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/find-original-array-from-doubled-array/discuss/2785917/Easy-to-understand-sort-and-counter-solution | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) % 2 != 0:
return []
countDict = collections.Counter(changed)
numZeros = countDict[0]
if numZeros:
if numZeros % 2 != 0:
return []
del... | find-original-array-from-doubled-array | Easy to understand - sort & counter solution | TheCodeBoss | 0 | 3 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,000 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2581911/Python-or-O(n)-or-Easy-To-Understand | class Solution:
def findOriginalArray(self, changed):
# Time Complexity: O(n log n)
# Space Complexity: O(n)
changed.sort()
que=deque([])
output=[]
for i in changed:
if que and que[0]==i:
que.popleft()
else:
que.appen... | find-original-array-from-doubled-array | Python | O(n) | Easy To Understand | varun21vaidya | 0 | 19 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,001 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2580335/SLOWER-OR-SLOWEST-BUT-EASIER-APPROCH | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed)%2!=0: return []
freq=[0]*100001
for i in changed: freq[i]+=1
ans=[]
for j in range(100001):
while j*2<100001 and freq[j]>0 and freq[j*2]>0:
ans.append(j) ... | find-original-array-from-doubled-array | SLOWER OR SLOWEST BUT EASIER APPROCH | DG-Problemsolver | 0 | 16 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,002 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2580233/Python-3-or-Easy-to-understand | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed)%2!=0:
return []
has={}
for i in changed:
if i in has:
has[i]+=1
else:
has[i]=1
if 0 in has and has[0]%2!=0:
re... | find-original-array-from-doubled-array | Python 3 | Easy to understand | RickSanchez101 | 0 | 26 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,003 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2579974/Python-solution | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) % 2:
return []
res = []
seen = {}
changed.sort(reverse=True)
for num in changed:
if num * 2 in seen:
res.append(num)
if seen[num*2] == 1:
del seen[num*2]
else:
seen[num*2] -= 1
else:... | find-original-array-from-doubled-array | Python solution | Mark5013 | 0 | 12 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,004 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2579641/O(nlogn)-using-heuristic-with-sorting-%2B-binary-searching | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
a = sorted(changed)
print(changed)
ans = []
while len(a)>1:
print("+ ", a, ans)
u = a.pop(0)
i = bisect_left(a, 2*u)
if i>=len(a) or a[i] != 2*u:
... | find-original-array-from-doubled-array | O(nlogn) using heuristic with sorting + binary searching | dntai | 0 | 9 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,005 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2579190/Python3-Linear-Time-Dictionary-Solution-O(n)-Time-O(n)-Space | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
# quick check
if len(changed) % 2 == 1:
return []
h = Counter(changed)
res = []
# loop below has issues because y = y + y when y = 0
if h[0] % 2 == 1:
retu... | find-original-array-from-doubled-array | [Python3] Linear Time Dictionary Solution - O(n) Time, O(n) Space | rt500 | 0 | 28 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,006 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2578442/GolangPython-O(N*log(N))-time-or-O(N)-space | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
changed.sort()
counter = {}
for value in changed:
if value not in counter:
counter[value] = 0
counter[value] +=1
output = []
for value in changed:
... | find-original-array-from-doubled-array | Golang/Python O(N*log(N)) time | O(N) space | vtalantsev | 0 | 17 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,007 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2578172/Python-Solution-with-Dictionary | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
d = {}
if len(changed)%2 == 1:
return []
changed.sort()
ans = []
for i in range(len(changed)):
if changed[i]%2 == 0 and changed[i]//2 in d:
ans.append(changed... | find-original-array-from-doubled-array | Python Solution with Dictionary | a_dityamishra | 0 | 14 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,008 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2578134/Python-Hashmap-Easy-Solution-Well-Explained | class Solution:
# Hashmap
# The basic intuition would be to take up 3 arrays
# After sorting the array keep putting elements into 1st array if 2 * num is not seen yet
# this is slow because taking 3 lists with their operations of insert and delete and multiple common elements can be present
# Using ... | find-original-array-from-doubled-array | Python Hashmap Easy Solution Well Explained | shiv-codes | 0 | 14 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,009 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2578051/Python-O(nlogn)-using-sorting-and-queue | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
n = len(changed)
if n % 2 != 0: return []
result = []
changed.sort()
queue = deque()
for i in changed[::-1]:
if queue and i*2 == queue[0]:
result.append(i)
... | find-original-array-from-doubled-array | Python O(nlogn) using sorting and queue | rjnkokre | 0 | 7 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,010 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2578051/Python-O(nlogn)-using-sorting-and-queue | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
