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/longest-uncommon-subsequence-i/discuss/1177751/99.02-faster-python(3line)-submission | class Solution:
def findLUSlength(self, a: str, b: str) -> int:
if a!=b:
return max(len(a),len(b))
return -1 | longest-uncommon-subsequence-i | 99.02% faster python(3line) submission ; | _jorjis | 1 | 107 | longest uncommon subsequence i | 521 | 0.603 | Easy | 9,200 |
https://leetcode.com/problems/longest-uncommon-subsequence-i/discuss/2829499/Python-or-Accepted-Soln | class Solution:
def findLUSlength(self, a: str, b: str) -> int:
if a == b:
return -1
else:
return max(len(a), len(b)) | longest-uncommon-subsequence-i | Python | Accepted Soln | ajay_gc | 0 | 1 | longest uncommon subsequence i | 521 | 0.603 | Easy | 9,201 |
https://leetcode.com/problems/longest-uncommon-subsequence-i/discuss/2807587/Easy-1-line-solution-Python | class Solution:
def findLUSlength(self, a: str, b: str) -> int:
return -1 if a == b else max(len(a), len(b)) | longest-uncommon-subsequence-i | Easy 1 line solution Python | kanykeiat | 0 | 4 | longest uncommon subsequence i | 521 | 0.603 | Easy | 9,202 |
https://leetcode.com/problems/longest-uncommon-subsequence-i/discuss/2645477/Easy-and-optimal-Solution | class Solution:
def findLUSlength(self, a: str, b: str) -> int:
return max(len(a),len(b)) if a!=b else -1 | longest-uncommon-subsequence-i | Easy and optimal Solution | Raghunath_Reddy | 0 | 6 | longest uncommon subsequence i | 521 | 0.603 | Easy | 9,203 |
https://leetcode.com/problems/longest-uncommon-subsequence-i/discuss/2495758/Python-simple-solution | class Solution:
def findLUSlength(self, a: str, b: str) -> int:
return -1 if a == b else max(len(a), len(b)) | longest-uncommon-subsequence-i | Python simple solution | StikS32 | 0 | 23 | longest uncommon subsequence i | 521 | 0.603 | Easy | 9,204 |
https://leetcode.com/problems/longest-uncommon-subsequence-i/discuss/1617446/python3-1-line | class Solution:
def findLUSlength(self, a: str, b: str) -> int:
return -1 if a==b else max(len(a),len(b)) | longest-uncommon-subsequence-i | python3 1 line | mikekaufman4 | 0 | 150 | longest uncommon subsequence i | 521 | 0.603 | Easy | 9,205 |
https://leetcode.com/problems/longest-uncommon-subsequence-i/discuss/1298880/Python3-dollarolution | class Solution:
def findLUSlength(self, a: str, b: str) -> int:
if a == b:
return -1
else:
if len(a) > len(b):
return len(a)
return len(b) | longest-uncommon-subsequence-i | Python3 $olution | AakRay | 0 | 119 | longest uncommon subsequence i | 521 | 0.603 | Easy | 9,206 |
https://leetcode.com/problems/longest-uncommon-subsequence-i/discuss/870787/Python3-1-line | class Solution:
def findLUSlength(self, a: str, b: str) -> int:
return max(len(a), len(b)) if a != b else -1 | longest-uncommon-subsequence-i | [Python3] 1-line | ye15 | 0 | 54 | longest uncommon subsequence i | 521 | 0.603 | Easy | 9,207 |
https://leetcode.com/problems/longest-uncommon-subsequence-i/discuss/481980/Python3%3A-simple-and-fastest | class Solution:
def findLUSlength(self, a: str, b: str) -> int:
if len(a) != len(b):
return max(len(a), len(b))
return -1 if a == b else len(a) | longest-uncommon-subsequence-i | Python3: simple and fastest | andnik | 0 | 208 | longest uncommon subsequence i | 521 | 0.603 | Easy | 9,208 |
https://leetcode.com/problems/longest-uncommon-subsequence-ii/discuss/380412/Solution-in-Python-3-(beats-~100) | class Solution:
def findLUSlength(self, S: List[str]) -> int:
C = collections.Counter(S)
S = sorted(C.keys(), key = len, reverse = True)
for i,s in enumerate(S):
if C[s] != 1: continue
b = True
for j in range(i):
I, c = -1, True
for i in s:
I = S[j].find(i,I+1)... | longest-uncommon-subsequence-ii | Solution in Python 3 (beats ~100%) | junaidmansuri | 2 | 648 | longest uncommon subsequence ii | 522 | 0.404 | Medium | 9,209 |
https://leetcode.com/problems/longest-uncommon-subsequence-ii/discuss/831309/Python3-find-the-longest-unique-str-and-not-a-subseq-of-any-other-longer-one-LUS-II | class Solution:
def findLUSlength(self, strs: List[str]) -> int:
def isSubseq(a, b):
j = 0
for i in range(len(b)):
if a[j] == b[i]:
j += 1
if j == len(a):
return True
return False
c = ... | longest-uncommon-subsequence-ii | Python3 find the longest unique str and not a subseq of any other longer one - LUS II | r0bertz | 1 | 439 | longest uncommon subsequence ii | 522 | 0.404 | Medium | 9,210 |
https://leetcode.com/problems/longest-uncommon-subsequence-ii/discuss/2768354/Python-straightforward-solution | class Solution:
def findLUSlength(self, strs: List[str]) -> int:
mem = dict()
for s in strs:
if len(s) not in mem:
mem[len(s)] = list()
mem[len(s)].append(s)
mem = sorted(list(mem.items()), key=lambda x: x[0], reverse=True)
def check(s1, s2):... | longest-uncommon-subsequence-ii | Python, straightforward solution | yiming999 | 0 | 7 | longest uncommon subsequence ii | 522 | 0.404 | Medium | 9,211 |
https://leetcode.com/problems/longest-uncommon-subsequence-ii/discuss/871246/Python3-check-for-subsequence | class Solution:
def findLUSlength(self, strs: List[str]) -> int:
def fn(p, s):
"""Return True if p is a subsequence of s."""
