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/average-salary-excluding-the-minimum-and-maximum-salary/discuss/2470889/python3-using-remove-and-sum | class Solution:
def average(self, salary: List[int]) -> float:
salary.remove(min(salary))
salary.remove(max(salary))
return sum(salary)/len(salary) | average-salary-excluding-the-minimum-and-maximum-salary | [python3] using remove and sum | hhlinwork | 0 | 28 | average salary excluding the minimum and maximum salary | 1,491 | 0.627 | Easy | 22,200 |
https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/discuss/2462497/Python3-solutions-Easy-with-POP | class Solution:
def average(self, salary: List[int]) -> float:
salary.sort(reverse=False)
salary.pop()
salary.pop(0)
sal = 0
for i in salary:
sal = sal+i
average = sal/len(salary)
return average | average-salary-excluding-the-minimum-and-maximum-salary | Python3 solutions -Easy with POP | prakashpandit | 0 | 38 | average salary excluding the minimum and maximum salary | 1,491 | 0.627 | Easy | 22,201 |
https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/discuss/2436764/Average-Salary-Excluding-the-Minimum-and-Maximum-Salary | class Solution:
def merge(self,arr,L,R,m):
i= j= k = 0
while i < len(L) and j<len(R):
if L[i]>R[j]:
arr[k]= R[j]
j+=1
else:
arr[k] =L[i]
i+=1
k+=1
while i<len(L):
arr[k]= L[i]
i+=1
k+=1
while j<len(R):
arr[k]= R[j]
j+=1
k+=1
def mergeSort(self,arr):
if len(arr)>1:
m = len(arr)//2
L = arr[:m]
R = arr[m:]
self.mergeSort(L)
self.mergeSort(R)
self.merge(arr,L,R,m)
def average(self, salary: List[int]) -> float:
self.mergeSort(salary)
print(salary)
neededArray =salary[1:-1]
x = sum(neededArray)
avg = x/len(neededArray)
return avg | average-salary-excluding-the-minimum-and-maximum-salary | Average Salary Excluding the Minimum and Maximum Salary | dhananjayaduttmishra | 0 | 23 | average salary excluding the minimum and maximum salary | 1,491 | 0.627 | Easy | 22,202 |
https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/discuss/2436764/Average-Salary-Excluding-the-Minimum-and-Maximum-Salary | class Solution:
def average(self, salary: List[int]) -> float:
salary.sort()
neededArray =salary[1:-1]
x = sum(neededArray)
avg = x/len(neededArray)
return avg | average-salary-excluding-the-minimum-and-maximum-salary | Average Salary Excluding the Minimum and Maximum Salary | dhananjayaduttmishra | 0 | 23 | average salary excluding the minimum and maximum salary | 1,491 | 0.627 | Easy | 22,203 |
https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/discuss/2436764/Average-Salary-Excluding-the-Minimum-and-Maximum-Salary | class Solution:
def average(self, salary: List[int]) -> float:
maximum = 999
minimun = 10**6 +1
for i in salary:
if i > maximum :
maximum = i
if i < minimun :
minimun = i
x = 0
count = 0
for i in salary :
if i != maximum and i != minimun :
x+=i
count+=1
avg = x/count
return avg | average-salary-excluding-the-minimum-and-maximum-salary | Average Salary Excluding the Minimum and Maximum Salary | dhananjayaduttmishra | 0 | 23 | average salary excluding the minimum and maximum salary | 1,491 | 0.627 | Easy | 22,204 |
https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/discuss/2329188/One-line-python | class Solution:
def average(self, salary: List[int]) -> float:
return (sum(salary)-min(salary)-max(salary))/(len(salary)-2) | average-salary-excluding-the-minimum-and-maximum-salary | One line python | sunakshi132 | 0 | 79 | average salary excluding the minimum and maximum salary | 1,491 | 0.627 | Easy | 22,205 |
https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/discuss/2316058/Just-another-one-liner | class Solution:
def average(self, salary: List[int]) -> float:
return sum(sorted(salary)[1:-1]) / (len(salary) - 2) | average-salary-excluding-the-minimum-and-maximum-salary | Just another one liner | pcdean2000 | 0 | 20 | average salary excluding the minimum and maximum salary | 1,491 | 0.627 | Easy | 22,206 |
https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/discuss/2307123/Python-oror-One-Line | class Solution:
def average(self, salary: List[int]):
return (sum(salary) - max(salary) - min(salary)) / (len(salary) - 2) | average-salary-excluding-the-minimum-and-maximum-salary | Python || One Line | Aman-Mahore | 0 | 47 | average salary excluding the minimum and maximum salary | 1,491 | 0.627 | Easy | 22,207 |
https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/discuss/2296893/ONE-LINER-oror-Beats-63-oror-Easiest-Solution | '''class Solution:
def average(self, salary: List[int]) -> float:
return (sum(salary)-(min(salary)+max(salary)))/(len(salary)-2)''' | average-salary-excluding-the-minimum-and-maximum-salary | ONE LINER || Beats 63% || Easiest Solution | keertika27 | 0 | 47 | average salary excluding the minimum and maximum salary | 1,491 | 0.627 | Easy | 22,208 |
https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/discuss/2296308/python3-or-Solution | class Solution:
def average(self, salary: List[int]) -> float:
salary.remove(min(salary))
salary.remove(max(salary))
avg = 0
for i in range(0,len(salary)):
avg += salary[i]
return float(avg / len(salary)) | average-salary-excluding-the-minimum-and-maximum-salary | python3 | Solution | arvindchoudhary33 | 0 | 42 | average salary excluding the minimum and maximum salary | 1,491 | 0.627 | Easy | 22,209 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1352274/Python3-simple-solution-using-list | class Solution:
def kthFactor(self, n: int, k: int) -> int:
factors = []
for i in range(1,n+1):
if n % i == 0:
factors.append(i)
if k > len(factors):
return -1
else:
return factors[k-1] | the-kth-factor-of-n | Python3 simple solution using list | EklavyaJoshi | 1 | 139 | the kth factor of n | 1,492 | 0.624 | Medium | 22,210 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1257792/Python-oror-clean-and-concise-oror | class Solution:
def kthFactor(self, n: int, k: int) -> int:
u=[]
for i in range(1,int(n**(0.5))+1):
if n%i==0:
u.append(i)
if n//i != i:
u.append(n//i)
u.sort()
if k<=len(u):
return u[k-1]
return -1 | the-kth-factor-of-n | Python || clean and concise || | chikushen99 | 1 | 261 | the kth factor of n | 1,492 | 0.624 | Medium | 22,211 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/2823278/Easy-Python3-SolutionororO(N)-oror-7-Line-Code | class Solution:
def kthFactor(self, n: int, k: int) -> int:
count = 0
for i in range(1, n + 1):
if n % i == 0:
count += 1
if count == k:
return i
return -1 | the-kth-factor-of-n | Easy Python3 Solution||O(N) || 7 Line Code | ghanshyamvns7 | 0 | 2 | the kth factor of n | 1,492 | 0.624 | Medium | 22,212 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/2822865/Python3-solution | class Solution:
def kthFactor(self, n: int, k: int) -> int:
factors = []
for i in range(1, n+1):
if n % i == 0:
factors.append(i)
try:
return factors[k-1]
except:
return -1 | the-kth-factor-of-n | Python3 solution | SupriyaArali | 0 | 2 | the kth factor of n | 1,492 | 0.624 | Medium | 22,213 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/2759752/python-Simple-code | class Solution:
def kthFactor(self, n: int, k: int) -> int:
l = []
for i in range(1,n+1):
if(n%i == 0):
l.append(i)
if k <= len(l):
return l[k-1]