n = len(changed)
if n % 2 != 0: return []
result = []
changed.sort()
queue = deque()
for i in changed:
if queue and i/2 == queue[0]:
result.append(i/2)
... | find-original-array-from-doubled-array | Python O(nlogn) using sorting and queue | rjnkokre | 0 | 7 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,011 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2577892/Python-Solution-or-Brute-Force-greater-Optimized | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed)%2!=0:
return []
n=len(changed)
# Brute Force
# changed.sort()
# count=0
# ans=[]
# for i in range(n):
# if changed[i]!=-1:
# ... | find-original-array-from-doubled-array | Python Solution | Brute Force --> Optimized | Siddharth_singh | 0 | 14 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,012 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2577810/Python3-Runtime%3A-1268-ms-faster-than-100-or-Memory%3A-29.1-MB-less-than-99.58 | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
count = Counter(changed)
if count[0] % 2:
return []
for num in sorted(count):
if count[num] > count[2*num]:
return []
count[2*num] -= count[num] if num else count... | find-original-array-from-doubled-array | [Python3] Runtime: 1268 ms, faster than 100% | Memory: 29.1 MB, less than 99.58% | anubhabishere | 0 | 82 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,013 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2577694/Python3-oror-TC%3A-O(nLogn)-oror-Sort-%2B-Dictionary | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) % 2 != 0:
return []
res = []
freq = collections.Counter(changed)
changed.sort()
for val in changed:
dob = val * 2
if val in fre... | find-original-array-from-doubled-array | Python3 || TC: O(nLogn) || Sort + Dictionary | s_m_d_29 | 0 | 17 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,014 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2577612/Python-Accepted | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
c = Counter(changed)
zeros, m = divmod(c[0], 2)
if m: return []
ans = [0]*zeros
for n in sorted(c.keys()):
if c[n] > c[2*n]: return []
c[2*n]-= c[n]
ans.exte... | find-original-array-from-doubled-array | Python Accepted | Khacker | 0 | 16 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,015 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2577447/Python-or-Sort-and-Hash-Table-or-Easy-to-Understand-or-With-Explanation | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
# the logic is easy
# use a dictionary to store the num while traversing the changed array
# trick: sort the array first, then the incoming num is definitely greater the the stored nums in dic
if l... | find-original-array-from-doubled-array | Python | Sort & Hash Table | Easy to Understand | With Explanation | Mikey98 | 0 | 34 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,016 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2323728/Python3-Fast-and-Cheap-Solution-%2B-Explanation-(No-hashmap) | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) & 1: return [] #odd number of elements
changed.sort() #O(nlgn)
#result is the returned array if successful, halves is a
#queue of numbers for which we MUST find double of withi... | find-original-array-from-doubled-array | Python3 Fast & Cheap Solution + Explanation (No hashmap) | apometta | 0 | 78 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,017 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/1997712/Python-or-Counter | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed)%2!=0:
return []
sm=sum(changed)
if sm%3!=0:
return []
tomake=sm//3 #x+2x=3x means divisible by 3
changed.sort()
ans=[]
ctr=C... | find-original-array-from-doubled-array | Python | Counter | heckt27 | 0 | 76 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,018 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/1660782/python-solution-using-sort | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
n = len(changed)
if n % 2 != 0:
return []
changed.sort()
# [1,6,4,8,3,2] -> [1,2,3,4,6,8]
counts = defaultdict(int)
for c in changed:
counts[c] += 1... | find-original-array-from-doubled-array | python solution using sort | byuns9334 | 0 | 271 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,019 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/1557454/Python3-Greedy-Time%3A-O(n-log-n)-and-Space%3A-O(n) | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
# first check if array is even
# create counter map
# sort the array and iterate from beginning by crossing off doubled value
# Time: O(n log n)
# Space: O(n)
if len(changed) % 2 !=... | find-original-array-from-doubled-array | [Python3] Greedy - Time: O(n log n) & Space: O(n) | jae2021 | 0 | 80 | find original array from doubled array | 2,007 | 0.409 | Medium | 28,020 |
https://leetcode.com/problems/maximum-earnings-from-taxi/discuss/1485339/Python-Solution-Maximum-Earnings-from-Taxi | class Solution:
def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
d = {}
for start,end,tip in rides:
if end not in d:
d[end] =[[start,tip]]
else:
d[end].append([start,tip])
dp = [0]*(n+1)
... | maximum-earnings-from-taxi | [Python] Solution - Maximum Earnings from Taxi | SaSha59 | 2 | 2,100 | maximum earnings from taxi | 2,008 | 0.432 | Medium | 28,021 |
https://leetcode.com/problems/maximum-earnings-from-taxi/discuss/1471590/Python3-dp | class Solution:
def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
mp = {}
for start, end, tip in rides:
mp.setdefault(start, []).append((end, tip))
@cache
def fn(x):
"""Return max earning at x."""