ss = iter(s)
return all(ch in ss for ch in p)
ans = -1
for i, s in enumerate(strs):
for ii in range... | longest-uncommon-subsequence-ii | [Python3] check for subsequence | ye15 | 0 | 312 | longest uncommon subsequence ii | 522 | 0.404 | Medium | 9,212 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/1582670/Python-Easy-Solution-or-Brute-Force-and-Optimal-Approach | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
# Brute Force: O(𝑛^2) - TLE
count = 0
for i in range(len(nums)):
sum = 0
for j in range(i, len(nums)):
sum += nums[j]
if sum % k == 0:
return True
return False
class Solution:
def checkSubarraySum(self, nums: L... | continuous-subarray-sum | Python Easy Solution | Brute Force and Optimal Approach | leet_satyam | 5 | 603 | continuous subarray sum | 523 | 0.285 | Medium | 9,213 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/1568794/Python-Simple-Solution-with-Detailed-Comments-(PrefixSum-Hashmap) | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
prefix_sums = defaultdict(lambda:float(inf))
#Key is prefix_sum%k, value is earliest occurence of the same prefix_sum
running_sum = 0
for i, n in enumerate(nums):
running_sum ... | continuous-subarray-sum | [Python] Simple Solution with Detailed Comments (PrefixSum Hashmap) | neurologicalstyle | 5 | 861 | continuous subarray sum | 523 | 0.285 | Medium | 9,214 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2744854/Python-Simple-and-Easy-Way-to-Solve-or-99-Faster | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
lookup = {0:-1}
curr_sum = 0
for i, n in enumerate(nums):
if k != 0:
curr_sum = (curr_sum + n) % k
else:
curr_sum += n
if curr_sum not in ... | continuous-subarray-sum | ✔️ Python Simple and Easy Way to Solve | 99% Faster 🔥 | pniraj657 | 4 | 526 | continuous subarray sum | 523 | 0.285 | Medium | 9,215 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/1562568/Simple-Python-solution-accumulative-mod | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
myDict={0:-1} #For edge cases where the first index is included in the solution ex: [2,4] k=3
total=0
for idx,n in enumerate(nums):
total+=n
if total%k not in m... | continuous-subarray-sum | Simple Python 🐍 solution accumulative mod | InjySarhan | 3 | 428 | continuous subarray sum | 523 | 0.285 | Medium | 9,216 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2073306/Python-or-O(N)-solution-using-hashmap | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
# ////// TC O(N) //////
remainder = { 0 : -1} # mapping remainder with last index
total = 0
for i,n in enumerate(nums):
total += n
rem = total % k
if rem not in ... | continuous-subarray-sum | Python | O(N) solution using hashmap | __Asrar | 2 | 603 | continuous subarray sum | 523 | 0.285 | Medium | 9,217 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/1604929/Python-VERY-simple-o(n)-compute-o(min(k-n))-space | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
sumsVisitedSoFar = set([0])
curSum, prevSum = 0, 0
for i,n in enumerate(nums):
curSum += n
curSum = curSum % k
if i!=0 and curSum in sumsVisitedSoFar:
return True
... | continuous-subarray-sum | Python, VERY simple, o(n) compute, o(min(k, n)) space | ko082528 | 1 | 403 | continuous subarray sum | 523 | 0.285 | Medium | 9,218 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/1233340/Timeout-Brute-Force-Recursive-Approach | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
n = len(nums)
if n <= 1:
return False
s = nums[0]
for i in range(1,n):
s += nums[i]
if s % k == 0:
return True
return self.checkSubarraySum(nums[1:], k) | continuous-subarray-sum | [Timeout] Brute Force ｜ Recursive Approach | user6148Qk | 1 | 180 | continuous subarray sum | 523 | 0.285 | Medium | 9,219 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/870762/Python3-prefix-modulo | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
prefix = 0 # prefix modulo
seen = {0: -1}
for i, x in enumerate(nums):
prefix += x
if k: prefix %= k
if prefix in seen and i - seen[prefix] >= 2: return True
seen.s... | continuous-subarray-sum | [Python3] prefix modulo | ye15 | 1 | 304 | continuous subarray sum | 523 | 0.285 | Medium | 9,220 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2839794/A-follow-up-question-Find-the-maximal-subarray-sum-which-is-a-multiple-of-k | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
presum = 0
presum_list = []
d = dict()
d[0] = -1
ans = -1 # default answer, return it if there is no subarray sum that is a multiple of k
for i, n_i in enumerate(nums):
presum +=... | continuous-subarray-sum | A follow up question -- Find the maximal subarray sum which is a multiple of k | MACHMichael | 0 | 2 | continuous subarray sum | 523 | 0.285 | Medium | 9,221 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2808104/O(n)-solution-using-Python | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
d = {0:-1}
_sum = 0
for i in range(len(nums)):
_sum += nums[i]
temp = _sum%k
if temp in d:
if i-d[temp] >= 2:
return True
else:
... | continuous-subarray-sum | O(n) solution using Python | dushyantRathore | 0 | 6 | continuous subarray sum | 523 | 0.285 | Medium | 9,222 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2807117/python-hashtable-prefix-sum | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
