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 |
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https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/discuss/944714/Python-3-or-Sort-%2B-Greedy-and-Sort-%2B-Binary-Search-or-Explanation | class Solution:
def minimumEffort(self, tasks: List[List[int]]) -> int:
tasks.sort(key=lambda x: x[0]-x[1])
def ok(mid):
for actual, minimum in tasks:
if minimum > mid or actual > mid: return False
if minimum <= mid: mid -= actual
return True
l, r = 0, 10 ** 9
while l <= r:
mid = (l+r) // 2
if ok(mid): r = mid - 1
else: l = mid + 1
return l | minimum-initial-energy-to-finish-tasks | Python 3 | Sort + Greedy & Sort + Binary Search | Explanation | idontknoooo | 2 | 173 | minimum initial energy to finish tasks | 1,665 | 0.562 | Hard | 24,100 |
https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/discuss/944714/Python-3-or-Sort-%2B-Greedy-and-Sort-%2B-Binary-Search-or-Explanation | class Solution:
def minimumEffort(self, tasks: List[List[int]]) -> int:
tasks.sort(key=lambda x: x[0]-x[1])
ans = cur = 0
for cost, minimum in tasks:
ans = min(cur-minimum, ans)
cur -= cost
return -ans | minimum-initial-energy-to-finish-tasks | Python 3 | Sort + Greedy & Sort + Binary Search | Explanation | idontknoooo | 2 | 173 | minimum initial energy to finish tasks | 1,665 | 0.562 | Hard | 24,101 |
https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/discuss/2551554/Python-solution-easy-without-heap-or-binary-search-just-sorting | class Solution:
def minimumEffort(self, tasks: List[List[int]]) -> int:
t=tasks
t.sort(key=lambda x:x[1]-x[0])
n=len(t)
m=t[0][1]
for i in range(n-1):
m=max(m+t[i+1][0],t[i+1][1])
return m | minimum-initial-energy-to-finish-tasks | Python solution easy without heap or binary search just sorting | Gunbnelch | 0 | 22 | minimum initial energy to finish tasks | 1,665 | 0.562 | Hard | 24,102 |
https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/discuss/2424494/Python3-or-Binary-Search-%2B-Sorting-or-With-Intuition | class Solution:
def isPossible(self,time,tasks):
for f,r in tasks:
if time<r:
return False
else:
time-=f
return time>=0
def minimumEffort(self, tasks: List[List[int]]) -> int:
tasks.sort(key=lambda x:(x[1]-x[0]),reverse=True)
sumTime=0
maxReq=0
for task in tasks:
sumTime+=task[1]
maxReq=max(maxReq,task[1])
low,high=maxReq,sumTime
ans=0
while low<=high:
mid=(low+high)>>1
if self.isPossible(mid,tasks):
ans=mid
high=mid-1
else:
low=mid+1
return ans | minimum-initial-energy-to-finish-tasks | [Python3] | Binary Search + Sorting | With Intuition | swapnilsingh421 | 0 | 43 | minimum initial energy to finish tasks | 1,665 | 0.562 | Hard | 24,103 |
https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/discuss/1533683/Python3-solution-comments | class Solution:
def minimumEffort(self, tasks: List[List[int]]) -> int:
def cmp_items(a): #creating a custom sorting function (we want to sort by difference ~ from the biggest to the smallest)
return a[0]-a[1]
tasks.sort(key=cmp_items)
my_energy = tasks[0][1] # how much we need to have at the start
current_task = tasks[0][1] - tasks[0][0] # how much remains after we finish the first task
for i in range(1, len(tasks)):
if current_task < tasks[i][1]:
my_energy += tasks[i][1] - current_task # how much we need to add in order to get the minimum to start the current task
current_task += tasks[i][1] - current_task
current_task -= tasks[i][0] # how much we have after finishing the current task
return my_energy | minimum-initial-energy-to-finish-tasks | Python3 solution comments | FlorinnC1 | 0 | 65 | minimum initial energy to finish tasks | 1,665 | 0.562 | Hard | 24,104 |
https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/discuss/1145204/Python3-Simple-Solution-faster-than-100 | class Solution:
def minimumEffort(self, tasks: List[List[int]]) -> int:
rem_energy = min_energy = sum(task[0] for task in tasks) #At first, minimum total energy=total actual energy
tasks = sorted(tasks, key =lambda x:x[0]-x[1]) #Reverse sort the tasks based on the difference between the minimum and actual
for task in tasks:
if task[1]>rem_energy: #The minimum energy of the task is greater than the total remaining energy
min_energy+=task[1]-rem_energy #So we need more energy
rem_energy+=task[1]-rem_energy
rem_energy-= task[0] #Some energy is spent to do the task, so reduce the amount of total remaining energy
return min_energy | minimum-initial-energy-to-finish-tasks | Python3 Simple Solution, faster than 100% | bPapan | 0 | 112 | minimum initial energy to finish tasks | 1,665 | 0.562 | Hard | 24,105 |