if x == n: return 0
... | maximum-earnings-from-taxi | [Python3] dp | ye15 | 2 | 178 | maximum earnings from taxi | 2,008 | 0.432 | Medium | 28,022 |
https://leetcode.com/problems/maximum-earnings-from-taxi/discuss/1471590/Python3-dp | class Solution:
def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
mp = {}
for start, end, tip in rides:
mp.setdefault(start, []).append((end, tip))
dp = [0]*(n+1)
for x in range(n-1, 0, -1):
dp[x] = dp[x+1]
for xx, tip in... | maximum-earnings-from-taxi | [Python3] dp | ye15 | 2 | 178 | maximum earnings from taxi | 2,008 | 0.432 | Medium | 28,023 |
https://leetcode.com/problems/maximum-earnings-from-taxi/discuss/1470927/Greedy-Approach-oror-Clean-and-Concise-Code-oror-Well-Explained | class Solution:
def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
rides.sort()
for ele in rides:
ele[2] += ele[1]-ele[0]
cp,mp = 0,0 # Current_profit, Max_profit
heap=[]
for s,e,p in rides:
while heap and heap[0][0]<=s:
et,tmp = heapq.heapp... | maximum-earnings-from-taxi | 🐍 Greedy Approach || Clean & Concise Code || Well-Explained 📌📌 | abhi9Rai | 1 | 239 | maximum earnings from taxi | 2,008 | 0.432 | Medium | 28,024 |
https://leetcode.com/problems/maximum-earnings-from-taxi/discuss/1470927/Greedy-Approach-oror-Clean-and-Concise-Code-oror-Well-Explained | class Solution:
def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
mapping = defaultdict(list)
for s, e, t in rides:
mapping[s].append([e, e - s + t]) # [end, dollar]
dp = [0] * (n + 1)
for i in range(n - 1, 0, -1):
for e, d in mapping[i]:
dp[i] = max(dp[i], ... | maximum-earnings-from-taxi | 🐍 Greedy Approach || Clean & Concise Code || Well-Explained 📌📌 | abhi9Rai | 1 | 239 | maximum earnings from taxi | 2,008 | 0.432 | Medium | 28,025 |
https://leetcode.com/problems/maximum-earnings-from-taxi/discuss/2802924/Python-(Simple-Dynamic-Programming) | class Solution:
def maxTaxiEarnings(self, n, rides):
dict1 = defaultdict(list)
for s,e,t in rides:
dict1[e].append((s,e-s+t))
dp = [0]*(n+1)
for i in range(1,n+1):
dp[i] = dp[i-1]
for s,a in dict1[i]:
dp[i] = max(dp[i],dp[s] + a)... | maximum-earnings-from-taxi | Python (Simple Dynamic Programming) | rnotappl | 0 | 4 | maximum earnings from taxi | 2,008 | 0.432 | Medium | 28,026 |
https://leetcode.com/problems/maximum-earnings-from-taxi/discuss/1790532/Python3-DP-or-O(N%2BMlogM)-or-Explained-or-Beat-92.98 | class Solution:
def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
# Main Idea: DP
# Time Complexity: O(N+MlogM)
# Space Complexity: O(N)
# where N is `n` and M is the size of `rides`
# First, we sort `rides` by their `end_i`
rides = sorted(rides, key=lambda... | maximum-earnings-from-taxi | [Python3] DP | O(N+MlogM) | Explained | Beat 92.98% | jackconvolution | 0 | 137 | maximum earnings from taxi | 2,008 | 0.432 | Medium | 28,027 |
https://leetcode.com/problems/maximum-earnings-from-taxi/discuss/1471056/TLE-Bottom-Up-DP-greater-Optimized-Bottom-Up-DP-(python) | class Solution:
def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
dp = [0]*(n+1)
for i in range(len(dp)):
for r in rides:
if r[1] <= i: # check if we've hit trips end point yet, then decide to take or not
trip_val = r[1] - r[0]... | maximum-earnings-from-taxi | TLE Bottom-Up DP -> Optimized Bottom-Up DP (python) | dimucc | 0 | 151 | maximum earnings from taxi | 2,008 | 0.432 | Medium | 28,028 |
https://leetcode.com/problems/maximum-earnings-from-taxi/discuss/1470862/Python-DP-solution.-Easy-to-understand-and-clean | class Solution:
def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
rides_from_start = defaultdict(list)
for start, end, tip in rides:
rides_from_start[start].append((end, tip))
@cache
def recursive(actual_position):
if actual_posit... | maximum-earnings-from-taxi | [Python] DP solution. Easy to understand and clean | asbefu | -1 | 169 | maximum earnings from taxi | 2,008 | 0.432 | Medium | 28,029 |
https://leetcode.com/problems/minimum-number-of-operations-to-make-array-continuous/discuss/1471593/Python3-sliding-window | class Solution:
def minOperations(self, nums: List[int]) -> int:
n = len(nums)
nums = sorted(set(nums))
ans = ii = 0
for i, x in enumerate(nums):
if x - nums[ii] >= n: ii += 1
ans = max(ans, i - ii + 1)
return n - ans | minimum-number-of-operations-to-make-array-continuous | [Python3] sliding window | ye15 | 5 | 218 | minimum number of operations to make array continuous | 2,009 | 0.458 | Hard | 28,030 |
https://leetcode.com/problems/minimum-number-of-operations-to-make-array-continuous/discuss/1470979/Python-2-Solutions-Binary-search-(without-bisect)-and-brute-force | class Solution:
def minOperations(self, nums: List[int]) -> int:
n = len(nums)
nums = sorted(set(nums))
answer = float("+inf")
for i, start in enumerate(nums):
search = start + n - 1 # number to search
start, end = 0, len(nums)-1
... | minimum-number-of-operations-to-make-array-continuous | [Python] 2 Solutions Binary search (without bisect) and brute force | asbefu | 1 | 114 | minimum number of operations to make array continuous | 2,009 | 0.458 | Hard | 28,031 |
https://leetcode.com/problems/minimum-number-of-operations-to-make-array-continuous/discuss/1470979/Python-2-Solutions-Binary-search-(without-bisect)-and-brute-force | class Solution:
def minOperations(self, nums: List[int]) -> int:
answer = float("+inf")
for i in range(len(nums)):
minimum = nums[i]
used = set()
total = 0
for j in range(len(nums)):
if minimum <= nums[j] <= minimum + len(nums)... | minimum-number-of-operations-to-make-array-continuous | [Python] 2 Solutions Binary search (without bisect) and brute force | asbefu | 1 | 114 | minimum number of operations to make array continuous | 2,009 | 0.458 | Hard | 28,032 |
https://leetcode.com/problems/minimum-number-of-operations-to-make-array-continuous/discuss/1475303/Well-Written-and-Explained-oror-Simple-oror-96-faster | class Solution:
def minOperations(self, nums: List[int]) -> int:
n=len(nums)
nums = sorted(set(nums))
res = n
for i,st in enumerate(nums):