return -1 | the-kth-factor-of-n | python Simple code | manishx112 | 0 | 11 | the kth factor of n | 1,492 | 0.624 | Medium | 22,214 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/2562310/Python-or-T-O(n-log(n))-or-93.39-faster | class Solution:
def kthFactor(self, n: int, k: int) -> int:
fs =list(self.factors(n)) # to make it subscriptable
fs.sort()
return fs[k-1] if len(fs) >= k else -1
def factors(self, n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
``` | the-kth-factor-of-n | Python | T-O(n log(n)) | 93.39% faster | Yusuf0 | 0 | 61 | the kth factor of n | 1,492 | 0.624 | Medium | 22,215 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/2532245/easy-python-solution | class Solution:
def kthFactor(self, n: int, k: int) -> int:
counter = 1
for i in range(1, n+1) :
if n%i == 0 :
if counter == k :
return i
else :
counter += 1
return -1 | the-kth-factor-of-n | easy python solution | sghorai | 0 | 26 | the kth factor of n | 1,492 | 0.624 | Medium | 22,216 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/2015395/Python-solution-faster-than-99 | class Solution:
def kthFactor(self, n: int, k: int) -> int:
factors = []
for i in range(1, n+1):
if n % i == 0:
factors.append(i)
if k <= len(factors):
return factors[k-1]
return -1 | the-kth-factor-of-n | Python solution faster than 99% | alishak1999 | 0 | 143 | the kth factor of n | 1,492 | 0.624 | Medium | 22,217 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1982600/Python-3-Solution | class Solution:
def kthFactor(self, n: int, k: int) -> int:
f = []
for i in range(1,n+1):
if n % i == 0:
f.append(i)
try:
return f[k-1]
except IndexError:
return -1 | the-kth-factor-of-n | Python 3 Solution | DietCoke777 | 0 | 68 | the kth factor of n | 1,492 | 0.624 | Medium | 22,218 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1952456/Python-(Simple-Approach-and-Beginner-Friendly) | class Solution:
def kthFactor(self, n: int, k: int) -> int:
arr = []
for i in range(1, n+1):
if n%i == 0:
arr.append(i)
print(arr)
return arr[k-1] if (k <= len(arr)) else -1 | the-kth-factor-of-n | Python (Simple Approach and Beginner-Friendly) | vishvavariya | 0 | 60 | the kth factor of n | 1,492 | 0.624 | Medium | 22,219 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1856811/time-o(sqrt(n))-space-o(1)-one-pass-self-explain-easy-to-understand | class Solution:
def kthFactor(self, n: int, k: int) -> int:
if not n or not k: return 0
cnt = 0
for i in range(1, int(math.sqrt(n)) + 1):
if n % i == 0:
cnt += 1
if k == cnt:
return i
k -= cnt
for i in range(int(math.sqrt(n - 1)), 0, -1):
if n % i == 0:
k -= 1
if k == 0:
return n // i
return -1 | the-kth-factor-of-n | time o(sqrt(n)) space o(1) one pass self-explain easy to understand | BichengWang | 0 | 118 | the kth factor of n | 1,492 | 0.624 | Medium | 22,220 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1856781/o(sqrt(n))-one-pass | class Solution:
def kthFactor(self, n: int, k: int) -> int:
if not n or not k: return 0
small_factors = []
large_factors = []
for i in range(1, int(math.sqrt(n)) + 1):
if n % i == 0:
small_factors.append(i)
large_factors.append(n // i)
if len(small_factors) == k: return small_factors[-1]
if small_factors[-1] == large_factors[-1]: del large_factors[-1]
if k - len(small_factors) > len(large_factors): return -1
return large_factors[-(k - len(small_factors))] | the-kth-factor-of-n | o(sqrt(n)) one pass | BichengWang | 0 | 57 | the kth factor of n | 1,492 | 0.624 | Medium | 22,221 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1800250/Python3-accepted-solution | class Solution:
def kthFactor(self, n: int, k: int) -> int:
count=0
for i in range(1,n+1):
if(n%i == 0):
count+=1
if(count==k):
return i
return -1 | the-kth-factor-of-n | Python3 accepted solution | sreeleetcode19 | 0 | 44 | the kth factor of n | 1,492 | 0.624 | Medium | 22,222 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1781477/python3-32ms-dictionary-implementation-or-faster-than-88-uses-lesser-memory-than-90 | class Solution:
def kthFactor(self, n: int, k: int) -> int:
factors = dict() #taking a dictionary here so that the factors do not get repeated.
count = 0
for i in range(1, int(n**0.5)+1):
if n%i == 0:
if not factors.get(i): #enter in the dictionary only if the factor hasn't already been added to the dict, you can put any value corresponding to the factor(key) in the dictionary.
factors[i] = 1
for i in range(int(n**0.5), 0, -1):
if n%i == 0:
if not factors.get(n//i):
count += 1
factors[n//i] = count
factors = list(factors.keys())
if len(factors) < k:
return -1
return factors[k-1]
``` | the-kth-factor-of-n | python3 - 32ms dictionary implementation | faster than 88% uses lesser memory than 90% | kgrv | 0 | 48 | the kth factor of n | 1,492 | 0.624 | Medium | 22,223 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1669060/Python3-Simple-solution-faster-than-97.08-of-submissions.-Time-and-space-complexity | class Solution:
def kthFactor(self, n: int, k: int) -> int:
"""Return the kth factor of an array n"""
narr = ([])
for i in range(n, 0, -1):
if n % i == 0:
narr.append(i)
# Then I only change n if n % i-1 == 0 or better still, n % i==0
if n % i-1 == 0:
n = n//i-1
narr.sort()
print(narr)
for i in range(len(narr)):
if i + 1 == k:
return narr[i]
return -1 | the-kth-factor-of-n | Python3 Simple solution faster than 97.08% of submissions. Time and space complexity? | Osereme | 0 | 60 | the kth factor of n | 1,492 | 0.624 | Medium | 22,224 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1650551/Simple-Python-O(n)-Time-O(n)-Space | class Solution:
def kthFactor(self, n: int, k: int) -> int:
resulting_array = []
for i in range(1, n+1):
if n % i == 0:
resulting_array.append(i)
if len(resulting_array) < k:
return -1
else:
return resulting_array[k-1] | the-kth-factor-of-n | Simple Python O(n) Time O(n) Space | KoalaKeys | 0 | 65 | the kth factor of n | 1,492 | 0.624 | Medium | 22,225 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1649868/Python-3-No-lists-or-dictionaries-solution | class Solution:
def kthFactor(self, n: int, k: int) -> int:
curi = 0
for i in range(1, n+1):
if n%i == 0:
curi += 1
if curi == k:
return i
return -1 | the-kth-factor-of-n | [Python 3] No lists or dictionaries solution | lpswheatley | 0 | 84 | the kth factor of n | 1,492 | 0.624 | Medium | 22,226 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1594342/Python-Two-Approaches-or-No-Extra-Space | class Solution:
def kthFactor(self, n: int, k: int) -> int:
# 1st Approach
# Time: O(sqrt(n))
# Space: O(n)
res = []
ans = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
if n // i == i:
res.append(i)
else:
res.append(i)
ans.insert(0, n//i)
res.extend(ans)
if len(res) >= k:
return res[k-1]
return -1
# 2nd Approach:
# Time: O(n)
# Space: O(1)
for i in range(1, n+1):
if n % i == 0:
k -= 1
if k == 0:
return i
return -1 | the-kth-factor-of-n | Python Two Approaches | No Extra Space ✔ | leet_satyam | 0 | 155 | the kth factor of n | 1,492 | 0.624 | Medium | 22,227 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1543541/Python-solution-oror-traversing-till-N2 | class Solution:
def kthFactor(self, n: int, k: int) -> int:
res = [1]
for i in range(2,int(ceil(n/2))+1):
if n % i == 0:
res.append(i)
res.append(n)
if k > len(res):
return -1