# Time complexity: O(nums.length).
# We perform O(nums.length) operations with a hash map, each taking O(1) time on average.
# Space complexity: O(min{nums.length,k}).
# The size of a hash map does not exc... | continuous-subarray-sum | python hashtable, prefix sum | sahilkumar158 | 0 | 4 | continuous subarray sum | 523 | 0.285 | Medium | 9,223 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2800630/Python-official-solution | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
# initialize the hash map with index 0 for sum 0
hash_map = {0: 0}
s = 0
for i in range(len(nums)):
s += nums[i]
# if the remainder s % k occurs for the first time
if s % ... | continuous-subarray-sum | Python official solution | subidit | 0 | 4 | continuous subarray sum | 523 | 0.285 | Medium | 9,224 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2786672/Sub-array-sum-easy-understanding | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
#take a hash Map which stores the prefix sum of the remainder with their index
#suppose if we get the same same remainder, then we got a sub array in the middele which is divisible by k we can return true
#one more ... | continuous-subarray-sum | Sub array sum-easy understanding | RaviShanker__Thadishetti | 0 | 6 | continuous subarray sum | 523 | 0.285 | Medium | 9,225 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2752674/Python-Optimal-Solution-w-ONE-BRANCH-only | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
s, sum_loc_dict = 0, {0: -1}
for i, n in enumerate(nums):
s = (s + n) % k
if (pre_i := sum_loc_dict.get(s, i)) < i - 1:
return True
sum_loc_dict[s] = sum_loc_dict.get(s, p... | continuous-subarray-sum | Python Optimal Solution w/ ONE BRANCH only | bylin | 0 | 4 | continuous subarray sum | 523 | 0.285 | Medium | 9,226 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2752203/Python-or-Optimized-Solution-or | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
n = len(nums)
# initialize hash map with remainder 0 and 0 elements..
hash_map = {0:0}
s = 0
for i in range(n):
s += nums[i]
# check if remainder not exist in hash map..
... | continuous-subarray-sum | Python | Optimized Solution | | quarnstric_ | 0 | 2 | continuous subarray sum | 523 | 0.285 | Medium | 9,227 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2747076/Modular-Arithmetic-Dictionary-or-Python | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
if len(nums)<2:
return False
dic = {0:-1}
total = 0
for ind, val in enumerate(nums):
total = (total + val)%k
if total in dic:
if ind - dic[t... | continuous-subarray-sum | Modular Arithmetic, Dictionary | Python | Abhi_-_- | 0 | 2 | continuous subarray sum | 523 | 0.285 | Medium | 9,228 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2746862/Prefix-Sum-Approach-Fast-and-Easy-Solution | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
hashmap = {0 : -1}
prefix_sum = 0
for ptr, val in enumerate(nums):
prefix_sum += val
remainder = prefix_sum % k
if remainder not in hashmap:
hashmap[remainder] = p... | continuous-subarray-sum | Prefix Sum Approach - Fast and Easy Solution | user6770yv | 0 | 9 | continuous subarray sum | 523 | 0.285 | Medium | 9,229 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2746382/Python-Solution-orT%3A-O(n)-orS%3AO(n) | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
remainder = {0:-1}
total = 0
for i, n in enumerate(nums):
total+=n
rem= total%k
if rem not in remainder:
remainder[rem] = i
elif ... | continuous-subarray-sum | Python Solution |T: O(n) |S:O(n) | pradeep288 | 0 | 8 | continuous subarray sum | 523 | 0.285 | Medium | 9,230 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2746370/Python!-8-Line-solution.-O(N) | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
presum = defaultdict(int)
for i, a in enumerate(nums):
presum[i] = (presum[i-1] + a) % k
presum[-1] = 0
indices = defaultdict(list)
for i, a in presum.items():
indi... | continuous-subarray-sum | 😎Python! 8 Line solution. O(N) | aminjun | 0 | 20 | continuous subarray sum | 523 | 0.285 | Medium | 9,231 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2746263/Modulus-beating-96-in-time-and-98-in-space | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
st = set()
n = len(nums)
cumsum = 0
# create a prev cumsum to enhance a lag
prev_cumsum = 0
for i in range(n):
if i == 0:
cumsum += nums[i]
cumsum %= k
... | continuous-subarray-sum | Modulus - beating 96% in time and 98% in space | leonine9 | 0 | 6 | continuous subarray sum | 523 | 0.285 | Medium | 9,232 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2746137/python-easy-solution-with-good-runtime | class Solution():
def checkSubarraySum(self, nums, k):
dic = {0:-1}
summ = 0
for i, n in enumerate(nums):
if k != 0:
summ = (summ + n) % k
else:
summ += n
if summ not in dic:
dic[summ] = i
... | continuous-subarray-sum | python easy solution with good runtime | V_Bhavani_Prasad | 0 | 8 | continuous subarray sum | 523 | 0.285 | Medium | 9,233 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2745998/Python-or-HashMap-(storing-remainder-and-the-index) | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
remainderDict = {0:-1}
total = 0
for i, n in enumerate(nums):
total += n
remainder = total % k
if remainder not in remainderDict:
remainderDict[remainder] = i
... | continuous-subarray-sum | Python | HashMap (storing remainder and the index) | KevinJM17 | 0 | 9 | continuous subarray sum | 523 | 0.285 | Medium | 9,234 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2745951/PYTHON3-Well-explained-Dict-based-solution | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
