] not in users:
users[el[0]] = set([el[1]])
else:
users[el[0]].update([el[1]])
for user in users.keys():
uam = len(users[user])
ans[uam-1] += 1
return ans | finding-the-users-active-minutes | Faster than 88% python | ketan12sharma | 0 | 38 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,959 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/2123917/python-3-oror-simple-hash-map-and-hash-set-solutionoror-O(n)O(n) | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
userMinutes = collections.defaultdict(set)
for user, minute in logs:
userMinutes[user].add(minute)
answer = [0] * k
for uam in map(len, userMinutes.values()):
if 1 <= uam <= k:
answer[uam - 1] += 1
return answer | finding-the-users-active-minutes | python 3 || simple hash map and hash set solution|| O(n)/O(n) | dereky4 | 0 | 84 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,960 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1729560/Python-Easy-Solution | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
temp=defaultdict(set)
res=[0]*k
for log in logs:
temp[log[0]].add(log[1])
for v in temp.values():
res[len(v)-1]+=1
return res | finding-the-users-active-minutes | Python Easy Solution | shandilayasujay | 0 | 107 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,961 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1708665/Understandable-code-for-beginners-like-me-in-python-!! | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
answer=[0]*k
userIdmap={}
for log in logs:
userid=log[0]
activeminute=log[1]
if userid not in userIdmap:
userIdmap[userid]=list()
userIdmap[userid].append(activeminute)
else:
if(activeminute not in userIdmap[userid]):
userIdmap[userid].append(activeminute)
for userid,activeminute in userIdmap.items():
userIdmap[userid]=len(activeminute)
for userid,count in userIdmap.items():
answer[count-1]+=1
return answer | finding-the-users-active-minutes | Understandable code for beginners like me in python !! | kabiland | 0 | 62 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,962 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1673056/Python3-oror-Hashmap-oror-O(n)-time-oror-O(n)-space | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
log_dict = {}
for idx, time in logs:
log_dict.setdefault(idx,set()).add(time)
res = [0] * k
for key in log_dict:
log_dict[key] = len(log_dict[key])
res[log_dict[key]-1] += 1
return res | finding-the-users-active-minutes | Python3 || Hashmap || O(n) time || O(n) space | s_m_d_29 | 0 | 79 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,963 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1608022/Python3-Hashmap-%2B-hashset-solution | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
user2uams = collections.defaultdict(set)
for user_id, minute in logs:
user2uams[user_id].add(minute)
res = [0] * k
for key in user2uams:
res[len(user2uams[key]) - 1] += 1
return res | finding-the-users-active-minutes | [Python3] Hashmap + hashset solution | maosipov11 | 0 | 67 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,964 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1371236/Python3-Easy-to-Understand | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
returned_counters = [0 for i in range(0, k)]
refer_dict = {}
for array in logs:
refer_dict[array[0]] = []
for array in logs:
if array[1] not in refer_dict[array[0]]:
refer_dict[array[0]].append(array[1])
for key, val in refer_dict.items():
uam = len(val)
returned_counters[uam - 1] += 1
return returned_counters | finding-the-users-active-minutes | Python3 Easy to Understand | RobertObrochta | 0 | 65 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,965 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1211433/Simplest-python-3-approach-using-dictionary-(-hash-map-)-or-faster-than-97 | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
userDict = dict()
"""
below we create a dictionary where the user ID is the key and the value
is a python set of the minutes the user was active
we create a set because we only want unique minutes to be counted
"""
for log in logs:
if(log[0] not in userDict):
userDict[log[0]] = set()
userDict[log[0]].add(log[1])
"""
for the eg; logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5
userDict will look like
{
0: {2, 5},
1: {2, 3}
}
which means user-0 was active for minutes 2,5 and so on.