return res[k-1] | the-kth-factor-of-n | Python solution || traversing till N/2 | s_m_d_29 | 0 | 97 | the kth factor of n | 1,492 | 0.624 | Medium | 22,228 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1442805/Python3Python-Simple-solution-using-is_integer-method-w-comments | class Solution:
def kthFactor(self, n: int, k: int) -> int:
# A function to check if a number is integer or not
def is_integer(n):
if isinstance(n, int):
return True
if isinstance(n, float):
return n.is_integer()
return False
# Get all the factors of the number
factors = []
for i in range(1,n+1):
if is_integer(n/i):
factors.append(i)
# k is less or equal to no. of factors, return
if k <= len(factors):
return factors[k-1]
return -1 | the-kth-factor-of-n | [Python3/Python] Simple solution using is_integer method w/ comments | ssshukla26 | 0 | 268 | the kth factor of n | 1,492 | 0.624 | Medium | 22,229 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1110058/Python-or-Beats-99-Memory-92-or-Simple | class Solution:
def kthFactor(self, n: int, k: int) -> int:
i=1
while k:
if n%i==0:
k-=1
if i>n:
return -1
i+=1
return i-1 | the-kth-factor-of-n | Python | Beats 99% Memory 92% | Simple | rajatrai1206 | 0 | 123 | the kth factor of n | 1,492 | 0.624 | Medium | 22,230 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/1070258/Python-Easy-O(n) | class Solution:
def kthFactor(self, n: int, k: int) -> int:
count = 0
for i in range(1,n+1):
if n % i == 0:
count+= 1
if count == k:
return i
return -1 | the-kth-factor-of-n | Python Easy O(n) | Akarsh_B | 0 | 56 | the kth factor of n | 1,492 | 0.624 | Medium | 22,231 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/960417/Python-7-liner-easy-solution | class Solution(object):
def kthFactor(self, n, k):
res= []
for i in range(1,n+1):
if n%i==0:
res.append(i)
if len(res)==k:
break
return res[-1] if len(res)==k else -1 | the-kth-factor-of-n | Python 7 liner easy solution | rachitsxn292 | 0 | 182 | the kth factor of n | 1,492 | 0.624 | Medium | 22,232 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/959780/Python3-O(N)-and-O(logN)-solns | class Solution:
def kthFactor(self, n: int, k: int) -> int:
for i in range(1, n+1):
if not n%i: k -= 1
if not k: return i
return -1 | the-kth-factor-of-n | [Python3] O(N) & O(logN) solns | ye15 | 0 | 61 | the kth factor of n | 1,492 | 0.624 | Medium | 22,233 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/959780/Python3-O(N)-and-O(logN)-solns | class Solution:
def kthFactor(self, n: int, k: int) -> int:
for i in range(1, int(sqrt(n))+1): # forward pass
if not n%i: k -= 1
if not k: return i
while i > 0: # backward pass
if i * i < n:
if not n%i: k -= 1
if not k: return n//i
i -= 1
return -1 | the-kth-factor-of-n | [Python3] O(N) & O(logN) solns | ye15 | 0 | 61 | the kth factor of n | 1,492 | 0.624 | Medium | 22,234 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/959662/Simple-7-Line-Python-Iterative | class Solution:
def kthFactor(self, n: int, k: int) -> int:
count = 0
for i in range(1,n+1):
if n % i == 0:
count += 1
if count == k:
return i
return -1 | the-kth-factor-of-n | Simple 7 Line Python Iterative | Xianggg | 0 | 24 | the kth factor of n | 1,492 | 0.624 | Medium | 22,235 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/713713/Intuitive-approach-by-iteration | class Solution:
def kthFactor(self, n: int, k: int) -> int:
factors = [1]
if n > 3:
factors = factors + [i for i in range(2, n//2+1) if n % i == 0]
factors.append(n)
return -1 if k > len(factors) else factors[k-1] | the-kth-factor-of-n | Intuitive approach by iteration | puremonkey2001 | 0 | 22 | the kth factor of n | 1,492 | 0.624 | Medium | 22,236 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/708468/Python3-fast-yet-space-efficient-solution-The-kth-Factor-or-n | class Solution:
def kthFactor(self, n: int, k: int) -> int:
front = []
for i in range(1, int(n**0.5) + 1):
if n % i == 0:
if len(front) == k - 1:
return i
front.append(i)
if k > len(front) * 2 - int(front[-1]**2 == n):
return -1
return n // front[-k+len(front)] | the-kth-factor-of-n | Python3 fast yet space efficient solution - The kth Factor or n | r0bertz | 0 | 80 | the kth factor of n | 1,492 | 0.624 | Medium | 22,237 |
https://leetcode.com/problems/the-kth-factor-of-n/discuss/708262/PythonPython3-The-kth-Factor-of-n | class Solution:
def kthFactor(self, n: int, k: int) -> int:
try:
return list(filter(lambda x:n%x==0,range(1,n+1)))[k-1]
except:
return -1 | the-kth-factor-of-n | [Python/Python3] The kth Factor of n | newborncoder | 0 | 76 | the kth factor of n | 1,492 | 0.624 | Medium | 22,238 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/708121/Easy-Solution-without-DP-Simple-Pictorial-Explanation-or-Python-Solution. | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
m=0
l=len(nums)
one=True
for i in range(0,l):
if nums[i]==0:
one=False
left=i-1
right=i+1
ones=0
while left>=0:
if nums[left]==1:
ones=ones+1
left=left-1
else:
break
while right<l:
if nums[right]==1:
ones=ones+1
right=right+1
else:
break
if ones>m:
m=ones
if one:
return l-1
return m | longest-subarray-of-1s-after-deleting-one-element | [Easy Solution without DP] Simple Pictorial Explanation | Python Solution. | lazerx | 12 | 1,100 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,239 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/708513/Python3-groupby-Longest-Subarray-of-1's-After-Deleting-One-Element | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
groups = [[k, len(list(g))] for k, g in itertools.groupby(nums)]
if len(groups) == 1:
return groups[0][1] - 1 if groups[0][0] else 0
ans = 0
for i in range(len(groups)):
k, klen = groups[i]
if k:
ans = max(ans, klen)
elif i not in [0, len(groups)-1] and klen == 1:
ans = max(ans, groups[i-1][1] + groups[i+1][1])
return ans | longest-subarray-of-1s-after-deleting-one-element | Python3 groupby - Longest Subarray of 1's After Deleting One Element | r0bertz | 2 | 162 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,240 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1694401/Python-3-Sliding-window-one-loop | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
start, max_window, ones_count = 0,0,0
for end in range(len(nums)):
ones_count += nums[end]
if (end - start + 1 - ones_count) > 1:
ones_count -= nums[start]
start += 1
max_window = max(max_window, end - start + 1)
return max_window - 1 | longest-subarray-of-1s-after-deleting-one-element | Python 3 Sliding window one loop | aver18 | 1 | 115 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,241 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1583932/Python-Sliding-window-Solution | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
windowStart = 0
hashmap = {x: 0 for x in nums}
max_length = 0
if 0 not in hashmap.keys():
return len(nums) - 1
for windowEnd in range(len(nums)):
hashmap[nums[windowEnd]] += 1
if hashmap[0] > 1:
hashmap[nums[windowStart]] -= 1
windowStart += 1
max_length = max(max_length, windowEnd - windowStart)
return (max_length) | longest-subarray-of-1s-after-deleting-one-element | [Python] - Sliding window Solution | TheBatmanNinja | 1 | 111 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,242 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1561514/Easy-understand-and-less-logical-solution | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
zeroPos = []
n = len( nums)
ct = 0