https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/discuss/1100447/Python3-sorting | class Solution:
def minimumEffort(self, tasks: List[List[int]]) -> int:
ans = val = 0
for x, y in sorted(tasks, key=lambda x: x[0]-x[1]):
if val < y:
ans += y - val
val = y
val -= x
return ans | minimum-initial-energy-to-finish-tasks | [Python3] sorting | ye15 | 0 | 58 | minimum initial energy to finish tasks | 1,665 | 0.562 | Hard | 24,106 |
https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/discuss/1100447/Python3-sorting | class Solution:
def minimumEffort(self, tasks: List[List[int]]) -> int:
pq = [(x-y, y) for x, y in tasks]
heapify(pq)
ans = val = 0
while pq:
x, y = heappop(pq)
if val < y:
ans += y - val
val = y
val -= x+y
return ans | minimum-initial-energy-to-finish-tasks | [Python3] sorting | ye15 | 0 | 58 | minimum initial energy to finish tasks | 1,665 | 0.562 | Hard | 24,107 |
https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/discuss/1068459/Python3-solution-faster-than-89.86-and-less-than-89.86 | class Solution:
def minimumEffort(self, tasks):
tasks.sort(key = lambda x:x[1]-x[0], reverse = True)
energy = tasks[0][1]
remain = tasks[0][1] - tasks[0][0]
for cost, threshold in tasks[1:]:
if remain < threshold:
energy += (threshold - remain)
remain = threshold - cost
else:
remain -= cost
return energy | minimum-initial-energy-to-finish-tasks | Python3 solution, faster than 89.86% and less than 89.86% | danieltseng | 0 | 100 | minimum initial energy to finish tasks | 1,665 | 0.562 | Hard | 24,108 |
https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/discuss/945060/Python-very-Concise-Solution-oror-Dhruv-Vavliya | class Solution:
def minimumEffort(self, tasks: List[List[int]]) -> int:
energy=0
remain=0
ar=[]
for task in range(len(tasks)):
ar.append((tasks[task][1]-tasks[task][0],task))
ar.sort()
for task in ar[::-1]:
if tasks[task[1]][1]>remain:
energy+=(tasks[task[1]][1]-remain)
remain=task[0]
else:
remain-=tasks[task[1]][1]
remain+=task[0]
return energy | minimum-initial-energy-to-finish-tasks | Python -- very Concise Solution || Dhruv Vavliya | risky_coder | 0 | 49 | minimum initial energy to finish tasks | 1,665 | 0.562 | Hard | 24,109 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1400359/Python3-Simple-Solution | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
i = 0
while word*(i+1) in sequence:
i+=1
return i | maximum-repeating-substring | Python3, Simple Solution | Flerup | 4 | 306 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,110 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/2192570/Python-simple-solution | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
for i in range(len(sequence)//len(word)+1,0,-1):
if i*word in sequence:
return i
else:
return 0 | maximum-repeating-substring | Python simple solution | StikS32 | 1 | 134 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,111 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1798591/Python3-solution-or-87-lesser-memory | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
c = 0
for i in range(1,len(sequence)//len(word)+1):
if word*i in sequence:
c += 1
continue
break
return c | maximum-repeating-substring | ✔Python3 solution | 87% lesser memory | Coding_Tan3 | 1 | 203 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,112 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1722457/Python-3-Short-and-Clean-using-find | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
n=len(sequence)
m=len(word)
c=0
for i in range(1,(n//m)+1):
s=word*i
if sequence.find(s)!=-1:
c=i
return c | maximum-repeating-substring | [Python 3] Short and Clean using find | akashmaurya001 | 1 | 95 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,113 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1357665/KMP-True-O(n) | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
cnt = len(sequence)//len(word)
repeat = word * cnt
newstr = repeat + '#' + sequence
lps = [0]
maxi = 0
for i in range(1, len(newstr)):
x = lps[-1]
while newstr[x] != newstr[i]:
if x == 0:
x = -1
break
x = lps[x-1]
lps.append(x+1)
if i>=len(repeat):
maxi = max(maxi, x+1)
return maxi//len(word) | maximum-repeating-substring | KMP True O(n) | deleted_user | 1 | 174 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,114 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1260385/python3-simple-solution | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