Then below we iterate over the dict and count the len of set for each key
which gives us the total no of *UNIQUE* minutes the user was active for
eg; the user-0 was active for two unique minutes cuz the len of its set is 2
then we increment the corresponding index i in our answer array by 1 which indicates
that 1 user was active for i minutes
eg; our answer array will be [0,2,0,0,0] which tells us there are 2 users
that were active for 2 *unique* minutes
"""
answer = [0]*k
for user in userDict:
answer[len(userDict[user])-1] += 1
return answer | finding-the-users-active-minutes | Simplest python 3 approach using dictionary ( hash map ) | faster than 97% | thecoder_elite | 0 | 103 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,966 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1166855/Python3-Easy-Solution-with-Comments-Used-Set-and-Map | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
'''
1. Use a dictionary where each key represents a user and the value is the set of minutes when the user was active
2. Initialize the 1-indexed resultant array with k zeros
3. For each logs[i] in list logs
-add time_i to the set s_i of key ID_i
-update the resultant array's index x where
x = no. of elements in s_i - 1, as the array is actually 0 indexed
'''
cnt_dict = {} #dictionary of set of unique active minutes for each user
res = [0]*k #resultant ouput array, initialized by zeros
for log in logs:
if log[0] in cnt_dict.keys(): #a key in the dictionary is already present for this user id
res[len(cnt_dict[log[0]])-1]-=1 #as the active minute is going to be added to the set, decrease the count of the index (i = no. of elements in set -1) in the array res
cnt_dict[log[0]].add(log[1]) #add this time to the set for this user in the dictionary
res[len(cnt_dict[log[0]])-1]+=1 #as a new user of this count may be obtained, increase the count of the index (i = no. of elements in set -1) in the array res
else:
cnt_dict[log[0]]=set() #no key present for this user in the dictionary, initialize empty set
cnt_dict[log[0]].add(log[1]) #add this time to the set for this user in the dictionary
res[len(cnt_dict[log[0]])-1]+=1 #as a new user of this count may be obtained, increase the count of the index (i = no. of elements in set -1) in the array res
return res | finding-the-users-active-minutes | Python3 Easy Solution with Comments, Used Set and Map | bPapan | 0 | 106 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,967 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1141812/Python-Solution | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
users = {}
answer = [0] * k
for idd, time in logs:
if idd not in users:
users[idd] = set()
users[idd].add(time)
else:
users[idd].add(time)
for user in users:
answer[len(users[user]) - 1] += 1
return answer | finding-the-users-active-minutes | Python Solution | mariandanaila01 | 0 | 75 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,968 |
https://leetcode.com/problems/finding-the-users-active-minutes/discuss/1141458/Simple-Python-3 | class Solution:
def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
log = defaultdict(set)
for x, y in logs:
log[x].add(y)
output = [0]*k
for i in log.values():
output[len(i)-1] += 1
return output | finding-the-users-active-minutes | Simple, Python 3 | VijayantShri | 0 | 44 | finding the users active minutes | 1,817 | 0.807 | Medium | 25,969 |
https://leetcode.com/problems/minimum-absolute-sum-difference/discuss/1715575/Python-%2B-Fully-Explained-%2B-Best-Solution | class Solution:
def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
n = len(nums1)
diff = []
sum = 0
for i in range(n):
temp = abs(nums1[i]-nums2[i])
diff.append(temp)
sum += temp
nums1.sort()
best_diff = []
for i in range(n):
idx = bisect.bisect_left(nums1, nums2[i])
if idx != 0 and idx != n:
best_diff.append(
min(abs(nums2[i]-nums1[idx]), abs(nums2[i]-nums1[idx-1])))
elif idx == 0:
best_diff.append(abs(nums2[i]-nums1[idx]))
else:
best_diff.append(abs(nums2[i]-nums1[idx-1]))
saved = 0
for i in range(n):
saved = max(saved, diff[i]-best_diff[i])
return (sum-saved) % ((10**9)+(7)) | minimum-absolute-sum-difference | [Python] + Fully Explained + Best Solution ✔ | leet_satyam | 10 | 540 | minimum absolute sum difference | 1,818 | 0.302 | Medium | 25,970 |
https://leetcode.com/problems/minimum-absolute-sum-difference/discuss/1141447/Python-Binary-Search | class Solution:
def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
nums1, nums2 = zip(*sorted(zip(nums1, nums2)))
mad = [abs(nums1[i] - nums2[i]) for i in range(len(nums1))]
M = sum(mad)
MOD = 10**9 + 7
best = 0