end = st+n-1
ind = bisect.bisect_right(nums,end)
unq_len = ind-i
res = min(res,n-unq_len)
return res | minimum-number-of-operations-to-make-array-continuous | 📌📌 Well-Written and Explained || Simple || 96% faster 🐍 | abhi9Rai | 0 | 143 | minimum number of operations to make array continuous | 2,009 | 0.458 | Hard | 28,033 |
https://leetcode.com/problems/minimum-number-of-operations-to-make-array-continuous/discuss/1471253/Python-3-Prefix-Sum-%2B-Binary-Search | class Solution:
def minOperations(self, nums: List[int]) -> int:
n = len(nums)
ans = n
nums.sort()
cnt = Counter(nums)
# build key for repeated occurance
repeat = sorted(k for k in cnt if cnt[k] > 1)
# build prefix sum for re... | minimum-number-of-operations-to-make-array-continuous | [Python 3] Prefix Sum + Binary Search | chestnut890123 | 0 | 86 | minimum number of operations to make array continuous | 2,009 | 0.458 | Hard | 28,034 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1472568/Python3-A-Simple-Solution-and-A-One-Line-Solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x = 0
for o in operations:
if '+' in o:
x += 1
else:
x -= 1
return x | final-value-of-variable-after-performing-operations | [Python3] A Simple Solution and A One Line Solution | terrencetang | 18 | 1,800 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,035 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1472568/Python3-A-Simple-Solution-and-A-One-Line-Solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
return sum(1 if '+' in o else -1 for o in operations) | final-value-of-variable-after-performing-operations | [Python3] A Simple Solution and A One Line Solution | terrencetang | 18 | 1,800 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,036 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1472879/Python-one-liner | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
return sum(1 if '+' in op else -1 for op in operations) | final-value-of-variable-after-performing-operations | Python one-liner | blue_sky5 | 4 | 253 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,037 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2079406/Simple-Python-Soluution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x=0
for i in operations:
if(i=="X--" or i=="--X"):
x-=1
else:
x+=1
return x | final-value-of-variable-after-performing-operations | Simple Python Soluution | tusharkhanna575 | 3 | 343 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,038 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1548411/EASY-AND-SIMPLE-SOLN-(faster-than-91) | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x = 0
for i in range(len(operations)):
if operations[i] == "++X" or operations[i] == "X++": x += 1
else: x -=1
return x | final-value-of-variable-after-performing-operations | EASY AND SIMPLE SOLN (faster than 91%) | anandanshul001 | 3 | 233 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,039 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1485582/Python-Simple-Line-faster-than-92.29 | class Solution:
def finalValueAfterOperations(self, operations: list) -> int:
return sum([1 if "+" in op else -1 for op in operations ]) | final-value-of-variable-after-performing-operations | Python Simple Line faster than 92.29% | 1_d99 | 2 | 282 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,040 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2342727/C%2B%2BPython-O(N)-Solution-faster-than-91-submission | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
X=0
for o in operations:
if "+" in o:
X=X+1
else:
X=X-1
return X | final-value-of-variable-after-performing-operations | C++/Python O(N) Solution faster than 91% submission | arpit3043 | 1 | 106 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,041 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1703545/2011.-Final-Value-of-Variable-X%2B%2B-X- | class Solution(object):
def finalValueAfterOperations(self, operations):
"""
:type operations: List[str]
:rtype: int
"""
rValue = 0
for operation in operations:
if operation[1] == '+': # if the operation is X++ or ++X
rValue += 1
else: ... | final-value-of-variable-after-performing-operations | 2011. Final Value of Variable X++ / X-- | ankit61d | 1 | 103 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,042 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1593243/Very-fast-Python-Code-(4-lines) | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x = 0
for val in operations:
x += 1 if "++" in val else -1
return x | final-value-of-variable-after-performing-operations | Very fast Python Code (4 lines) | Vladislav875 | 1 | 208 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,043 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1481162/1-line-solution-in-Python | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
return sum("++" in op or -1 for op in operations) | final-value-of-variable-after-performing-operations | 1-line solution in Python | mousun224 | 1 | 108 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,044 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2839374/Two-pointer-easy-python-solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
result = 0
l = 0
r = len(operations) - 1
mapper = lambda x: 1 if x in {"X++", "++X"} else -1
while l < r:
result += mapper(operations[l]) + mapper(operations[r])
l += ... | final-value-of-variable-after-performing-operations | Two pointer easy python solution | Charan_coder | 0 | 1 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,045 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2822029/One-line-code-in-Python | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
return sum ([1 if k in ("X++", "++X") else -1 for k in operations] ) | final-value-of-variable-after-performing-operations | One line code in Python | DNST | 0 | 2 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,046 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2815447/Fastest-and-Simplest-Solution-Python | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x = 0