# here, we store first indices for sequences that fulfill condition;
# remainder 0 (when prefix sum equals k) is the only
# one that is allowed to occur just once, thus, we add
# it to the map w... | continuous-subarray-sum | 🧙🏻♂️ PYTHON3 Well explained Dict based solution | karinadelcheva | 0 | 6 | continuous subarray sum | 523 | 0.285 | Medium | 9,235 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2745890/Fast-Python-Solution-(Hashmap)-Time%3A-O(N)-Space%3A-O(K) | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
# sum_nums = sum(nums)
# if nums == [23,2,6,4,7] and k ==13:
# return False
# else:
# if sum_nums %k ==0:
# for i in nums:
# for j in n... | continuous-subarray-sum | Fast Python Solution (Hashmap) ; Time: O(N), Space: O(K) | avs-abhishek123 | 0 | 9 | continuous subarray sum | 523 | 0.285 | Medium | 9,236 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2745887/Python-Solution-or-95-Faster-or-Prefix-Running-Sum-DIVMOD-Based | class Solution:
# @lru_cache(None)
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
runningSum = 0
store = {
runningSum: -1
}
for idx, num in enumerate(nums):
runningSum += num
runningSum %= k
... | continuous-subarray-sum | Python Solution | 95% Faster | Prefix Running Sum DIVMOD Based | Gautam_ProMax | 0 | 8 | continuous subarray sum | 523 | 0.285 | Medium | 9,237 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2745884/Python-solution | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
remainder = {}
for idx, num in enumerate(nums):
if idx > 0:
nums[idx] += nums[idx-1]
num = nums[idx]
if num%k == 0 and idx >= 1:
return True
if... | continuous-subarray-sum | Python solution | tesfish | 0 | 5 | continuous subarray sum | 523 | 0.285 | Medium | 9,238 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2745773/Python.-Simple-solution-using-dictionary.-O(n) | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
s, mem = 0, {0: -1} #key - residue, value - index of the first instance
for i, n in enumerate(nums):
m = (n + s) % k
#if you find current residue in the dictionary
#and the distance betw... | continuous-subarray-sum | Python. Simple solution using dictionary. O(n) | nonchalant-enthusiast | 0 | 3 | continuous subarray sum | 523 | 0.285 | Medium | 9,239 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2745476/Python-or-Simple-prefix-sum-solution | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
div_rem = defaultdict(list)
carry = 0
prefix = [0]*len(nums)
for i in range(len(nums)):
carry += nums[i]
prefix[i] = carry
div_rem[carry % k].append(i)
# Simple ca... | continuous-subarray-sum | Python | Simple prefix sum solution | LordVader1 | 0 | 29 | continuous subarray sum | 523 | 0.285 | Medium | 9,240 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2745072/Python-or-O(k)-space-or-Pigeon-Hole-Principle-or-Hashmap-or-Prefix-Sum | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
n = len(nums)
# making prefix of the array % k
nums[0] %= k
for i in range(1, n):
nums[i] = (nums[i - 1] + nums[i]) % k
# hashmap to store all the indices of a particular number (0 to k -... | continuous-subarray-sum | Python | O(k) space | Pigeon Hole Principle | Hashmap | Prefix Sum | deepaklaksman | 0 | 28 | continuous subarray sum | 523 | 0.285 | Medium | 9,241 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2744617/Help-me-with-the-time-Python3-solution. | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
d , s = {0:-1} , 0
for i, n in enumerate(nums):
if k != 0:
s = (s + n) % k
else:
s += n
if s not in d:
d[s] = i
else:
... | continuous-subarray-sum | Help me with the time Python3 solution. | rupamkarmakarcr7 | 0 | 7 | continuous subarray sum | 523 | 0.285 | Medium | 9,242 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2744489/Solution-Using-Python3 | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
dic = {0:-1}
n = len(nums)
prefix = 0
for i in range(n):
prefix = (prefix + nums[i])%k
if prefix not in dic:
dic[prefix] = i
else:
if (i - ... | continuous-subarray-sum | Solution Using Python3 | dnvavinash | 0 | 15 | continuous subarray sum | 523 | 0.285 | Medium | 9,243 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2744479/Python-O(n)-solution-faster | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
rem={0:-1}
total=0
for i,j in enumerate(nums):
total+=j
r=total%k
if( r not in rem):
rem[r]=i
elif(i-rem[r]>1):
return True
re... | continuous-subarray-sum | Python O(n) solution faster | Raghunath_Reddy | 0 | 21 | continuous subarray sum | 523 | 0.285 | Medium | 9,244 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2744408/Python3-solution-O(n)-space-and-time | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
modArray = {0:-1}
s = 0
for i in range(len(nums)):
s += nums[i]
if s%k in modArray:
if i - modArray[s%k] >= 2:
return True
else:
... | continuous-subarray-sum | Python3 solution O(n) space and time | slavaheroes | 0 | 13 | continuous subarray sum | 523 | 0.285 | Medium | 9,245 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2744264/Super-Simple-Python-Solution!-O(n)-Space-and-Time! | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
seen, prev, curr = set(), 0, 0
for n in nums:
prev, curr = curr, (curr + n) % k
if curr in seen: return True
seen.add(prev)
return False | continuous-subarray-sum | 🔥 🐍 ✅ Super Simple Python Solution! O(n) Space and Time! | Cavalier_Poet | 0 | 14 | continuous subarray sum | 523 | 0.285 | Medium | 9,246 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2665749/Solution-in-python-using-dictionary | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
#prefix sum
for i in range(1,len(nums)):
nums[i]=nums[i]+nums[i-1]
#finding remainder
remainder=[i%k for i in nums]
#value with sum zero has index -1
d={0:-1}
... | continuous-subarray-sum | Solution in python using dictionary | harshmishra0014 | 0 | 6 | continuous subarray sum | 523 | 0.285 | Medium | 9,247 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2664312/Python3-Solution-oror-O(N)-Time-and-Space-Complexity | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
prefixSum=0
dic={}
count=0
for i in range(len(nums)):
prefixSum+=nums[i]
rem=prefixSum%k
if rem not in dic:
dic[rem]=0
if rem==0:
... | continuous-subarray-sum | Python3 Solution || O(N) Time & Space Complexity | akshatkhanna37 | 0 | 15 | continuous subarray sum | 523 | 0.285 | Medium | 9,248 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/2576940/Python-runtime-O(n)-memory-O(min(nk)) | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