for i in range(len(nums1)):
if nums1[i] != nums2[i]:
j = bisect.bisect_left(nums1, nums2[i])
if j == len(nums1):
best = max(best, mad[i] - abs(nums1[-1] - nums2[i]))
elif j == 0:
best = max(best, mad[i] - abs(nums1[0] - nums2[i]))
else:
new = min(abs(nums1[j] - nums2[i]), abs(nums1[j-1] - nums2[i]))
best = max(best, mad[i] - new)
return (M - best) % MOD | minimum-absolute-sum-difference | [Python] Binary Search | rowe1227 | 4 | 384 | minimum absolute sum difference | 1,818 | 0.302 | Medium | 25,971 |
https://leetcode.com/problems/minimum-absolute-sum-difference/discuss/1141371/Python3-greedy | class Solution:
def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
s1 = sorted(nums1)
ans = bnft = 0
for x, y in zip(nums1, nums2):
ans += abs(x - y)
k = bisect_left(s1, y)
if k < len(s1): bnft = max(bnft, abs(x - y) - (s1[k] - y)) # benefit of replacing x to s1[k]
if 0 < k: bnft = max(bnft, abs(x - y) - (y - s1[k-1])) # benefit of replacing x to s1[k-1]
return (ans - bnft) % 1_000_000_007 | minimum-absolute-sum-difference | [Python3] greedy | ye15 | 3 | 280 | minimum absolute sum difference | 1,818 | 0.302 | Medium | 25,972 |
https://leetcode.com/problems/minimum-absolute-sum-difference/discuss/1665756/Python-O(nlogn)-time-O(n)-space-solution-using-sorting-%2B-binary-search | class Solution:
def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
n = len(nums1)
total = 0
for i in range(n):
total += abs(nums1[i] - nums2[i])
s = sorted(nums1)
def minimize(x):
left, right = 0, n-1
while left < right:
mid = (left + right) // 2
if s[mid] >= x:
right = mid
else:
left = mid + 1
if left == 0:
return s[left]-x
else:
return min(abs(s[left]-x), abs(x-s[left-1]))
res = float('inf')
for i in range(n):
t = total - abs(nums1[i]-nums2[i])
c = minimize(nums2[i])
res = min(res, t + c)
return min(res, total) % (10**9 + 7) | minimum-absolute-sum-difference | Python O(nlogn) time, O(n) space solution using sorting + binary search | byuns9334 | 1 | 150 | minimum absolute sum difference | 1,818 | 0.302 | Medium | 25,973 |
https://leetcode.com/problems/minimum-absolute-sum-difference/discuss/2260640/Python3-or-binary-search-or-commented | class Solution:
def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
nums3 = sorted(nums1)
n = [-1 for i in range(len(nums1))] # i = index in nums1 | n[i] = index in nums3 | the best diff for every nums1[i] is nums3[n[i]] - nums2[i]
pos = -math.inf
m_diff, s = 0, 0
for i in range(len(nums2)):
m1 = bisect.bisect_left(nums3, nums2[i]) # m1 = upper and m2 = lower
if m1 >= len(nums1): # in case m1 is outside the array's range
m1 = len(nums1)-1
m2 = m1 - 1
val = abs(nums1[i] - nums2[i])
if abs(nums3[m1] - nums2[i]) < val:
val = abs(nums3[m1] - nums2[i])
n[i] = m1
if m2 >= 0:
if abs(nums3[m2] - nums2[i]) < val:
n[i] = m2
for i in range(len(nums3)): #get the maximum change in difference and the position to change in order to get that
currentDiff = abs(nums1[i] - nums2[i])
newDiff = abs(nums3[n[i]] - nums2[i])
if n[i] != -1 and currentDiff > newDiff and currentDiff - newDiff > m_diff:
m_diff = currentDiff - newDiff
pos = i
if pos != -math.inf: nums1[pos] = nums3[n[pos]] # no change to be made if pos == -inf
for i in range(len(nums1)):
s += abs(nums1[i] - nums2[i])
return s % (10**9 + 7) | minimum-absolute-sum-difference | Python3 | binary search | commented | FlorinnC1 | 0 | 174 | minimum absolute sum difference | 1,818 | 0.302 | Medium | 25,974 |
https://leetcode.com/problems/minimum-absolute-sum-difference/discuss/1351088/Bisect-for-distinct-97-speed | class Solution:
def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
distinct1 = sorted(set(nums1))
len1 = len(distinct1)
max_diff = sum_abs_diff = 0
for a, b in zip(nums1, nums2):
abs_diff = abs(a - b)
sum_abs_diff += abs_diff
idx = bisect_left(distinct1, b)
if idx < len1:
max_diff = max(max_diff, abs_diff - abs(distinct1[idx] - b))
if idx > 0:
max_diff = max(max_diff, abs_diff - abs(distinct1[idx - 1] - b))
return (sum_abs_diff - max_diff) % 1_000_000_007 | minimum-absolute-sum-difference | Bisect for distinct, 97% speed | EvgenySH | 0 | 167 | minimum absolute sum difference | 1,818 | 0.302 | Medium | 25,975 |
https://leetcode.com/problems/minimum-absolute-sum-difference/discuss/1142385/python3-Solution-for-ref-with-sort-greedy-algorithm-%2B-binary-search | class Solution:
def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
## Sort the array to prepare for binary search
snums1 = sorted(nums1)
## Binary search solution for finding the closest value based on absolute difference is picked up as block from here (since already exists).