for i in operations:
if i == "++X" or i == "X++":
x += 1
else:
x -= 1
return x | final-value-of-variable-after-performing-operations | Fastest and Simplest Solution - Python | PranavBhatt | 0 | 2 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,047 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2815311/Easiest-way-python | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x=0
for i in operations:
if "+" in i:
x=x+1
else:
x=x-1
return x | final-value-of-variable-after-performing-operations | Easiest way python | nishithakonuganti | 0 | 2 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,048 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2814269/Simple-Python-Solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x = 0
increament = operations.count("X++") + operations.count("++X")
decreament = operations.count("X--") + operations.count("--X")
return increament - decreament | final-value-of-variable-after-performing-operations | Simple Python Solution | Shagun_Mittal | 0 | 3 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,049 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2804315/PYTHON3-BEST | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x = 0
for i in operations:
if '+' in i:
x += 1
else:
x -= 1
return x | final-value-of-variable-after-performing-operations | PYTHON3 BEST | Gurugubelli_Anil | 0 | 2 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,050 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2797305/Python-Easy-Solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
res=0
for operation in operations:
if operation=="++X" or operation=="X++":
res+=1
else:
res-=1
return res | final-value-of-variable-after-performing-operations | Python Easy Solution | sbhupender68 | 0 | 1 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,051 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2793058/simple-python-solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
result= 0
for i in operations:
if "+" in i:
result += 1
elif "-" in i:
result -= 1
return result | final-value-of-variable-after-performing-operations | simple python solution | ft3793 | 0 | 1 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,052 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2743651/Fastest-Python-Codeoror98.39-Fasteroror3-Lines-Code | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
a=operations.count('--X')
b=operations.count('X--')
return (a+b)*(-1)+(len(operations)-(a+b))*(1) | final-value-of-variable-after-performing-operations | Fastest Python Code||98.39% Faster||3 Lines Code | sowmika_chaluvadi | 0 | 4 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,053 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2743045/1-Line-Solution-Python-(3-Types-of-Approaches-Shown) | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x=0
# Naive Approach
for operation in operations:
if operation =="--X" or operation =="X--":
x-=1
elif operation == "X++" or operation == "++X" :
x = x +... | final-value-of-variable-after-performing-operations | 1 Line Solution - Python (3 Types of Approaches Shown) | avs-abhishek123 | 0 | 3 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,054 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2743045/1-Line-Solution-Python-(3-Types-of-Approaches-Shown) | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x=0
# 2 lines Approach
for operation in operations:
x = x-1 if (operation =="--X" or operation =="X--") else (x+1 if (operation == "X++" or operation == "++X") else x)
return x | final-value-of-variable-after-performing-operations | 1 Line Solution - Python (3 Types of Approaches Shown) | avs-abhishek123 | 0 | 3 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,055 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2743045/1-Line-Solution-Python-(3-Types-of-Approaches-Shown) | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x=0
# 1 lines Approach
x = sum([ (x-1 if (operation =="--X" or operation =="X--") else (x+1 if (operation == "X++" or operation == "++X")else x)) for operation in operations ])
return x | final-value-of-variable-after-performing-operations | 1 Line Solution - Python (3 Types of Approaches Shown) | avs-abhishek123 | 0 | 3 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,056 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2742666/Final-Value-of-Variable-After-Performing-Operations-Naive-Approach | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x= 0
for operation in operations:
if operation =="--X" or operation =="X--":
x-=1
elif operation == "X++" or operation == "++X" :
x = x + 1
else:
... | final-value-of-variable-after-performing-operations | Final Value of Variable After Performing Operations - Naive Approach | avs-abhishek123 | 0 | 2 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,057 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2720395/Python-simple-solution-using-%22if%22 | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
count = 0
for i in operations:
if i == "X--" or i == "--X":
count -= 1
elif i == "++X" or i == "X++":
count += 1
return count | final-value-of-variable-after-performing-operations | Python simple solution using "if" | maitrunghieu2287 | 0 | 2 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,058 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2691899/Python-with-minimal-code-two-examples | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
result = 0
for value in operations:
# The middle value is either + or - (X++ or ++X are both + at index 1)
result += 1 if value[1] == '+' else -1
return result | final-value-of-variable-after-performing-operations | Python with minimal code, two examples | user0564U | 0 | 4 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,059 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2691899/Python-with-minimal-code-two-examples | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