x = 0
while True:
if word*(x+1) in sequence:
x += 1
else:
return x | maximum-repeating-substring | python3 simple solution | EklavyaJoshi | 1 | 131 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,115 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/2726421/Python3-Linear-Masochist-Solution | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
# record the index of every match using kmp
# then find the longest consecutive sequence (with diff = len(word))
# this is not an easy problem
def kmp_failure(p):
m = len(p)
f = [0] * m
i = 1
j = 0
while i < m:
if p[j] == p[i]:
f[i] = j+1
j += 1
i += 1
elif j > 0:
j = f[j-1]
else:
f[i] = 0
i += 1
return f
if len(word) > len(sequence):
return 0
m = len(word)
n = len(sequence)
f = kmp_failure(word)
i = 0
j = 0
res = 0
matches = dict()
while i < n:
if word[j] == sequence[i]:
if j == m-1:
if i-m in matches:
matches[i] = matches[i-m] + 1
else:
matches[i] = 1
if matches[i] > res:
res = matches[i]
# note that you shouldn't just go j += 1 here!
# because you might skip a match
if j > 0:
j = f[j-1]
else:
i += 1
continue
j += 1
i += 1
elif j > 0:
j = f[j-1]
else:
i += 1
return res | maximum-repeating-substring | [Python3] Linear Masochist Solution | rt500 | 0 | 27 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,116 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/2643440/4-lines-python3-solution | class Solution:
def maxRepeating(self, s: str, w: str) -> int:
n =len(s)//len(w)
for i in range(n,-1,-1):
if w*i in s:
return i | maximum-repeating-substring | 4 lines python3 solution | abhayCodes | 0 | 20 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,117 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/2601589/python-or-explanation-in-comments | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
seq_len = len(sequence)
word_len = len(word)
res, i = 0, 0
# return the number of times 'word' appears in continuation from index position 'idx'
def get_k_reps(idx):
acc = 1
while sequence[idx: idx + word_len * acc] == word * acc:
acc += 1
return acc - 1
while i < seq_len - word_len + 1:
# found match of word in the sliding window
if sequence[i: i + word_len] == word:
count = get_k_reps(i)
res = max(count, res)
i += 1
return res | maximum-repeating-substring | python | explanation in comments | sproq | 0 | 28 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,118 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/2140008/Python-KMP-Simple-Solution | class Solution:
def maxRepeating(self, sequence, word):
maxCount_prefix = self.findMaxSubstring(word, sequence)
maxCount_suffix = self.findMaxSubstring(word[::-1], sequence[::-1])
return max(maxCount_prefix, maxCount_suffix)
def findMaxSubstring(self, pattern, text):
lps = self.get_lps(pattern);
currentSubstring, longestSubstring = 0, 0
i, j = 0, 0
while i < len(text):
if text[i] == pattern[j]:
i, j = i + 1, j + 1
if j == len(pattern):
currentSubstring += 1
longestSubstring = max(currentSubstring, longestSubstring)
j = 0
else:
currentSubstring = 0
if j == 0:
i += 1
else:
j = lps[j - 1]
return longestSubstring;
def get_lps(self, pattern):
lps = [0] * len(pattern) # first lps val will always be one
prev_lps, i = 0, 1
while i < len(pattern):
if pattern[i] == pattern[prev_lps]:
lps[i] = prev_lps + 1
prev_lps, i = prev_lps + 1, i + 1
else:
if prev_lps == 0:
lps[i] = 0
i += 1
else:
prev_lps = lps[prev_lps - 1]
return lps | maximum-repeating-substring | Python - KMP Simple Solution | ErickMwazonga | 0 | 84 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,119 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/2070486/Python-Two-Scans-Forward-And-Backward | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
def forward():
l, mx, current, i = len(word), 0, 0, 0
while i < len(sequence):
if sequence[i : i + l] == word:
current += 1
i += l - 1
elif current > 0:
mx = max(mx, current)
current = 0
i += 1
return max(mx, current)
def backward():
l, mx, current, i = len(word), 0, 0, len(sequence) - 1
while i >= 0:
w = sequence[i - l + 1 : i + 1]
if w == word:
current += 1
i -= l - 1
else:
if current > 0:
mx = max(mx, current)
current = 0
i -= 1
return max(mx, current)
return max(forward(), backward()) | maximum-repeating-substring | Python Two Scans Forward And Backward | Hejita | 0 | 86 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,120 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1483628/While-loop-96-speed | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
count = 1