dp = []
for i in range ( n):
if nums[i] == 1:
ct += 1
dp.append(ct)
else:
ct = 0
dp.append(ct)
zeroPos.append(i)
zn = len( zeroPos)
if zn == 0:
return n-1
if zn == n:
return 0
realMax = 0
for i in range ( zn):
val1 = dp[zeroPos[i]]
val2 = dp[zeroPos[i]]
if zeroPos[i] > 0:
val1 = dp[zeroPos[i]-1]
val2 = 0
if zeroPos[i]+1 < n:
val2 = dp[zeroPos[i]+1]
if i < zn-1:
val2 = dp[zeroPos[i+1]-1]
if i == zn-1:
val2 = dp[n-1]
realMax = max( realMax, val1+val2)
return realMax
``` | longest-subarray-of-1s-after-deleting-one-element | Easy understand and less logical solution | krunalk013 | 1 | 78 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,243 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1111101/Python3-sliding-window | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
ans = 0
queue = deque([-1])
for i, x in enumerate(nums):
if not x: queue.append(i)
if len(queue) > 2: queue.popleft()
ans = max(ans, i - queue[0] - 1)
return ans | longest-subarray-of-1s-after-deleting-one-element | [Python3] sliding window | ye15 | 1 | 93 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,244 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/708198/PythonPython3-Longest-Subarray-of-1's-After-Deleting-One-Element | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
if all(map(lambda x:x==1,nums)):
return sum(nums)-1
i = 0
j = 0
zero = []
sum_z = 0
while j<len(nums):
if nums[j] == 0 and not zero:
zero.append(j)
elif nums[j] == 0:
k = sum(nums[i:j])
sum_z = k if k>sum_z else sum_z
i, zero[0] = zero[0]+1, j
j += 1
k = sum(nums[i:j])
sum_z = k if k>sum_z else sum_z
return sum_z | longest-subarray-of-1s-after-deleting-one-element | [Python/Python3] Longest Subarray of 1's After Deleting One Element | newborncoder | 1 | 119 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,245 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2814605/python-super-easy-sliding-window | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
l = 0
r = 0
ans = 0
current_s = nums[l]
while l < len(nums) and r < len(nums):
if current_s + 1 >= (r-l+1):
ans = max(ans, r-l)
r += 1
if r < len(nums):
current_s += nums[r]
else:
current_s -= nums[l]
l +=1
return ans | longest-subarray-of-1s-after-deleting-one-element | python super easy sliding window | harrychen1995 | 0 | 1 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,246 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2812144/Python-Simple-sliding-window-solution-O(n)-time-and-O(1)-space | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
left, right = 0 , 0
count = 0
maxi = 0
while right < len(nums):
if nums[right] == 0:
count += 1
while count > 1:
if nums[left] == 0:
count -= 1
left += 1
maxi = max(maxi, right - left + 1)
right += 1
return maxi - 1 | longest-subarray-of-1s-after-deleting-one-element | Python Simple sliding window solution O(n) time and O(1) space | vijay_2022 | 0 | 1 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,247 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2810180/EASIEST-SELF-EXPLAINER-PYTHON-CODE-SINGLE-ITERATION | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
k = 1
i,j = 0,0
ans = 0
while(i <= j and j < len(nums)):
if(nums[j] == 0):
k -= 1
if(k < 0):
while(i <= j and k != 0):
if(nums[i] == 0):
k += 1
i += 1
ans = max(ans,j-i+1)
j += 1
return ans-1 | longest-subarray-of-1s-after-deleting-one-element | EASIEST SELF EXPLAINER PYTHON CODE SINGLE ITERATION | MAYANK-M31 | 0 | 4 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,248 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2740404/Simple-Python-or-O(n)-time-O(1)-space-dynamic-programming | class Solution:
def longestSubarray(self, xs: List[int]) -> int:
ones1 = 0
ones2 = 0
ans = 0
for x in xs:
if x == 1:
ones2 += 1
else:
ans = max(ans, ones1 + ones2)
ones1, ones2 = ones2, 0
ans = max(ans, ones1 + ones2)
if ans == len(xs):
return ans - 1
return ans | longest-subarray-of-1s-after-deleting-one-element | Simple Python | O(n) time, O(1) space, dynamic programming | on_danse_encore_on_rit_encore | 0 | 7 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,249 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2652849/Python%3A-Beats-95-Time-90-Space-O(n)-Solution | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
res, lp, k = 0, 0, 1
for rp in range(len(nums)):
if nums[rp] == 0:
k -= 1
if k < 0:
if nums[lp] == 0:
k += 1
lp += 1
res = max(res, rp-lp)
return res | longest-subarray-of-1s-after-deleting-one-element | Python: Beats 95% Time 90% Space O(n) Solution | shnartho | 0 | 55 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,250 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2645816/Python3-Solution-oror-O(N)-Time-and-O(1)-Space-Complexity | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
count_0=0
n=len(nums)
maxSize=0
currSize=0
left=0
for i in range(n):
currSize+=1
if nums[i]==0:
count_0+=1
if count_0>1:
while count_0>1 and left<=i:
if nums[left]==0:
count_0-=1
currSize-=1
left+=1
if currSize>maxSize and count_0<=1:
maxSize=currSize
return maxSize-1 | longest-subarray-of-1s-after-deleting-one-element | Python3 Solution || O(N) Time & O(1) Space Complexity | akshatkhanna37 | 0 | 3 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,251 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2410858/Python-Solution-or-Classic-Two-Pointer-Sliding-Window-or-Count-0s | class Solution:
def longestSubarray(self, arr: List[int]) -> int:
ans = 0
start, end = 0, 0
currZs = 0
while end < len(arr):
if arr[end] == 0:
currZs += 1
while currZs > 1:
if arr[start] == 0:
currZs -= 1
start += 1
ans = max(ans,end-start+1)
end += 1
return ans-1 | longest-subarray-of-1s-after-deleting-one-element | Python Solution | Classic Two Pointer - Sliding Window | Count 0s | Gautam_ProMax | 0 | 23 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,252 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2346793/Python-easy-to-read-and-understand-or-prefix-suffix | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
n = len(nums)
pre, suf = [1]*n, [1]*n
if nums[0] == 0:pre[0] = 0
if nums[-1] == 0:suf[-1] = 0
for i in range(1, n):
if nums[i] == 1 and nums[i-1] == 1:
pre[i] = pre[i-1] + 1
elif nums[i] == 0:
pre[i] = 0
for i in range(n-2, -1, -1):
if nums[i] == 1 and nums[i+1] == 1:
suf[i] = suf[i+1] + 1
elif nums[i] == 0:
suf[i] = 0
ans = 0
for i in range(n):
if i == 0:
ans = max(ans, suf[i+1])
elif i == n-1:
ans = max(ans, pre[i-1])
else:
ans = max(ans, pre[i-1] + suf[i+1])
return ans | longest-subarray-of-1s-after-deleting-one-element | Python easy to read and understand | prefix-suffix | sanial2001 | 0 | 39 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,253 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2301377/Python-Solution-or-Using-Sliding-Window-Approach | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
L = 0
R = 0
ans = 0
count_of_zeros = 0
#start the sliding window process!
while R < len(nums):
if(nums[R] == 0):
count_of_zeros += 1
#stopping condition!
while count_of_zeros == 2:
ans = max(ans, R - L - 1)
#potentially before contracting window, check if left is pointing to a zero!
if(nums[L] == 0):
count_of_zeros -= 1
L += 1
#continue expanding sliding window!
R += 1
#edge case: if count_of_zeros of input array is ever less than or equal to 1, then
#recompute it!