d = {0:-1}
s = 0
for i in range(len(nums)):
s += nums[i]
if s%k not in d:
d[s%k] = i
elif s%k in d and d[s%k] < i-1:
return True
return Fal... | continuous-subarray-sum | Python, runtime O(n), memory O(min(n,k)) | tsai00150 | 0 | 91 | continuous subarray sum | 523 | 0.285 | Medium | 9,249 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/1743652/WEEB-DOES-PYTHONC%2B%2B-PREFIX-SUM | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
dp = defaultdict(int)
dp[0] = -1 # key : remainder, value : leftmost idx
curSum = 0
for i in range(len(nums)):
curSum += nums[i]
if curSum % k in dp and i - dp[curSum % k] >= 2:
return True
if curSum % k not in dp:
... | continuous-subarray-sum | WEEB DOES PYTHON/C++ PREFIX SUM | Skywalker5423 | 0 | 222 | continuous subarray sum | 523 | 0.285 | Medium | 9,250 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/765441/Python3%3A-Brute-solution-quadratic-time-complexity | class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
k = abs(k)
required_sum = 0
for outer_idx, i in enumerate(nums[0:-1]):
local_sum = i
for inner_idx, j in enumerate(nums[outer_idx+1:]):
local_sum+=j
if k>0 and... | continuous-subarray-sum | Python3: Brute solution, quadratic time complexity | bhattacharyya_d | 0 | 149 | continuous subarray sum | 523 | 0.285 | Medium | 9,251 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/511295/Python3-hash-table-O(n) | class Solution:
def checkSubarraySum(self, nums, k):
cum_sum_mod = []
cur_sum = 0
for i in range(len(nums)):
cur_sum = cur_sum + nums[i]
cum_sum_mod.append(cur_sum % k if k != 0 else cur_sum)
hash_table = {0:-1}
for i in range(len(cum_sum_mod)): ... | continuous-subarray-sum | Python3, hash table, O(n) | complexfilter | 0 | 341 | continuous subarray sum | 523 | 0.285 | Medium | 9,252 |
https://leetcode.com/problems/continuous-subarray-sum/discuss/389183/Solution-in-Python-3-(beats-~99)-(Using-Set)-(No-Dictionary) | class Solution:
def checkSubarraySum(self, N: List[int], K: int) -> bool:
L, S, s = len(N), set([0]), 0
for i in range(L-1):
if (N[i]+N[i+1]) == 0 or (K and not (N[i]+N[i+1]) % K): return True
if K == 0: return False
for n in N:
a, s = s, (s + n) % K
if s in S and a != s: retur... | continuous-subarray-sum | Solution in Python 3 (beats ~99%) (Using Set) (No Dictionary) | junaidmansuri | -7 | 949 | continuous subarray sum | 523 | 0.285 | Medium | 9,253 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/1077760/Python.-very-clear-and-simplistic-solution. | class Solution:
def findLongestWord(self, s: str, d: List[str]) -> str:
def is_subseq(main: str, sub: str) -> bool:
i, j, m, n = 0, 0, len(main), len(sub)
while i < m and j < n and n - j >= m - i:
if main[i] == sub[j]:
i += 1
j += 1... | longest-word-in-dictionary-through-deleting | Python. very clear and simplistic solution. | m-d-f | 6 | 879 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,254 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/871111/Python3-scan | class Solution:
def findLongestWord(self, s: str, d: List[str]) -> str:
def fn(word):
"""Return True if word matches a subsequence of s."""
ss = iter(s)
return all(c in ss for c in word)
return next((w for w in sorted(d, key=lambda x: (-len(x), ... | longest-word-in-dictionary-through-deleting | [Python3] scan | ye15 | 4 | 199 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,255 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/871111/Python3-scan | class Solution:
def findLongestWord(self, s: str, d: List[str]) -> str:
def fn(word):
"""Return True if word matches a subsequence of s."""
ss = iter(s)
return all(c in ss for c in word)
ans = ""
for w in d:
if fn(w) and (le... | longest-word-in-dictionary-through-deleting | [Python3] scan | ye15 | 4 | 199 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,256 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/871111/Python3-scan | class Solution:
def findLongestWord(self, s: str, d: List[str]) -> str:
def fn(word):
"""Return True if word is a subsequence of s."""
i = 0
for c in word:
while i < len(s) and s[i] != c: i += 1
if i == len(s): return False
... | longest-word-in-dictionary-through-deleting | [Python3] scan | ye15 | 4 | 199 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,257 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/871111/Python3-scan | class Solution:
def findLongestWord(self, s: str, d: List[str]) -> str:
def fn(word):
"""Return True if word is a subsequence of s."""
i = 0
for c in s:
if i < len(word) and word[i] == c: i += 1
return i == len(word)
... | longest-word-in-dictionary-through-deleting | [Python3] scan | ye15 | 4 | 199 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,258 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/871111/Python3-scan | class Solution:
def findLongestWord(self, s: str, d: List[str]) -> str:
ans = ""
for word in d:
ss = iter(s)
if all(c in ss for c in word) and (len(word) > len(ans) or len(word) == len(ans) and word < ans):
ans = word
return ans | longest-word-in-dictionary-through-deleting | [Python3] scan | ye15 | 4 | 199 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,259 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/2607247/Python-simple-straight-forward-solution | class Solution:
def findLongestWord(self, s: str, dictionary: list[str]) -> str:
def is_valid(string): # can we get the string by deleting some letters or not
i = 0
for letter in s:
if letter == string[i]:
i += 1
if i == len(string... | longest-word-in-dictionary-through-deleting | Python simple straight-forward solution | Mark_computer | 2 | 118 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,260 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/2087464/Clean-commented-Python-Faster-than-99.5-hashmap-one-pass | class Solution:
def findLongestWord(self, s: str, dictionary: List[str]) -> str:
ans = ""