while word * count in sequence:
count += 1
return count - 1 | maximum-repeating-substring | While loop, 96% speed | EvgenySH | 0 | 270 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,121 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1118730/Python-easy-to-understand-with-find() | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
result = 0
for i in range(1, int(len(sequence) / len(word)) + 1):
if sequence.find(word * i) != -1:
result += 1
else:
break
return result | maximum-repeating-substring | [Python] easy to understand with find() | cruim | 0 | 184 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,122 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1093627/Python3-brute-force | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
ans = 0
for k in range(101):
if word*k in sequence: ans = k
return ans | maximum-repeating-substring | [Python3] brute-force | ye15 | 0 | 125 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,123 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1093627/Python3-brute-force | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
if len(sequence) < len(word): return 0 # edge case
pattern = word * (len(sequence)//len(word))
lps = [0]
k = 0
for i in range(1, len(pattern)):
while k and pattern[k] != pattern[i]: k = lps[k-1]
if pattern[i] == pattern[k]: k += 1
lps.append(k)
ans = k = 0
for i in range(len(sequence)):
while k and pattern[k] != sequence[i]: k = lps[k-1]
if pattern[k] == sequence[i]: k += 1
ans = max(ans, k//len(word))
if k == len(pattern): return ans
return ans | maximum-repeating-substring | [Python3] brute-force | ye15 | 0 | 125 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,124 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/954431/Python-easy-and-efficient-solution | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
i = 0
n = len(word)
m = len(sequence)
c, k = 0, 0
# start traversing in the sequence
while i < m:
# if word found in sequence increase counter and increment loop by len(word)
if sequence[i:i+n] == word:
c += 1
i += n
# edge-case
# if end of sequence is reached and last letter is part of word assign k = max(k,c)
if i == m:
k = max(k,c)
else:
# assign k = max(previous max occurence of word , current max occurence)
k = max(k,c)
i += 1
c = 0
return k | maximum-repeating-substring | Python easy and efficient solution | kadarsh2k00 | 0 | 272 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,125 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1763973/Python-dollarolution | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
if word not in sequence:
return 0
i = 1
while len(word) * i < len(sequence)+1:
if word*i in sequence:
i += 1
else:
break
return i-1 | maximum-repeating-substring | Python $olution | AakRay | -1 | 103 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,126 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/1761432/Python3-92-faster | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
count = 1
while True:
if (word*count in sequence):
count += 1
else:
return count -1 | maximum-repeating-substring | Python3 92% faster | ABinfinity | -1 | 81 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,127 |
https://leetcode.com/problems/maximum-repeating-substring/discuss/955407/Newbie-Python3-Solution-%3A-24ms | class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
if word in sequence:
replaced = sequence.replace(word, '0')
else:
return 0
flag, ans = 0, 0
for char in replaced:
if char == '0':
flag += 1
else:
ans = max(ans, flag)
flag = 0
ans = max(ans, flag)
return ans | maximum-repeating-substring | [Newbie Python3 Solution : 24ms] | user6148Qk | -1 | 104 | maximum repeating substring | 1,668 | 0.396 | Easy | 24,128 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/1833998/Python3-oror-Explanation-and-Example | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
curr=list1
for count in range(b):
if count==a-1: # travel to a node and --> step 1
start=curr # then save pointer in start
curr=curr.next # continue travel to b node --> step 2
start.next=list2 # point start to list2 --> step3
while list2.next: # travel list2 --> step 4
list2=list2.next
list2.next=curr.next # map end of list2 to b
return list1 | merge-in-between-linked-lists | Python3 || Explanation & Example | rushi_javiya | 2 | 48 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,129 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/1387631/Python3-greater85-2-Pointer-Solution.-O(N)-O(1)-Space | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