if(count_of_zeros <= 1):
ans = max(ans, R - L - 1)
return ans | longest-subarray-of-1s-after-deleting-one-element | Python Solution | Using Sliding Window Approach | JOON1234 | 0 | 42 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,254 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2288009/Simple-python-solution | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
currentOnes = 0
prevOnes = 0
maxCount = 0
for i in range(len(nums)):
if nums[i] == 1:
currentOnes += 1
else:
maxCount = max(maxCount, currentOnes + prevOnes)
prevOnes = currentOnes
currentOnes = 0
maxCount = max(maxCount, currentOnes + prevOnes)
return maxCount-1 if maxCount == len(nums) else maxCount | longest-subarray-of-1s-after-deleting-one-element | Simple python solution | Gilbert770 | 0 | 41 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,255 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/2204765/Pyrhon-3-oror-Simple-Solution-oror-Sliding-Window-oror-O(n)-and-O(1) | class Solution:
def longestSubarray(self, nums) -> int:
left = 0
right = 0
zero = 0
while right < len(nums):
if nums[right] == 0:
zero += 1
if zero > 1:
if nums[left] == 0:
zero -= 1
left += 1
right += 1
return right - left - 1 | longest-subarray-of-1s-after-deleting-one-element | Pyrhon 3 || Simple Solution || Sliding Window || O(n) & O(1) | igalart2000 | 0 | 47 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,256 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1637687/Python3-Clean-and-Fast-Solution-with-Sliding-Window | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
idx_zeros = [-1, -1]
ans = 0
for i, num in enumerate(nums):
if num == 0:
ans = max(ans, i - idx_zeros[0] - 2)
idx_zeros[0], idx_zeros[1] = idx_zeros[1], i
if nums[-1]:
ans = max(ans, len(nums) - idx_zeros[0] - 2)
return ans | longest-subarray-of-1s-after-deleting-one-element | Python3 Clean and Fast Solution with Sliding Window | ynnekuw | 0 | 36 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,257 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1628075/python3-time%3A-O(n)-using-prefix-and-suffix | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
n = len(nums)
right = [num for num in nums]
for i in range(n - 1, -1, -1):
if nums[i] == 1:
right[i] += right[i + 1] if i + 1 < n else 0
ans = 0
prev = 0
for i in range(n):
valR = right[i + 1] if i + 1 < n else 0
ans = max(ans, prev + valR)
prev = prev + 1 if nums[i] == 1 else 0
return ans | longest-subarray-of-1s-after-deleting-one-element | python3 time: O(n) using prefix and suffix | reynaldocv | 0 | 29 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,258 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1617995/Most-intuitive-python3-solution | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
left=0
countZeros=0
res=0
n=len(nums)
for right in range(n):
if nums[right]==0:
countZeros+=1
while countZeros>1:
if nums[left]==0:
countZeros-=1
left+=1
res=max(res,right-left)
return res | longest-subarray-of-1s-after-deleting-one-element | Most intuitive python3 solution | Karna61814 | 0 | 46 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,259 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1521676/Python-5-lines | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
if 0 not in nums: return len(nums) - 1
# arrays = list of 1 subarrays [1,1,1,0,1,1] -> ['111', '11']
arrays = ''.join(map(str,nums)).split('0')
ans = float('-inf')
for i in range(1, len(arrays)): ans = max(ans, len(arrays[i-1] + arrays[i]))
return ans | longest-subarray-of-1s-after-deleting-one-element | Python 5 lines | SmittyWerbenjagermanjensen | 0 | 72 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,260 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1442568/Python-very-easy.-O(n)-time-O(1)-space.-No-DP-No-sliding-window | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
n = len(nums)
if nums.count(0) == 0: return n-1
if nums.count(1) == 0: return 0
nums.append(0)
zeroes = [-1]
ones = []
maxv = 0
for idx, num in enumerate(nums):
if num == 0:
if len(zeroes) == 1:
zeroes.append(idx)
else:
zeroes = [zeroes[1], idx]
one = zeroes[1]-zeroes[0] -1
if len(ones) <= 1:
ones.append(one)
elif len(ones) == 2:
ones = [ones[1], one]
if len(ones) == 2:
maxv = max(maxv, ones[1]+ones[0])
return maxv | longest-subarray-of-1s-after-deleting-one-element | Python very easy. O(n) time O(1) space. No DP, No sliding window | byuns9334 | 0 | 66 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,261 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1430281/Python3-solution-two-approaches | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
zeros_index = []
for i in range(len(nums)):
if nums[i] == 0:
zeros_index.append(i)
if len(zeros_index) in [0,1]:
return len(nums)-1
count = 0
for i in range(len(zeros_index)):
if i == 0:
count = zeros_index[i+1]-zeros_index[i] - 1 + zeros_index[i]
elif i == len(zeros_index) - 1:
count = max(count, zeros_index[i] - zeros_index[i-1] - 1 + len(nums)-zeros_index[i]-1)
else:
count = max(count, zeros_index[i] - zeros_index[i-1] - 1 + zeros_index[i+1] -zeros_index[i] - 1)
return count | longest-subarray-of-1s-after-deleting-one-element | Python3 solution two approaches | EklavyaJoshi | 0 | 29 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,262 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1430281/Python3-solution-two-approaches | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
i = j = 0
flag = False
res = 0
while j < len(nums):
if nums[j] == 0:
if flag == False:
flag = True
else:
res = max(j-i-1,res)
while i < j:
if nums[i] == 0:
break
i += 1
i = i + 1
j += 1
res = max(j-i-1,res)
return res | longest-subarray-of-1s-after-deleting-one-element | Python3 solution two approaches | EklavyaJoshi | 0 | 29 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,263 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1311395/Python-Iterative-easy-to-understand | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
# k is the number of possible replacements
k = 1
left = 0
result = 0
for right in range(len(nums)):
if nums[right] == 0:
k -= 1
while k < 0:
if nums[left] == 0:
k += 1
left += 1
result = max(result, right - left)
return result | longest-subarray-of-1s-after-deleting-one-element | [Python] Iterative easy to understand | dlog | 0 | 146 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,264 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/1218096/Python-Solution-Beats-99.51 | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
if 0 not in nums:
return len(nums)-1
if 1 not in nums:
return 0
a = []
cnt = 0
for i in nums:
if i == 0:
if cnt != 0:
a.append(cnt)
cnt = 0
a.append(i)
else:
cnt += 1
if cnt!=0:
a.append(cnt)
Max = 0
for i in range(len(a)):
if a[i] != 0:
continue
if a[i] == 0 and i == len(a)-1:
Max = max(Max,a[i-1])
elif a[i] == 0 and i == 0:
Max = max(Max,a[i+1])
elif a[i] == 0:
Max = max(Max,a[i+1]+a[i-1])
return Max | longest-subarray-of-1s-after-deleting-one-element | Python Solution Beats 99.51% | Piyush_0411 | 0 | 105 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,265 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/980453/Intuitive-approach-by-sliding-windows | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
ans, past_n, curr_n = 0, 0, 0
for n in nums:
if n == 0:
ans = max(past_n + curr_n, ans)
past_n = curr_n
curr_n = 0
else:
curr_n += 1
ans = max(past_n + curr_n, ans)
# if ans == len(nums) which means all numbers are 1. But we still have to delete one
# number which lead to final result as `ans - 1`
return ans if ans < len(nums) else ans - 1 | longest-subarray-of-1s-after-deleting-one-element | Intuitive approach by sliding windows | puremonkey2001 | 0 | 34 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,266 |
https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/discuss/713288/Python3-Better-100-Mem-75-Speed.-With-comments-(Game-Like) | class Solution:
def longestSubarray(self, nums: List[int]) -> int:
sums=0 #Summaries of our streaks before making two mistakes
sumlist=list() #list of that summaries. So we can find our biggest streak later with max()
if sum(nums)==len(nums): #First of all lets test the case there might be no penalties at all
return sum(nums)-1
s=0 #Penalty counter. Be careful, 2 penalties and streak is over!
i=0 #lets start our journey from beginning of our list.
while i< len(nums):
if nums[i]==0 and s==0: #First Penalty!(0) You can continue, for now...
s+=1 #Saving that you already done your first penalty
p=i #Saving your first penalty's location.
i+=1 #Go next one!
elif nums[i]==0 and s!=0: #Second Penalty(0) stop now!!! And tell everything you count by far to 'SUMLIST'
sumlist.append(sums) #saving your streak...
s=0 #new beginning no penalties...
sums=0 #but ofcourse youre starting from bottom
i=p #start from your first penalty's location!
i+=1 #actually next of it. Since its a penalty, its not fair to start with a penalty already
else:
sums+=1 #No penalties? Add 1 more to your pocket
i+=1 #Go next one
sumlist.append(sums) #There might be only one zero so lets append sums just in case, we will return max of that sumlist so, no problem.
return max(sumlist) #Returning maximum of our streaks with only one penalty! | longest-subarray-of-1s-after-deleting-one-element | [Python3] Better %100 Mem - %75 Speed. With comments (Game-Like) | justlookin | 0 | 35 | longest subarray of 1s after deleting one element | 1,493 | 0.602 | Medium | 22,267 |
https://leetcode.com/problems/parallel-courses-ii/discuss/1111255/Python3-dp | class Solution:
def minNumberOfSemesters(self, n: int, dependencies: List[List[int]], k: int) -> int:
pre = [0]*n # prerequisites
for u, v in dependencies:
pre[v-1] |= 1 << (u-1)
@cache
def fn(mask):
"""Return min semesters to take remaining courses."""