## https://tutorialspoint.dev/algorithm/divide-and-conquer/find-closest-number-array
def find_closest_abs_diff(arr, n, target):
if (target <= arr[0]):
return arr[0]
if (target >= arr[n - 1]):
return arr[n - 1]
i = 0; j = n; mid = 0
while (i < j):
mid = (i + j) // 2
if (arr[mid] == target):
return arr[mid]
if (target < arr[mid]) :
if (mid > 0 and target > arr[mid - 1]):
return getClosest(arr[mid - 1], arr[mid], target)
j = mid
else :
if (mid < n - 1 and target < arr[mid + 1]):
return getClosest(arr[mid], arr[mid + 1], target)
i = mid + 1
return arr[mid]
def getClosest(val1, val2, target):
if (target - val1 >= val2 - target):
return val2
else:
return val1
## Find absolute difference values of array's
absdiffarr = [abs(nums1[idx] - nums2[idx]) for idx in range(len(nums1))]
## Find the sum of absolute differences, we are trying to minimize this value.
absdiff = sum(absdiffarr)
output = float('inf')
## go over the arrays and find the value that can be replaced to minimize absolute difference.
for idx in range(len(nums1)):
change = find_closest_abs_diff(snums1, len(nums1), nums2[idx])
## remove the current absolute difference at index and check for minimized value,
## if we replace that with the closest value found in the array.
output = min(output, absdiff-absdiffarr[idx]+(abs(change - nums2[idx])))
return output % (10**9+7) | minimum-absolute-sum-difference | [python3] Solution for ref with sort, greedy algorithm + binary search | vadhri_venkat | 0 | 90 | minimum absolute sum difference | 1,818 | 0.302 | Medium | 25,976 |
https://leetcode.com/problems/minimum-absolute-sum-difference/discuss/1141379/Accepted-O(N)-Easy-to-understand-solution-or-Inline-comments-for-explanation | class Solution:
def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
n = len(nums1)
MOD = 10**9+7
# Maintain a MAX heap defined as structure - (abs_diffi, nums1[i], nums[j])
heap = []
abs_sum = 0
for i in range(n):
abs_diff = abs(nums1[i]-nums2[i])
abs_sum = (abs_sum+abs_diff)%MOD
heapq.heappush(heap,(-abs_diff, nums1[i], nums2[i])) # note the -ve sign to make it max_heap
print(heap)
# Get the element with max absolute difference from heap
top = heapq.heappop(heap)
abs_sum -= abs(top[0]) # subtract previous abs_diff as we'll update it below
k = top[2]
# find a new_val closest to k
# and min_diff is the new absoulte diff with k
min_diff = abs(top[0])
new_val = 0
for num1 in nums1:
new_diff = abs(num1-k)
if new_diff<min_diff:
min_diff = new_diff
new_val = num1
# add the new diff to calculate updated sum
abs_sum = (abs_sum+min_diff)%MOD
return abs_sum | minimum-absolute-sum-difference | [Accepted] O(N) Easy to understand solution | Inline comments for explanation | CaptainX | 0 | 206 | minimum absolute sum difference | 1,818 | 0.302 | Medium | 25,977 |
https://leetcode.com/problems/number-of-different-subsequences-gcds/discuss/1144445/Python3-enumerate-all-possibilities | class Solution:
def countDifferentSubsequenceGCDs(self, nums: List[int]) -> int:
nums = set(nums)
ans = 0
m = max(nums)
for x in range(1, m+1):
g = 0
for xx in range(x, m+1, x):
if xx in nums:
g = gcd(g, xx)
if g == x: ans += 1
return ans | number-of-different-subsequences-gcds | [Python3] enumerate all possibilities | ye15 | 4 | 234 | number of different subsequences gcds | 1,819 | 0.385 | Hard | 25,978 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/1152412/Python3-line-sweep | class Solution:
def arraySign(self, nums: List[int]) -> int:
ans = 1
for x in nums:
if x == 0: return 0
if x < 0: ans *= -1
return ans | sign-of-the-product-of-an-array | [Python3] line sweep | ye15 | 58 | 6,300 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,979 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/1499851/Solution-using-xor-and-only-a-bool-for-memory | class Solution:
def arraySign(self, nums: List[int]) -> int:
signs = False
for x in nums:
if x == 0 : return 0
signs = signs ^ (x < 0)
if signs : return -1
else: return 1 | sign-of-the-product-of-an-array | Solution using xor and only a bool for memory | somedev12 | 5 | 221 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,980 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2224875/Python3-simple-one-liner | class Solution:
def arraySign(self, nums: List[int]) -> int:
def signFun(x):
return 1 if x > 0 else -1 if x < 0 else 0
return signFun(math.prod(nums)) | sign-of-the-product-of-an-array | 📌 Python3 simple one liner | Dark_wolf_jss | 3 | 52 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,981 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/1200880/WEEB-DOES-PYTHON-(BEATS-99.32) | class Solution:
def arraySign(self, nums: List[int]) -> int:
nums, count = sorted(nums), 0
for i in range(len(nums)):
if nums[i] == 0:
return 0
if nums[i] > 0:
if count % 2 == 0: return 1
else: return -1
count+=1
if count % 2 == 0: # if all are negative
return 1
else: return -1 | sign-of-the-product-of-an-array | WEEB DOES PYTHON (BEATS 99.32%) | Skywalker5423 | 2 | 281 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,982 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2013972/Python-Simple-Solution-or-Time-%3A-O(n) | class Solution:
def arraySign(self, nums: List[int]) -> int:
negative = 1
for number in nums:
if number == 0:
return 0
elif number < 0:
negative *= -1
return negative | sign-of-the-product-of-an-array | Python Simple Solution | Time : O(n) | Call-Me-AJ | 1 | 70 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,983 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2013972/Python-Simple-Solution-or-Time-%3A-O(n) | class Solution:
def arraySign(self, nums: List[int]) -> int:
product = 1
for number in nums:
product *= number
return 0 if product == 0 else -1 if product < 0 else 1 | sign-of-the-product-of-an-array | Python Simple Solution | Time : O(n) | Call-Me-AJ | 1 | 70 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,984 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/1851671/Python3-straightforward-solution | class Solution:
def arraySign(self, nums: List[int]) -> int:
p = math.prod(nums)
if p > 0:
return 1
elif p < 0:
return -1
else:
return 0 | sign-of-the-product-of-an-array | Python3 straightforward solution | alishak1999 | 1 | 44 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,985 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/1851425/2-Lines-Python-Solution-oror-92-Faster-(60ms)-oror-Memory-less-than-65 | class Solution:
def arraySign(self, nums: List[int]) -> int:
p = reduce(mul, nums)
return 1 if p>0 else -1 if p<0 else 0 | sign-of-the-product-of-an-array | 2-Lines Python Solution || 92% Faster (60ms) || Memory less than 65% | Taha-C | 1 | 80 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,986 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/1830272/Python-or-Time%3A-O(n)-or-Space%3A-O(1) | class Solution:
def arraySign(self, nums: List[int]) -> int:
sign = 1
for n in nums:
if n == 0:
return 0
elif n < 0:
sign = -sign
return sign | sign-of-the-product-of-an-array | Python | Time: O(n) | Space: O(1) | nashvenn | 1 | 29 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,987 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/1363381/simple-and-easy-using-python | class Solution:
def arraySign(self, nums: List[int]) -> int:
t=1
for i in range(len(nums)):
t = t* nums[i]
if t> 0:
return 1
elif t<0:
return -1
else:
return 0 | sign-of-the-product-of-an-array | simple & easy using python | gulsan | 1 | 117 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,988 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/1175539/Sign-of-product-(Fast-and-easy-solution) | class Solution:
def arraySign(self, nums: List[int]) -> int:
lst = nums
prod = 1
for i in lst:
prod = prod * i
if prod > 0:
return 1
elif prod < 0:
return -1
elif prod == 0:
return 0 | sign-of-the-product-of-an-array | Sign of product (Fast and easy solution) | iamvatsalpatel | 1 | 90 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,989 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2793239/python-simplest | class Solution:
def arraySign(self, nums: List[int]) -> int:
x=1
for i in range(len(nums)):
x=x*nums[i]
if x==0:
return 0
elif x<0:
return -1
else:
return 1 | sign-of-the-product-of-an-array | [python]--simplest | user9516zM | 0 | 3 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,990 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2760779/Python-Solution-All-approach | class Solution:
def arraySign(self, nums: List[int]) -> int:
count_negative = 0
for ele in nums:
if ele<0:
count_negative+=1
elif ele==0:
return 0
if count_negative&1:
return -1
return 1
#ANother approach
''' product=1
for i in nums:
if i ==0:
return 0
elif i<0:
product*=-1
return product
# Brute FORCE
n=len(nums)
prod=1
for i in range (n):
prod=prod*nums[i]
if prod==0:
return 0
elif prod<0:
return -1
elif prod>0:
return 1'''
#Now lets's optimise this code
''' if 0 in nums:
return 0
count=0
for i in nums:
if i<0:
count+=1
if count%2==0:
return 1
else:
return -1''' | sign-of-the-product-of-an-array | Python Solution All approach | mritunjayyy | 0 | 3 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,991 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2759362/Simple-Python3-Solution-Better-Runtime | class Solution:
def arraySign(self, nums: List[int]) -> int:
if 0 in nums:
return 0
result = 0
for x in nums:
if x < 0:
result += 1
return 1 if result %2 == 0 else -1 | sign-of-the-product-of-an-array | Simple Python3 Solution Better Runtime | vivekrajyaguru | 0 | 1 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,992 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2711251/optimized-solution-python-faster-than-92-submissions | class Solution(object):
def arraySign(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
neg_num=0
for i in nums:
if i==0:
return 0;
elif i<1:
neg_num+=1
if neg_num %2 == 0:
return 1
return -1
#up vote is appreciated | sign-of-the-product-of-an-array | optimized solution python faster than 92% submissions | ojasgupta25 | 0 | 4 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,993 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2702115/Python-or-Less-memory-than-98.3-or-Without-calculating-product | class Solution:
def arraySign(self, nums: List[int]) -> int:
negative_count = 0
for n in nums:
if n == 0: return 0
elif n < 0: negative_count += 1
if negative_count % 2 == 0: return 1
else: return -1 | sign-of-the-product-of-an-array | ✅ Python | Less memory than 98.3% | Without calculating product | anandanshul001 | 0 | 6 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,994 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2692819/Easy-Solution | class Solution:
def arraySign(self, nums: List[int]) -> int:
res=1
for i in nums:
res*=i
if(res==0):
return 0
elif(res>0):
return 1
return -1 | sign-of-the-product-of-an-array | Easy Solution | Raghunath_Reddy | 0 | 4 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,995 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2692818/Easy-Solution | class Solution:
def arraySign(self, nums: List[int]) -> int:
res=1
for i in nums:
res*=i
if(res==0):
return 0
elif(res>0):
return 1
return -1 | sign-of-the-product-of-an-array | Easy Solution | Raghunath_Reddy | 0 | 2 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,996 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2682891/Python3-or-Easy | class Solution:
def arraySign(self, nums: List[int]) -> int:
p = 1 # product
if 0 in nums:
return 0
for i in nums:
if i < 0:
p *= -1
if p > 0:
return 1
else:
return -1 | sign-of-the-product-of-an-array | Python3 | Easy | AnzheYuan1217 | 0 | 17 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,997 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2667803/Python-or-O(n)-solution | class Solution:
def arraySign(self, nums: List[int]) -> int:
minus, zero = 0, False
for num in nums:
if num == 0:
zero = True
break
if num < 0:
minus += 1
if zero:
return 0
if minus % 2:
return -1
return 1 | sign-of-the-product-of-an-array | Python | O(n) solution | LordVader1 | 0 | 9 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,998 |
https://leetcode.com/problems/sign-of-the-product-of-an-array/discuss/2665144/Python3-Solution.-NO-need-to-calculate-Product | class Solution:
def arraySign(self, nums: List[int]) -> int:
product = 1
for x in nums:
if x == 0:
return 0
if x < 0:
product *= -1
return product | sign-of-the-product-of-an-array | Python3 Solution. [NO need to calculate Product] | Udaay | 0 | 10 | sign of the product of an array | 1,822 | 0.66 | Easy | 25,999 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.