# store for each starting letter the potential word. We don't store the word directly, but its index in the dictionary.
# In addition we store '0' as pointer to the starting character. We will... | longest-word-in-dictionary-through-deleting | Clean, commented Python, Faster than 99.5%, hashmap, one-pass | boris17 | 1 | 44 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,261 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/1705859/Python-simple-solution-using-hashmap-and-sorting-and-two-pointers | class Solution:
def findLongestWord(self, s: str, dictionary: List[str]) -> str:
def compare(str1, str2):
m, n = len(str1), len(str2)
i, j = 0, 0
while i < m and j < n:
if j == n-1 and str1[i] == str2[j]:
return True
... | longest-word-in-dictionary-through-deleting | Python simple solution using hashmap and sorting and two-pointers | byuns9334 | 1 | 153 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,262 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/1078966/Very-simple-Python-solution-O(mn) | class Solution:
def findLongestWord(self, s: str, d: List[str]) -> str:
res=""
maxLen=0
for word in d:
pW=0
pS=0
while pW<len(word) and pS<len(s):
if word[pW]==s[pS]:
pW+=1
pS+=1
... | longest-word-in-dictionary-through-deleting | Very simple Python 🐍 solution O(mn) | InjySarhan | 1 | 162 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,263 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/639917/Python-easy-to-understand-solution-with-isSubsequence() | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
index = 0
for i in range(0, len(t)):
if index == len(s):
return True
if s[index] == t[i]:
index += 1
return index == len(s)
def findLongestWord(self, s: str, d: L... | longest-word-in-dictionary-through-deleting | Python easy to understand solution with isSubsequence() | Ningmin | 1 | 251 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,264 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/639917/Python-easy-to-understand-solution-with-isSubsequence() | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
index = 0
for i in range(0, len(t)):
if index == len(s):
return True
if s[index] == t[i]:
index += 1
return index == len(s)
def findLongestWord(self, s: str, d: L... | longest-word-in-dictionary-through-deleting | Python easy to understand solution with isSubsequence() | Ningmin | 1 | 251 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,265 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/2779884/easy-python-solution | class Solution:
def checkPresence(self, word, target) :
index = 0
for i in range(len(target)) :
if target[i] == word[index] :
index += 1
if index == len(word) :
return True
return len(word) == index
... | longest-word-in-dictionary-through-deleting | easy python solution | sghorai | 0 | 4 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,266 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/2715642/Python-O(NM)-O(M) | class Solution:
def findLongestWord(self, s: str, dictionary: List[str]) -> str:
pointers = [0] * len(dictionary)
res = ""
for char in s:
for i in range(len(pointers)):
idx, word = pointers[i], dictionary[i]
if idx == len(word):
... | longest-word-in-dictionary-through-deleting | Python - O(NM), O(M) | Teecha13 | 0 | 5 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,267 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/2484070/Python3-or-Sorting | class Solution:
def isSubsequence(self,w,s):
p1,p2=0,0
while p1<len(w) and p2<len(s):
if w[p1]==s[p2]:
p1+=1
p2+=1
return p1==len(w)
def findLongestWord(self, s: str, dictionary: List[str]) -> str:
ans=''
dictionary.sort()
f... | longest-word-in-dictionary-through-deleting | [Python3] | Sorting | swapnilsingh421 | 0 | 19 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,268 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/2091793/Python-or-Sorting-and-Checking-isSubsequence-approach | class Solution:
def isSubsequence(self,s,word):
if len(s) < len(word):
return False
i,j = 0,0
while i < len(s) and j < len(word):
if s[i] == word[j]:
i += 1
j += 1
else:
i += 1
return len(word) == j
... | longest-word-in-dictionary-through-deleting | Python | Sorting and Checking isSubsequence approach | __Asrar | 0 | 47 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,269 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/1078290/python-ez-understanding-solution | class Solution:
def check_sub(self,target,s):
'''
helper function that determine whether the target word is a candidate or not
'''
if len(target) > len(s):
return False
i,j = 0,0
while i < len(target) and j < len(s):
if target[i] == s[j]:
... | longest-word-in-dictionary-through-deleting | python ez understanding solution | yingziqing123 | 0 | 72 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,270 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/1077970/Brute-force-easy-to-understand-Python-solution | class Solution:
def findLongestWord(self, s: str, d: List[str]) -> str:
le=len(s)
res=""
for word in d:
n=len(word)
i,j=0,0
count=0
while(i<n and j<le):
if s[j]==word[i]:
count+=1
i+=1
... | longest-word-in-dictionary-through-deleting | Brute force easy to understand Python solution | _Rehan12 | 0 | 44 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,271 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/1077633/Python-or-Explained-and-Easy-and-Fast-or-Beats-90 | class Solution:
def findLongestWord(self, s: str, d: List[str]) -> str:
def check(s: str, d: str) -> bool:
it = iter(s)
return all(l in it for l in d)
m = ""
for i in d:
if check(s, i) and (len(m) < len(i) or (len(m) == len(i) and m > i)):
... | longest-word-in-dictionary-through-deleting | Python | Explained & Easy & Fast | Beats 90% | SlavaHerasymov | 0 | 119 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,272 |
https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/discuss/715631/Python3-check-isSubsequence-using-str.find-Longest-Word-in-Dictionary-through-Deleting | class Solution:
def findLongestWord(self, s: str, d: List[str]) -> str:
def isSubsequence(a: str, b: str) -> bool:
i = 0
for c in a:
if (i := b.find(c, i)) == -1:
return False
i += 1
return True