if mask == (1 << n) - 1: return 0 # all courses taken
can = [] # available courses
for i in range(n):
if not mask & 1 << i and mask & pre[i] == pre[i]:
can.append(i)
ans = inf
for courses in combinations(can, min(k, len(can))):
temp = mask | reduce(lambda x, y: x | 1 << y, courses, 0)
ans = min(ans, 1 + fn(temp))
return ans
return fn(0) | parallel-courses-ii | [Python3] dp | ye15 | 2 | 171 | parallel courses ii | 1,494 | 0.311 | Hard | 22,268 |
https://leetcode.com/problems/parallel-courses-ii/discuss/2381477/faster-than-100.00-of-Python3 | class Solution:
def minNumberOfSemesters(self, n: int, relations: List[List[int]], k: int) -> int:
graph = [0] * n
out_degree = [0] * n
# -1 to fix 1-based indexing offset from prompt.
for pre_req, course in relations:
graph[course-1] += 1 << (pre_req-1)
out_degree[pre_req-1] += 1
# Just converts course to its shifted value
c2shift = [1<<course for course in range(n)]
start = 0
goal = 2**n-1 # will eq course_total once all have been taken.
queue = collections.deque([(start,0)])
seen = [0] * (2 ** n)
# Similar to Bellman-Ford
while queue:
# course_total is state. Each bit representing a taken course.
course_total, steps = queue.popleft()
available = []
for course_num in range(n):
if (course_total & graph[course_num] == graph[course_num]) \
and (course_total & c2shift[course_num] == 0):
available.append(course_num)
# pre_req courses can unlock others.
pre_reqs = [c2shift[course_num] for course_num in available if out_degree[course_num]]
leaves = [c2shift[course_num] for course_num in available if out_degree[course_num] == 0]
# We only include leaf courses when we have extra space
if len(pre_reqs) <= k:
course_total += sum(pre_reqs) + sum(leaves[:k-len(pre_reqs)])
if course_total == goal:
return steps + 1
if not seen[course_total]:
queue.append((course_total,steps+1))
seen[course_total] = 1
else:
# Trying every combination of the pre_reqs.
# comb is required here because we can't simply take them all (len(pre_reqs) > k)
for batch in itertools.combinations(pre_reqs, k):
diff = sum(batch)
t = course_total + diff
if t == goal:
return steps + 1
if not seen[t]:
queue.append((t, steps+1))
seen[t] = 1![Uploading file...]() | parallel-courses-ii | faster than 100.00% of Python3 | vimla_kushwaha | 1 | 153 | parallel courses ii | 1,494 | 0.311 | Hard | 22,269 |
https://leetcode.com/problems/parallel-courses-ii/discuss/1859239/Python-Topological-Sort-with-Bitmask-and-Backtracking | class Solution:
def minNumberOfSemesters(self, n: int, relations: List[List[int]], k: int) -> int:
adj = defaultdict(list)
in_deg = [0]*n
for prev, nxt in relations:
adj[prev-1].append(nxt-1)
in_deg[nxt-1]+=1
# bit_mask:
# 1 means not taken
# 0 mean taken
# initialize to 1<<n -1
@cache
def backtrack(bit_mask):
if not bit_mask:
return 0
catalog = [i for i in range(n) if in_deg[i] == 0 and bit_mask & 1<<i]
ret = float("inf")
for k_courses in combinations(catalog,min(k,len(catalog))):
nxt_bit_mask = bit_mask
for course in k_courses:
nxt_bit_mask ^= 1<<course
for parent in adj[course]:
in_deg[parent]-=1
ret = min( ret, 1 + backtrack( nxt_bit_mask ) )
for course in k_courses:
for parent in adj[course]:
in_deg[parent]+=1
return ret
return backtrack((1<<n) -1) | parallel-courses-ii | [Python] Topological Sort with Bitmask and Backtracking | haydarevren | 1 | 237 | parallel courses ii | 1,494 | 0.311 | Hard | 22,270 |
https://leetcode.com/problems/parallel-courses-ii/discuss/708303/Python3-BFS-with-bit-mask-(commented) | class Solution:
def minNumberOfSemesters(self, n: int, dependencies: List[List[int]], k: int) -> int:
graph = collections.defaultdict(int)
for x, y in dependencies:
graph[y] |= 1 << (x - 1) # dependency mask of course y
last_mask = (1 << n) - 1 # all course
queue = collections.deque([(0, 0)]) # start with zero course taken
visited = {0}
while queue:
mask, num_semesters = queue.popleft()
# mask & (1 << (x - 1)) == 0 course have not been taken
# mask & graph[x] == graph[x] but all its dependency course have been taken
free_courses = [x for x in range(1, n + 1) if (mask & (1 << (x - 1)) == 0) and mask & graph[x] == graph[x]]
for courses in itertools.combinations(free_courses, min(k, len(free_courses))):
new_mask = mask
for x in courses:
new_mask |= 1 << (x - 1)
if new_mask == last_mask:
return num_semesters + 1
if new_mask not in visited:
visited.add(new_mask)
queue.append((new_mask, num_semesters + 1)) | parallel-courses-ii | [Python3] BFS with bit mask (commented) | ChelseaChenC | 0 | 124 | parallel courses ii | 1,494 | 0.311 | Hard | 22,271 |
https://leetcode.com/problems/path-crossing/discuss/1132447/Python3-simple-solution-faster-than-99-users | class Solution:
def isPathCrossing(self, path: str) -> bool:
l = [(0,0)]
x,y = 0,0
for i in path:
if i == 'N':
y += 1
if i == 'S':
y -= 1
if i == 'E':
x += 1
if i == 'W':
x -= 1
if (x,y) in l:
return True
else:
l.append((x,y))
return False | path-crossing | Python3 simple solution faster than 99% users | EklavyaJoshi | 4 | 252 | path crossing | 1,496 | 0.558 | Easy | 22,272 |
https://leetcode.com/problems/path-crossing/discuss/709229/Python3-10-line-simulation | class Solution:
def isPathCrossing(self, path: str) -> bool:
x, y = 0, 0
seen = {(x, y)}
for move in path:
if move == "N": y += 1
elif move == "S": y -= 1
elif move == "E": x += 1
else: x -= 1
if (x, y) in seen: return True
seen.add((x, y))
return False | path-crossing | [Python3] 10-line simulation | ye15 | 4 | 381 | path crossing | 1,496 | 0.558 | Easy | 22,273 |
https://leetcode.com/problems/path-crossing/discuss/1194037/Python-Line-by-Line-Explanation(Using-dictionary)-Fast-and-efficient-solution | class Solution:
def isPathCrossing(self, path: str) -> bool:
#Store the directions(key) with their corresponding actions(values)
directions = {'N': [0,1], 'E':[1,0], 'W':[-1,0], 'S':[0,-1]}
#Keep the track of visited points
visited = set()
#Add the initial point from where you're starting
visited.add((0,0))
#Current trackers of x and y coordinates
curr_x,curr_y = 0,0
#Loop through all the path
for i in path:
curr_x += directions[i][0]
curr_y += directions[i][1]
#If visited for first time, add them to visited
if (curr_x,curr_y) not in visited:
visited.add((curr_x,curr_y))
else:
return True #Else return True
return False #Return True if there is no re-visiting | path-crossing | Python Line by Line Explanation(Using dictionary) - Fast and efficient solution | iamkshitij77 | 2 | 163 | path crossing | 1,496 | 0.558 | Easy | 22,274 |
https://leetcode.com/problems/path-crossing/discuss/2819934/Simple-Python-oror-Understandable-and-Clean | class Solution:
def isPathCrossing(self, path: str) -> bool:
coor = [0, 0]
coors = [[0, 0]]
for i in path:
if i == 'N': coor[1] += 1
elif i == 'E': coor[0] += 1
elif i == 'S': coor[1] -= 1
else: coor[0] -= 1