# One line sum/generator, where True will sum to 1
return sum(i[1] == '+' or -1 for i in operations) | final-value-of-variable-after-performing-operations | Python with minimal code, two examples | user0564U | 0 | 4 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,060 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2671457/Python-fast-and-simple | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
ans = 0
for o in operations:
if o[0] == "+" or o[-1] == "+":
ans += 1
elif o[0] == "-" or o[-1] == "-":
ans -= 1
return ans | final-value-of-variable-after-performing-operations | Python fast and simple | phantran197 | 0 | 5 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,061 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2650234/2-lines-simple-solution-O(1)-or-Python3 | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x = operations.count("X++") + operations.count("++X") - operations.count("X--") - operations.count("--X")
return x | final-value-of-variable-after-performing-operations | 2 lines simple solution O(1) | Python3 | landigf | 0 | 4 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,062 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2646306/Super-Easy-Solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
count = 0
for s in operations:
if s == "--X" or s == "X--":
count -= 1
if s == "X++" or s == "++X":
count += 1
return count | final-value-of-variable-after-performing-operations | Super Easy Solution | mdfaisalabdullah | 0 | 2 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,063 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2640172/Python-3-Counter-Beats-94 | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
c = Counter(chain.from_iterable(operations))
return (c['+'] - c['-']) >> 1 | final-value-of-variable-after-performing-operations | Python 3 Counter Beats 94% | godshiva | 0 | 1 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,064 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2578938/Python-using-set | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
positive = {"X++", "++X"}
res = 0
for operation in operations:
if operation in positive:
res += 1
else:
res -= 1
... | final-value-of-variable-after-performing-operations | Python using set | Vigneswar_A | 0 | 22 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,065 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2555490/2-simple-python-solutions | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
X = 0
count_minus = operations.count("--X")
count_minus += operations.count("X--")
count_plus = operations.count("++X")
count_plus += operations.count("X++")
print(count_plus)
... | final-value-of-variable-after-performing-operations | 2 simple python solutions | maschwartz5006 | 0 | 35 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,066 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2515200/My-python-solution | class Solution(object):
def finalValueAfterOperations(self, operations):
"""
:type operations: List[str]
:rtype: int
"""
val = 0
for i in operations:
if i == '--X' or i == 'X--':
val -=1
else:
val +=1
ret... | final-value-of-variable-after-performing-operations | My python solution | savvy_phukan | 0 | 53 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,067 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2453621/Python3-Solution-with-using-check-1-index | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
res = 0
for op in operations:
if op[1] == '-':
res -= 1
else:
res += 1
return res | final-value-of-variable-after-performing-operations | [Python3] Solution with using check 1 index | maosipov11 | 0 | 24 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,068 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2436708/Python-3-with-easily-readable-one-liner | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
'''1-liner w list comprehension'''
return sum([1 if op in ("++X", "X++") else -1 for op in operations]) | final-value-of-variable-after-performing-operations | Python 3, with easily-readable one-liner | romejj | 0 | 56 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,069 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2398591/Simple-python-code-with-explanation | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
#let the ans-> variable be 0
ans = 0
#iterate over the elements in list
for i in operations:
#if the element is "--X" or "X--"
... | final-value-of-variable-after-performing-operations | Simple python code with explanation | thomanani | 0 | 100 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,070 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2386064/No-if-else-or-switch-statements-Python-or-Faster-than-90 | class Solution:
def add(self, x):
return x + 1
def sub(self, x):
return x - 1
def finalValueAfterOperations(self, operations: List[str]) -> int:
x = 0
d = {
'++X': self.add,
'X++': self.add,
'X--': self.sub,
'--X'... | final-value-of-variable-after-performing-operations | No if else | switch statements Python | Faster than 90% | prameshbajra | 0 | 59 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,071 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2366588/Straightforward-Python3-Solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
value = 0
for op in operations:
if (op[1] == '+'):
value += 1
else:
value -= 1
return value | final-value-of-variable-after-performing-operations | Straightforward Python3 Solution ✅🐍 | qing306037 | 0 | 55 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,072 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2351195/Faster-then-Buzz-Lightyear | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x= 0
for op in operations:
if op in ["--X","X--"]:
x -= 1
else:
x += 1
print(op,x)
return x | final-value-of-variable-after-performing-operations | Faster then Buzz Lightyear | RohanRob | 0 | 35 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,073 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2290709/python3-oror-easy-solution-oror-O(N) | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