heap = [... | longest-word-in-dictionary-through-deleting | Python3 check isSubsequence using str.find - Longest Word in Dictionary through Deleting | r0bertz | 0 | 244 | longest word in dictionary through deleting | 524 | 0.512 | Medium | 9,273 |
https://leetcode.com/problems/contiguous-array/discuss/577489/Python-O(n)-by-partial-sum-and-dictionary.-90%2B-w-Visualization | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
partial_sum = 0
# table is a dictionary
# key : partial sum value
# value : the left-most index who has the partial sum value
table = { 0: -1}
max_length = 0
for idx, number in enume... | contiguous-array | Python O(n) by partial sum and dictionary. 90%+ [w/ Visualization] | brianchiang_tw | 12 | 1,700 | contiguous array | 525 | 0.468 | Medium | 9,274 |
https://leetcode.com/problems/contiguous-array/discuss/1743700/Python-3-(700ms)-or-O(n)-Dictionary-HashMap-Solution-or-One-Pass-Easy-to-Understand | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
m,c=0,0
d={0:-1}
for i in range(len(nums)):
if nums[i]==0:
c-=1
else:
c+=1
if c in d:
m=max(m,i-d[c])
else:
d[c]=i
... | contiguous-array | Python 3 (700ms) | O(n) Dictionary HashMap Solution | One-Pass Easy to Understand | MrShobhit | 4 | 375 | contiguous array | 525 | 0.468 | Medium | 9,275 |
https://leetcode.com/problems/contiguous-array/discuss/1951726/Python-solution-with-comment | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
dic = {0:-1} # dic record: count(the times we met 1 - the time we met 0) : idx
count = 0 # lets say count == 3, it means until this idx, if 0 appeared x time, 1 appeared ... | contiguous-array | Python solution with comment | byroncharly3 | 2 | 136 | contiguous array | 525 | 0.468 | Medium | 9,276 |
https://leetcode.com/problems/contiguous-array/discuss/1760153/WEEB-DOES-PYTHONC%2B%2B-PREFIX-SUM | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
dp = defaultdict(int)
dp[0] = -1
curSum = 0
result = 0
for i in range(len(nums)):
if nums[i] == 1:
curSum += 1
else:
curSum -= 1
if curSum not in dp: # store leftmost idx for longest subarray
dp[curSum] = i
else:
... | contiguous-array | WEEB DOES PYTHON/C++ PREFIX SUM | Skywalker5423 | 2 | 93 | contiguous array | 525 | 0.468 | Medium | 9,277 |
https://leetcode.com/problems/contiguous-array/discuss/577773/Python3-using-prefix-sum | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
seen = {0: -1}
ans = prefix = 0
for i, x in enumerate(nums):
prefix += 2*x-1
ans = max(ans, i - seen.setdefault(prefix, i))
return ans | contiguous-array | [Python3] using prefix sum | ye15 | 2 | 254 | contiguous array | 525 | 0.468 | Medium | 9,278 |
https://leetcode.com/problems/contiguous-array/discuss/1634324/Python-oror-Easy-Solution-oror-beat~99.6 | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
count, maxi = 0, 0
d = {0: -1}
for i, j in enumerate(nums):
if j == 0:
count += -1
else:
count += 1
try:
temp = i - d[count]
if maxi < temp:
maxi = temp
except:
d[count] = i
return maxi | contiguous-array | Python || Easy Solution || beat~99.6% | naveenrathore | 1 | 185 | contiguous array | 525 | 0.468 | Medium | 9,279 |
https://leetcode.com/problems/contiguous-array/discuss/1450366/Python-3-Solution-Beats-100-Of-Python-Submissions | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
n=len(nums)
d={}
best=0
summ=0
for i in range(n):
summ+=1 if nums[i]==1 else -1
if summ==0:
best=i+1
continue
if summ in d:
if ... | contiguous-array | Python 3 Solution Beats 100% Of Python Submissions | reaper_27 | 1 | 133 | contiguous array | 525 | 0.468 | Medium | 9,280 |
https://leetcode.com/problems/contiguous-array/discuss/2686576/Python3-O(n)-time | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
count = [[0, 0]]
c0, c1 = 0, 0
for i in nums:
if i == 0:
c0 += 1
else:
c1 += 1
count.append([c0, c1])
d = {}
res = 0
... | contiguous-array | Python3 O(n) time | ddawngbof | 0 | 2 | contiguous array | 525 | 0.468 | Medium | 9,281 |
https://leetcode.com/problems/contiguous-array/discuss/2408804/Fully-Detailed-Python-Explanation-for-Beginners-or-faster-89.16-or-Easy-to-understand | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
"""
Main ideas:
1. Let "count" be the sum (of list elements) with inital value of 0
2. Add 1 to "count" when we encounter 1 in list; subtract 1 from "count" when we encounter 0 in list
3. If we get the ... | contiguous-array | Fully Detailed Python Explanation for Beginners | faster 89.16% | Easy to understand | harishmanjunatheswaran | 0 | 40 | contiguous array | 525 | 0.468 | Medium | 9,282 |
https://leetcode.com/problems/contiguous-array/discuss/2206787/Simple-and-Easy-Approach-oror-Hashmap | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
count = 0
n = len(nums)
maxLen = 0
hashMap = {}
for i in range(n):
if nums[i] == 0:
count -= 1
else:
count += 1
if count == 0:
... | contiguous-array | Simple and Easy Approach || Hashmap | Vaibhav7860 | 0 | 74 | contiguous array | 525 | 0.468 | Medium | 9,283 |
https://leetcode.com/problems/contiguous-array/discuss/1818386/Python-easy-to-read-and-understand | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
sums = 0
ans = 0
d = {0: -1}
for i in range(len(nums)):
sums += -1 if nums[i] == 0 else 1
if sums in d:
ans = max(ans, i-d[sums])
else:
d[sums] = i
... | contiguous-array | Python easy to read and understand | sanial2001 | 0 | 163 | contiguous array | 525 | 0.468 | Medium | 9,284 |
https://leetcode.com/problems/contiguous-array/discuss/1776518/Python3-linear-time-solution-prefixsum | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
#replace the 0s with -1s, so that sum of equal no. of 0s and 1s becomes zero
for i in range(len(nums)):
if nums[i] == 0:
nums[i] = -1
prefixSumNums = [0]*len(nums)
my_dict =... | contiguous-array | Python3 - linear time solution - prefixsum | sourav-saha | 0 | 58 | contiguous array | 525 | 0.468 | Medium | 9,285 |
https://leetcode.com/problems/contiguous-array/discuss/1755027/Python-Easy-and-clean-solution-O(n)-(beats-93.61) | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
n = len(nums)
d = dict()
d[0] = -1
pre_sum, ans = 0, 0
for i in range(n):
pre_sum += 1 if nums[i] else -1
if pre_sum in d:
ans = max(ans, i - d[pre_sum])
... | contiguous-array | [Python] Easy and clean solution O(n) (beats 93.61%) | nomofika | 0 | 154 | contiguous array | 525 | 0.468 | Medium | 9,286 |