if coor in coors: return True
coors.append(coor[:])
return False | path-crossing | Simple Python || Understandable & Clean | qiy2019 | 1 | 24 | path crossing | 1,496 | 0.558 | Easy | 22,275 |
https://leetcode.com/problems/path-crossing/discuss/1878635/Python-Simple-and-Elegant!-Tuple-Set-and-Match-case! | class Solution:
def isPathCrossing(self, path):
point, points = (0,0), {(0,0)}
for dir in path:
match dir:
case "N": point = (point[0]+1, point[1])
case "S": point = (point[0]-1, point[1])
case "E": point = (point[0], point[1]+1)
case "W": point = (point[0], point[1]-1)
if point in points: return True
else: points.add(point)
return False | path-crossing | Python - Simple and Elegant! Tuple, Set, and Match case! | domthedeveloper | 1 | 69 | path crossing | 1,496 | 0.558 | Easy | 22,276 |
https://leetcode.com/problems/path-crossing/discuss/1856205/Simple-Python-Solution-oror-95-Faster-oror-Memory-less-than-75 | class Solution:
def isPathCrossing(self, path: str) -> bool:
points=[(0,0)] ; i=0 ; j=0
for p in path:
if p=='N': j+=1
if p=='S': j-=1
if p=='E': i+=1
if p=='W': i-=1
if (i,j) in points: return True
else: points.append((i,j))
return False | path-crossing | Simple Python Solution || 95% Faster || Memory less than 75% | Taha-C | 1 | 51 | path crossing | 1,496 | 0.558 | Easy | 22,277 |
https://leetcode.com/problems/path-crossing/discuss/1392134/Python-3-Easy-array-solution-or-32-ms-(58.69)-14.3-mb-(80.98) | class Solution:
def isPathCrossing(self, path: str) -> bool:
x, y = 0, 0 # Setup initial points
points = [(0, 0)] # Add initial points to visited array
for c in path: # Iterate through characters in string
# Depending on direction, change x and y value
if c == "N":
y += 1
if c == "E":
x += 1
if c == "S":
y -= 1
if c == "W":
x -= 1
if (x, y) in points: # Check if tuple of x and y is in points, if so, return true
return True
points.append((x, y)) # Add tuple of points to visited array
# Return false beacuse we couldn't find any duplicate
return False | path-crossing | [Python 3] Easy array solution | 32 ms (58.69%), 14.3 mb (80.98%) | SunCrusader | 1 | 85 | path crossing | 1,496 | 0.558 | Easy | 22,278 |
https://leetcode.com/problems/path-crossing/discuss/709606/PYTHON-36ms-solution-using-set | class Solution:
def isPathCrossing(self, path: str) -> bool:
xCoordinate = 0
yCoordinate = 0
pointsCovered = [(0,0)]
for direction in path:
if direction == 'N':
yCoordinate+=1
elif direction == 'S':
yCoordinate-=1
elif direction == 'E':
xCoordinate +=1
else:
xCoordinate-=1
pointsCovered.append((xCoordinate,yCoordinate))
return not len(pointsCovered) == len(set(pointsCovered)) | path-crossing | PYTHON - 36ms solution using set | amanpathak2909 | 1 | 82 | path crossing | 1,496 | 0.558 | Easy | 22,279 |
https://leetcode.com/problems/path-crossing/discuss/2741614/Python3-95-faster-with-explanation | class Solution:
def isPathCrossing(self, path: str) -> bool:
rmap = {}
x, y = 0,0
rmap[x] = {y: False}
for direction in path:
if direction == 'N':
y += 1
elif direction == 'W':
x -= 1
elif direction == 'S':
y -= 1
elif direction == 'E':
x += 1
else:
print('not valid')
if x in rmap:
if y in rmap[x]:
return True
else:
rmap[x][y] = False
else:
rmap[x] = {y :False}
return False | path-crossing | Python3, 95% faster with explanation | cvelazquez322 | 0 | 16 | path crossing | 1,496 | 0.558 | Easy | 22,280 |
https://leetcode.com/problems/path-crossing/discuss/2727109/Easy-understanding-Quick-Python-Solution | class Solution:
def isPathCrossing(self, path: str) -> bool:
i = 0
j = 0
hist = [[0,0]]
for a in path:
if a == 'N':
i += 1
if a == 'S':
i -= 1
if a == 'E':
j += 1
if a == 'W':
j -= 1
if [i,j] not in hist:
hist.append([i,j])
else:
return True
return False | path-crossing | Easy-understanding Quick Python Solution | EdenXiao | 0 | 1 | path crossing | 1,496 | 0.558 | Easy | 22,281 |
https://leetcode.com/problems/path-crossing/discuss/2723970/python-easy-solution-and-99-faster | class Solution:
def isPathCrossing(self, path: str) -> bool:
arr = []
x = 0
y = 0
for d in path:
arr.append((x,y))
if d == "N":
y+=1
elif d== "S":
y-=1
elif d == "E":
x+=1
elif d== "W":
x-=1
if (x,y) in arr:
return True
return False | path-crossing | python easy solution and 99% faster | VikramKumarcoder | 0 | 6 | path crossing | 1,496 | 0.558 | Easy | 22,282 |
https://leetcode.com/problems/path-crossing/discuss/2606044/Python-Set-Solution-or-Beats-90-time-or-O(N) | class Solution:
def isPathCrossing(self, path: str) -> bool:
x, y = 0, 0
visited = set()
visited.add((x,y))
for d in path:
if d == 'N':
y += 1
elif d == 'S':
y -= 1
elif d == 'E':
x += 1
elif d == 'W':
x -= 1
if (x, y) in visited:
return True
else:
visited.add((x,y))
return False | path-crossing | Python Set Solution | Beats 90% time | O(N) | dos_77 | 0 | 9 | path crossing | 1,496 | 0.558 | Easy | 22,283 |
https://leetcode.com/problems/path-crossing/discuss/2408018/Python3-Easy-Approach-using-Sets-and-Dictionaries | class Solution:
def isPathCrossing(self, path: str) -> bool:
x, y = 0, 0
visited = {(x, y)}
for move in path:
if move == 'N': y += 1
elif move == 'S': y -= 1
elif move == 'E': x += 1
else: x -= 1
if (x, y) in visited: return True
visited.add((x, y))
return False | path-crossing | Python3 Easy Approach using Sets and Dictionaries | vjgadre21 | 0 | 22 | path crossing | 1,496 | 0.558 | Easy | 22,284 |
https://leetcode.com/problems/path-crossing/discuss/2315103/Python-fast-Hashset | class Solution:
def isPathCrossing(self, path: str) -> bool:
c = set()
x,y = 0,0
c.add((x,y))
for i in path:
if i == 'N':
y+=1
elif i == 'E':
x+=1
elif i == 'W':
x-=1
else:
y-=1
if (x,y) in c:
return True
else:
c.add((x,y))
return False | path-crossing | Python fast Hashset | sunakshi132 | 0 | 30 | path crossing | 1,496 | 0.558 | Easy | 22,285 |
https://leetcode.com/problems/path-crossing/discuss/2315098/Faster-than-98-Python-hashmap | class Solution:
def isPathCrossing(self, path: str) -> bool:
c = {(0,0):1}
x,y = 0,0
for i in path:
if i == 'N':
y+=1
elif i == 'E':
x+=1
elif i == 'W':
x-=1
else:
y-=1
if (x,y) in c:
return True
else:
c[(x,y)]=1
return False | path-crossing | Faster than 98% Python hashmap | sunakshi132 | 0 | 25 | path crossing | 1,496 | 0.558 | Easy | 22,286 |
https://leetcode.com/problems/path-crossing/discuss/2289403/PythonorEasy-and-Detailed | class Solution:
def isPathCrossing(self, path: str) -> bool:
ns=0#this is for north and south direction
ew=0#this is for east and north directions
li=[[ns,ew]]#visited paths
for i in path:
if i=='N':
ns+=1
elif i=='S':
ns-=1
elif i=='E':
ew+=1
else:
ew-=1
if [ns,ew] not in li:
li.append([ns,ew])
else:
return True
return False | path-crossing | Python|Easy and Detailed | ajay_keelu | 0 | 20 | path crossing | 1,496 | 0.558 | Easy | 22,287 |