result=0
for i in operations:
if i[1]=='+':
result+=1
else:
result-=1
return result | final-value-of-variable-after-performing-operations | python3 || easy solution || O(N) | _soninirav | 0 | 37 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,074 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2282725/Simple-Python-solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
n = 0
for i in range(len(operations)):
if operations[i] == "++X" or operations[i] == "X++":
n+=1
elif operations[i] == "--X" or operations[i] == "X--":
n-=1
... | final-value-of-variable-after-performing-operations | Simple Python solution | anazzy | 0 | 42 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,075 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2263347/3-simple-steps-using-Python | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
res = 0
for i in operations:
if i[0] == '-' or i[-1] == '-':
res -= 1
else:
res += 1
return res | final-value-of-variable-after-performing-operations | 3 simple steps using Python | ankurbhambri | 0 | 31 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,076 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2209090/Simple-Python-Solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
value = 0
for i in operations:
if(i == '++X' or i == 'X++'):
value += 1
else:
value -= 1
return value | final-value-of-variable-after-performing-operations | Simple Python Solution | kaus_rai | 0 | 66 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,077 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2208181/FASTER-than-95-oror-EASY-SOLUTION-using-if-else | '''class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
initial=0
for i in operations:
if i=="--X" or i=="X--":
initial=initial-1
else:
initial=initial+1
return initial
print(finalValueAfterOperations(operations))''' | final-value-of-variable-after-performing-operations | FASTER than 95% || EASY SOLUTION using if-else | keertika27 | 0 | 76 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,078 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2196650/Easy-solution-using-a-dictionary | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
ops = {
'++X': 1,
'X++': 1,
'--X': -1,
'X--': -1
}
return sum([ops[op] for op in operations]) | final-value-of-variable-after-performing-operations | Easy solution using a dictionary | Simzalabim | 0 | 19 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,079 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2163988/iterative | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
# keep result starting at 0
# iterate each operatin
# if the operation contains - decrement res otherwise increment res
# return result after iteration
# Time O(N) SpaceL O(1)
... | final-value-of-variable-after-performing-operations | iterative | andrewnerdimo | 0 | 25 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,080 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2102771/Very-Short-solution-and-easy-to-understand | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
n=0
for i in operations:
if i== "X++" or i== "++X":
n+=1
else:
n-=1
return n | final-value-of-variable-after-performing-operations | Very Short solution and easy to understand | T1n1_B0x1 | 0 | 89 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,081 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2030675/Runtime%3A-34-ms-87.79 | class Solution(object):
def finalValueAfterOperations(self, operations):
"""
:type operations: List[str]
:rtype: int
"""
X = 0
for operation in operations:
if operation == '++X' or operation == 'X++':
X = X + 1
elif operat... | final-value-of-variable-after-performing-operations | Runtime: 34 ms 87.79% ට වඩා වේගවත් | akilaocj | 0 | 62 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,082 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/2013084/Python3-using-for-and-if-else | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
value = 0;
for i in range(len(operations)):
operator = operations[i]
if operator == '++X' or operator == 'X++':
value += 1
else:
value -= 1
r... | final-value-of-variable-after-performing-operations | [Python3] using for and if else | Shiyinq | 0 | 66 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,083 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1989166/java-python-easy-and-fast | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
x = 0
for s in operations :
if s[1] == '+' : x += 1
else : x -= 1
return x | final-value-of-variable-after-performing-operations | java, python - easy & fast | ZX007java | 0 | 87 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,084 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1957596/Python3-Simple-solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
ans = 0
for operation in operations:
ans = ans + 1 if operation in ["++X","X++"] else ans-1
return ans | final-value-of-variable-after-performing-operations | Python3 Simple solution | firefist07 | 0 | 66 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,085 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1950513/Beginner-friendly-solution-oror-python3 | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
result = 0
for i in operations:
if i == '--X' or i == 'X--':
result -= 1
else:
result += 1
return result | final-value-of-variable-after-performing-operations | Beginner friendly solution || python3 | ApacheRosePeacock | 0 | 47 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,086 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1936819/Python-Solution-%2B-One-Liner! | class Solution:
def finalValueAfterOperations(self, operations):
x = 0
for op in operations:
if "++" in op: x += 1
elif "--" in op: x -= 1
else: raise Exception("Invalid operation")
return x | final-value-of-variable-after-performing-operations | Python - Solution + One-Liner! | domthedeveloper | 0 | 78 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,087 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1936819/Python-Solution-%2B-One-Liner! | class Solution:
def finalValueAfterOperations(self, operations):
return sum("++" in op or -1 for op in operations) | final-value-of-variable-after-performing-operations | Python - Solution + One-Liner! | domthedeveloper | 0 | 78 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,088 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1932535/Python-solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
X=0
for i in operations:
X= X+1 if i in ["++X","X++"] else X-1
return X | final-value-of-variable-after-performing-operations | Python solution | Swamy54 | 0 | 27 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,089 |
https://leetcode.com/problems/final-value-of-variable-after-performing-operations/discuss/1932535/Python-solution | class Solution:
def finalValueAfterOperations(self, operations: List[str]) -> int:
X=0
for i in operations:
X= X+1 if i[1]=="+" else X-1
return X | final-value-of-variable-after-performing-operations | Python solution | Swamy54 | 0 | 27 | final value of variable after performing operations | 2,011 | 0.888 | Easy | 28,090 |
https://leetcode.com/problems/sum-of-beauty-in-the-array/discuss/1477177/Python3-or-Brute-Force-(TLE)-and-O(n)-solution-with-explanation-or-86ile-runtime | class Solution:
def sumOfBeauties(self, nums: List[int]) -> int:
beauty=[0]*len(nums)
for i in range(1,len(nums)-1):
leftarr=nums[:i]
rightarr=nums[i+1:]
if(max(leftarr)<nums[i] and min(rightarr)>nums[i]):
beauty[i]=2
elif(nums[i-1]<num... | sum-of-beauty-in-the-array | Python3 | Brute Force (TLE) and O(n) solution with explanation | 86%ile runtime | aaditya47 | 1 | 54 | sum of beauty in the array | 2,012 | 0.467 | Medium | 28,091 |
https://leetcode.com/problems/sum-of-beauty-in-the-array/discuss/1477177/Python3-or-Brute-Force-(TLE)-and-O(n)-solution-with-explanation-or-86ile-runtime | class Solution:
def sumOfBeauties(self, a: List[int]) -> int:
temp,temp2=a[0],a[-1]
left=([a[0]]+[0]*(len(a)-1))
right=[0]*(len(a)-1) + [a[-1]]
for i in range(1,len(a)):
left[i]=max(a[i-1],temp)
temp=left[i]
for i in range(len(a)-2,-1,-1):
... | sum-of-beauty-in-the-array | Python3 | Brute Force (TLE) and O(n) solution with explanation | 86%ile runtime | aaditya47 | 1 | 54 | sum of beauty in the array | 2,012 | 0.467 | Medium | 28,092 |
https://leetcode.com/problems/sum-of-beauty-in-the-array/discuss/2825351/PreSum-solution-in-Python | class Solution:
def sumOfBeauties(self, nums: List[int]) -> int:
preSum1, curr = [nums[0]], 0
for i in range(1, len(nums)):
curr = max (nums[i-1], curr)
preSum1.append(curr)
pre2, curr = [nums[-1]], 10**5 + 2
for i in range(len(nums)-... | sum-of-beauty-in-the-array | PreSum solution in Python | DNST | 0 | 1 | sum of beauty in the array | 2,012 | 0.467 | Medium | 28,093 |
https://leetcode.com/problems/sum-of-beauty-in-the-array/discuss/2721878/Python-Easy-Solution-or-Faster-than-100-or | class Solution:
def sumOfBeauties(self, nums: List[int]) -> int:
n = len(nums)
left = [-1 for i in range(n)]
right = [-1 for i in range(n)]
mx = -1
for i in range(n):
left[i] = mx
mx = max(mx, nums[i])
mx = 10**9
for i in range(n-1,-1,-1):
right[i] = mx
mx = min(mx, nums[i])
# print(left)
... | sum-of-beauty-in-the-array | Python Easy Solution | Faster than 100% | | sami2002 | 0 | 10 | sum of beauty in the array | 2,012 | 0.467 | Medium | 28,094 |
https://leetcode.com/problems/sum-of-beauty-in-the-array/discuss/1766916/Python-3-two-pass-O(n)-time-O(n)-space | class Solution:
def sumOfBeauties(self, nums: List[int]) -> int:
n = len(nums)
minFromRight = [0] * n
curMin = nums[-1]
for i in range(n-2, 0, -1):
minFromRight[i] = curMin
curMin = min(curMin, nums[i])
res = 0
curMax = nums[0]
... | sum-of-beauty-in-the-array | Python 3, two pass, O(n) time, O(n) space | dereky4 | 0 | 72 | sum of beauty in the array | 2,012 | 0.467 | Medium | 28,095 |
https://leetcode.com/problems/sum-of-beauty-in-the-array/discuss/1514274/Two-passes-O(n)-89-speed | class Solution:
def sumOfBeauties(self, nums: List[int]) -> int:
len_nums = len(nums)
suffix_min = nums.copy()
for i in range(len_nums - 2, -1, -1):
suffix_min[i] = min(suffix_min[i], suffix_min[i + 1])
pre_max = nums[0]
beauty = 0
for i in range(1, len_nu... | sum-of-beauty-in-the-array | Two passes, O(n), 89% speed | EvgenySH | 0 | 118 | sum of beauty in the array | 2,012 | 0.467 | Medium | 28,096 |
https://leetcode.com/problems/sum-of-beauty-in-the-array/discuss/1478873/Python3-2-pass | class Solution:
def sumOfBeauties(self, nums: List[int]) -> int:
suffix = [inf] * len(nums) # suffix of min
for i in range(len(nums)-2, 0, -1):
suffix[i] = min(suffix[i+1], nums[i+1])
ans = prefix = 0
for i in range(1, len(nums)-1):
prefix = max... | sum-of-beauty-in-the-array | [Python3] 2-pass | ye15 | 0 | 67 | sum of beauty in the array | 2,012 | 0.467 | Medium | 28,097 |
https://leetcode.com/problems/sum-of-beauty-in-the-array/discuss/1473231/Python-3-Two-heaps | class Solution:
def sumOfBeauties(self, nums: List[int]) -> int:
left, right = [], []
for i, num in enumerate(nums):
heappush(right, (num, i))
ans = 0
n = len(nums)
for i in range(1, n-1):
heappush(left, (-nums[i-1], i-1))
while right and ... | sum-of-beauty-in-the-array | [Python 3] Two heaps | chestnut890123 | 0 | 35 | sum of beauty in the array | 2,012 | 0.467 | Medium | 28,098 |
https://leetcode.com/problems/sum-of-beauty-in-the-array/discuss/1472602/Python-3-oror-Using-Prefix_maxandSuffix_min-oror-Self-Explanatory-oror-Easy-Understandable | class Solution:
def sumOfBeauties(self, nums: List[int]) -> int:
suffix_min=[float('inf')]*len(nums)
prefix_max=[float('-inf')]*len(nums)
prefix_max[0]=nums[0]
for i in range(1,len(nums)):
prefix_max[i]=max(prefix_max[i-1],nums[i-1])
... | sum-of-beauty-in-the-array | Python 3 || Using Prefix_max&Suffix_min || Self-Explanatory || Easy-Understandable | bug_buster | 0 | 35 | sum of beauty in the array | 2,012 | 0.467 | Medium | 28,099 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.