https://leetcode.com/problems/contiguous-array/discuss/1743983/Easiest-Python-Solution-Using-Dictionary | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
d={}
maxlen=0
prefix_sum=0
n=len(nums)
for i in range(n):
prefix_sum = prefix_sum-1 if nums[i]==0 else prefix_sum+1
if prefix_sum==0:
maxlen=max(maxlen,i+1)
el... | contiguous-array | 📍Easiest Python Solution Using Dictionary | AdityaTrivedi88 | 0 | 16 | contiguous array | 525 | 0.468 | Medium | 9,287 |
https://leetcode.com/problems/contiguous-array/discuss/1743669/Python3-Hashmap | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
dic = {0 : -1}
accumulate, res = 0, 0
for i,num in enumerate(nums):
accumulate += 1 if num == 1 else -1
if accumulate in dic:
res = max(res, i - dic[accumulate])
else:
... | contiguous-array | [Python3] Hashmap | Rainyforest | 0 | 12 | contiguous array | 525 | 0.468 | Medium | 9,288 |
https://leetcode.com/problems/contiguous-array/discuss/1743521/Python-Solutionor-T%3AO(n)orS%3AO(n)orHashmap | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
max_len, sum = 0, 0
hash_map = {}
for i in range(len(nums)):
if nums[i] == 0:
sum -= 1
else:
sum += 1
if sum == 0:
max_len = i + 1
... | contiguous-array | Python Solution| T:O(n)|S:O(n)|Hashmap | pradeep288 | 0 | 40 | contiguous array | 525 | 0.468 | Medium | 9,289 |
https://leetcode.com/problems/contiguous-array/discuss/1743453/Python-Simple-and-Clean-Python-Solution-Using-Dictionary | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
count=0
max_len=0
d={}
d[0]=-1
for i in range(len(nums)):
if nums[i]==1:
count = count + 1
else:
count = count - 1
if count not in d:
d[count]=i
else:
max_len=max(max_len,i-d[count])
return ma... | contiguous-array | [ Python ]✔✔ Simple and Clean Python Solution Using Dictionary 🔥✌🔥 | ASHOK_KUMAR_MEGHVANSHI | 0 | 85 | contiguous array | 525 | 0.468 | Medium | 9,290 |
https://leetcode.com/problems/contiguous-array/discuss/1743269/525.-Contiguous-Array | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
lookup = {0: -1}
length = 0
val = 0
for i, num in enumerate(nums):
if num == 1:
val += 1
else:
val -= 1
if val in lookup:
length = ma... | contiguous-array | 525. Contiguous Array | prasunbhunia | 0 | 59 | contiguous array | 525 | 0.468 | Medium | 9,291 |
https://leetcode.com/problems/contiguous-array/discuss/1622406/Build-Logic-step-by-step-greater-Brute-Force-to-Efficient-Solution | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
#Brute Force Solution --- Throws Time Limit Exceeded
#What the code is doing :
#First of all, we will fix ith element and then traverse from ith element till the last element
#Using two separate variables, o... | contiguous-array | Build Logic step by step --> Brute Force to Efficient Solution | aarushsharmaa | 0 | 72 | contiguous array | 525 | 0.468 | Medium | 9,292 |
https://leetcode.com/problems/contiguous-array/discuss/1582745/Python-Easy-Solution-or-O(n)-Approach | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
hashMap = {0: -1}
add = 0
maxLen = 0
for i in range(len(nums)):
if nums[i] == 0:
add += -1
else:
add += 1
if add in hashMap:
maxLen = max(maxLen, i-hashMap[add])
else:
hashMap[add] = i
return maxLen | contiguous-array | Python Easy Solution | O(n) Approach | leet_satyam | 0 | 119 | contiguous array | 525 | 0.468 | Medium | 9,293 |
https://leetcode.com/problems/contiguous-array/discuss/1070206/Easy-to-understand-Python3-solution. | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
counts = {0: [-1]}
count = 0
length = 0
for i in range(len(nums)):
if nums[i] == 0:
count -= 1
else:
... | contiguous-array | Easy to understand Python3 solution. | RuslanKalashnikov | 0 | 178 | contiguous array | 525 | 0.468 | Medium | 9,294 |
https://leetcode.com/problems/contiguous-array/discuss/653476/Python-O(n)-7-lines | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
level, max_value, d = 0, 0, {0: -1}
for i, n in enumerate(nums):
level += (-1, 1)[n]
max_value = max(max_value, i - d.setdefault(level, i))
return max_value | contiguous-array | Python O(n), 7 lines | dliamuziki2017 | 0 | 126 | contiguous array | 525 | 0.468 | Medium | 9,295 |
https://leetcode.com/problems/contiguous-array/discuss/577533/Python3-very-detailed-explanation-easy-to-understand-contrary-to-all-other-posts-here | class Solution:
def findMaxLength(self, nums: List[int]) -> int:
acc = 0 # Stores the value for the current position in the nums
# Why we have diff_dict?
# we need to know how many elements before we had the difference value matching the current value at question
# Example:
... | contiguous-array | Python3 very detailed explanation, easy to understand, contrary to all other posts here | ozyinc | 0 | 134 | contiguous array | 525 | 0.468 | Medium | 9,296 |
https://leetcode.com/problems/beautiful-arrangement/discuss/1094146/Python-Backtracking-the-more-intuitive-way | class Solution:
def countArrangement(self, n: int) -> int:
self.count = 0
self.backtrack(n, 1, [])
return self.count
def backtrack(self, N, idx, temp):
if len(temp) == N:
self.count += 1
return
for i in range(1, N+1):
... | beautiful-arrangement | Python Backtracking, the more intuitive way | IamCookie | 8 | 872 | beautiful arrangement | 526 | 0.646 | Medium | 9,297 |
https://leetcode.com/problems/beautiful-arrangement/discuss/1908252/python-permutation-solution | class Solution:
def countArrangement(self, n: int) -> int:
ans=[]
def permute(l,p,index):
if l==[]:
ans.append(p)
for i,v in enumerate(l):
if v%(index+1)==0 or (index+1)%v==0:
permute(l[:i]+l[i+1:],p+[v],index+1)
... | beautiful-arrangement | python permutation solution | prateek4463 | 2 | 172 | beautiful arrangement | 526 | 0.646 | Medium | 9,298 |
https://leetcode.com/problems/beautiful-arrangement/discuss/872354/Python3-backtracking | class Solution:
def countArrangement(self, N: int) -> int:
def fn(k):
"""Return number of beautiful arrangements."""
if k == N: return 1 # boundary condition
ans = 0
for kk in range(k, N):
if nums[kk] % (k+1) == 0 or (k+1) % nums[kk] =... | beautiful-arrangement | [Python3] backtracking | ye15 | 2 | 225 | beautiful arrangement | 526 | 0.646 | Medium | 9,299 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.