https://leetcode.com/problems/path-crossing/discuss/1986460/dictionary | class Solution:
def isPathCrossing(self, path: str) -> bool:
moves = {"N": 1, "S": -1, "E": 1, "W": -1}
seen = [(0, 0)]
for move in path:
if move in "EW":
location = (seen[-1][0] + moves[move], seen[-1][1])
else:
location = (seen[-1][0], seen[-1][1] + moves[move])
if location in seen: return True
else: seen.append(location)
return False | path-crossing | dictionary | andrewnerdimo | 0 | 44 | path crossing | 1,496 | 0.558 | Easy | 22,288 |
https://leetcode.com/problems/path-crossing/discuss/1905370/Python-easy-solution-faster-than-88-and-memory-less-than-95 | class Solution:
def isPathCrossing(self, path: str) -> bool:
x = 0
y = 0
check = [[0, 0]]
for i in path:
if i == 'N':
y += 1
elif i == 'S':
y -= 1
elif i == 'E':
x += 1
elif i == 'W':
x -= 1
if [x, y] in check:
return True
check.append([x, y])
return False | path-crossing | Python easy solution faster than 88% and memory less than 95% | alishak1999 | 0 | 86 | path crossing | 1,496 | 0.558 | Easy | 22,289 |
https://leetcode.com/problems/path-crossing/discuss/1753810/Python-dollarolution | class Solution:
def isPathCrossing(self, path: str) -> bool:
d = {'N':[1,1],'S':[1,-1],'W':[0,-1],'E':[0,1]}
xy = [0,0]
v = [[0,0]]
for i in path:
xy[d[i][0]] += d[i][1]
if xy in v:
return True
else:
v.append(xy[0:])
return False | path-crossing | Python $olution | AakRay | 0 | 54 | path crossing | 1,496 | 0.558 | Easy | 22,290 |
https://leetcode.com/problems/path-crossing/discuss/1420292/Python3-Faster-Than-84.25 | class Solution:
def isPathCrossing(self, path: str) -> bool:
x, y, v = 0, 0, set()
v.add((0, 0))
for i in path:
if i == 'N':
y += 1
elif i == 'S':
y -= 1
elif i == 'E':
x += 1
elif i == 'W':
x -= 1
l = len(v)
v.add((x, y))
if l == len(v):
return True
return False | path-crossing | Python3 Faster Than 84.25% | Hejita | 0 | 58 | path crossing | 1,496 | 0.558 | Easy | 22,291 |
https://leetcode.com/problems/path-crossing/discuss/780771/Intuitive-approach-by-holding-all-visited-cell-in-set | class Solution:
def isPathCrossing(self, path: str) -> bool:
pos = [0, 0]
pos_set = set([tuple(pos)])
for a in path:
if a == 'N':
pos[0] += 1
elif a == 'S':
pos[0] -= 1
elif a == 'W':
pos[1] -= 1
elif a == 'E':
pos[1] += 1
if tuple(pos) in pos_set:
return True
pos_set.add(tuple(pos))
return False | path-crossing | Intuitive approach by holding all visited cell in set | puremonkey2001 | 0 | 35 | path crossing | 1,496 | 0.558 | Easy | 22,292 |
https://leetcode.com/problems/path-crossing/discuss/716480/Python3-Intuitive-100-Memory-67-Runtime-O(N)-solution | class Solution:
def isPathCrossing(self, path: str) -> bool:
curr_point = (0,0)
visited = set()
for d in path:
if d == 'N':
curr_point = (curr_point[0], curr_point[1] + 1)
elif d == 'S':
curr_point = (curr_point[0], curr_point[1] - 1)
elif d == 'E':
curr_point = (curr_point[0] + 1, curr_point[1])
elif d == 'W':
curr_point = (curr_point[0] - 1, curr_point[1])
else:
raise AttributeError('Direction: {} in path in invalid'.format(d))
if curr_point == (0,0) or curr_point in visited:
return True
visited.add(curr_point)
return False | path-crossing | [Python3] - Intuitive 100% Memory 67% Runtime O(N) solution | skull_ | 0 | 102 | path crossing | 1,496 | 0.558 | Easy | 22,293 |
https://leetcode.com/problems/path-crossing/discuss/709373/Python3-complex-number-solution-Path-Crossing | class Solution:
def isPathCrossing(self, path: str) -> bool:
p = complex(0, 0)
points = set((p,))
m = {'N': 1j, 'S': -1j, 'E': 1, 'W': -1}
for d in path:
p += m[d]
if p in points:
return True
points.add(p)
return False | path-crossing | Python3 complex number solution - Path Crossing | r0bertz | 0 | 56 | path crossing | 1,496 | 0.558 | Easy | 22,294 |
https://leetcode.com/problems/path-crossing/discuss/709274/PythonPython3-Path-Crossing | class Solution:
def isPathCrossing(self, path: str) -> bool:
directions = {
'N': (0,1),
'E': (1,0),
'W': (-1,0),
'S': (0,-1)
}
s = (0, 0)
storage = [(0,0)]
for d in path:
x,y = s
i,j = directions[d]
x, y = x+i, y+j
s = x,y
if s in storage:
return True
else:
storage.append(s)
return False | path-crossing | [Python/Python3] Path Crossing | newborncoder | 0 | 109 | path crossing | 1,496 | 0.558 | Easy | 22,295 |
https://leetcode.com/problems/check-if-array-pairs-are-divisible-by-k/discuss/709252/Python3-3-line-frequency-table | class Solution:
def canArrange(self, arr: List[int], k: int) -> bool:
freq = dict()
for x in arr: freq[x%k] = 1 + freq.get(x%k, 0)
return all(freq[x] == freq.get(xx:=(k-x)%k, 0) and (x != xx or freq[x]%2 == 0) for x in freq) | check-if-array-pairs-are-divisible-by-k | [Python3] 3-line frequency table | ye15 | 3 | 255 | check if array pairs are divisible by k | 1,497 | 0.396 | Medium | 22,296 |
https://leetcode.com/problems/check-if-array-pairs-are-divisible-by-k/discuss/2667459/Python3-oror-Hashmap-Solution-oror-O(n)-oror-BGG | class Solution:
def canArrange(self, arr: List[int], k: int) -> bool:
h = {}
for i in arr:
r = i % k
if r not in h:
h[r] = 0
h[r]+=1
if h.get(0,0) %2 !=0:
return False
for r1 in h:
r2 = k - r1
if r1 == 0:
continue
if h[r1] != h.get(r2,0):
return False
return True
# Please upvote if you understand the solution | check-if-array-pairs-are-divisible-by-k | Python3 || Hashmap Solution || O(n) || BGG | hoo__mann | 0 | 18 | check if array pairs are divisible by k | 1,497 | 0.396 | Medium | 22,297 |
https://leetcode.com/problems/check-if-array-pairs-are-divisible-by-k/discuss/1756155/Python3oror-O(n)-time-O(n)-space | class Solution:
def canArrange(self, arr: List[int], k: int) -> bool:
if sum(arr) % k != 0:
return False
freq = [0] * k
for val in arr:
freq[val % k] += 1
if freq[0] % 2 != 0 or (k % 2 == 0 and freq[k//2] % 2 != 0):
return False
start, end = 1, k - 1
while start < end:
if freq[start] != freq[end]:
return False
start += 1
end -= 1
return True
#TC -> O(n) || SC -> O(n) | check-if-array-pairs-are-divisible-by-k | Python3|| O(n) time O(n) space | s_m_d_29 | 0 | 185 | check if array pairs are divisible by k | 1,497 | 0.396 | Medium | 22,298 |
https://leetcode.com/problems/check-if-array-pairs-are-divisible-by-k/discuss/1547778/Easy-for-beginers-just-use-list-fast-than-69 | class Solution:
def canArrange(self, arr: List[int], k: int) -> bool:
n = len(arr)
if n % 2 != 0:
return False
tmp = []
for i in range(n):
tmp.append(arr[i] % k)
zero = [n for n in tmp if n == 0]
unzero = [n for n in tmp if n != 0]
if len(zero) % 2 != 0:
return False
unzero.sort()
m = len(unzero)
for i in range(m):
if unzero[i] + unzero[m - 1 - i] != k:
return False
return True | check-if-array-pairs-are-divisible-by-k | Easy for beginers, just use list fast than 69% | zixin123 | 0 | 171 | check if array pairs are divisible by k | 1,497 | 0.396 | Medium | 22,299 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.