point1 = list1
point2 = list1
for _ in range(b - a):
point2 = point2.next
prev = None
for _ in range(a):
prev = point1
point1 = point1.next
point2 = point2.next
prev.next = list2
while list2.next:
list2 = list2.next
list2.next = point2.next
return list1 | merge-in-between-linked-lists | [Python3] >85% 2 Pointer Solution. O(N), O(1) Space | whitehatbuds | 2 | 185 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,130 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/2828012/python | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
dummynode = ListNode(-1, list1)
curr = dummynode
idx = 0
while curr.next:
if idx == a:
head = curr
curr = curr.next
head.next = list2
while list2.next:
list2 = list2.next
if idx == b:
list2.next = curr.next
break
curr = curr.next
idx += 1
return dummynode.next | merge-in-between-linked-lists | python | xy01 | 0 | 1 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,131 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/2764595/Python3-Solution | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
temp = list1
x = None
while a > 0:
x = temp
temp = temp.next
a -= 1
temp1 = list1
while b > 0:
temp1 = temp1.next
b -= 1
temp2 = list2
while temp2.next != None:
temp2 = temp2.next
if x is not None:
x.next = list2
else:
list1 = list2
temp2.next = temp1.next
return list1 | merge-in-between-linked-lists | Python3 Solution | mediocre-coder | 0 | 3 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,132 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/2597438/Python3-solution-O(N)-time-complexity-O(1)-space-complexity-99-less-memory | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
prev, first, second = None, list1, list1
#Moves the second pointer forward
while b - a > 0:
second = second.next
b -= 1
#Moves all pointers forward
while a > 0:
prev = first
first = first.next
second = second.next
a -= 1
#prev is the tail in the first part of list1, link the next pointer to list2 head
prev.next = list2
#Iterates through the second list
curr = list2
while curr.next:
curr = curr.next
#Attaches to the tail of list2 the second part of list1 (second is the last node to be removed)
curr.next = second.next
return list1 | merge-in-between-linked-lists | Python3 solution, O(N) time complexity, O(1) space complexity, 99% less memory | matteogianferrari | 0 | 15 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,133 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/2579606/Python-Readable-O(N%2BM)-pass | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
# keep the list head
head = list1
# get to the first node
for _ in range(a-1):
list1 = list1.next
# save the next node (to continue iterating list1)
curr = list1.next
# stitch together
list1.next = list2
# go right after the second node
for _ in range(b-a+1):
curr = curr.next
# get the end of list2
while list2.next is not None:
list2 = list2.next
# stitch together
list2.next = curr
return head | merge-in-between-linked-lists | [Python] - Readable O(N+M) pass | Lucew | 0 | 13 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,134 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/2519595/Python-or-Explained-2-pointer-Solution-or-Faster-than-99.87-or-Less-memory-than-93.79 | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
l1, l2 = list1, list2
for _ in range(a - 1):
l1 = l1.next
temp = l1.next
l1.next = list2
for _ in range(b - a + 1):
temp = temp.next
while l2.next:
l2 = l2.next
l2.next = temp
return list1 | merge-in-between-linked-lists | Python | Explained 2-pointer Solution | Faster than 99.87% | Less memory than 93.79% | ahmadheshamzaki | 0 | 26 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,135 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/2344596/Easy-python | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
#iterate on list2 to find last pointer
curr=list2
while curr:
if curr.next==None:
last2 = curr
curr=curr.next
c=0
#iterate on list1
prev=list1
curr=list1.next
while curr:
if c==a-1:
prev.next=list2 #point a-1th to list2
elif c==b:
last2.next = curr #point last pointer after bth node
prev=curr
c+=1
curr=curr.next
return list1 | merge-in-between-linked-lists | Easy python | sunakshi132 | 0 | 15 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,136 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/2274149/Simple-Python-Solution-Easy-to-Understand! | class Solution(object):
def mergeInBetween(self, list1, indexA, indexB, list2):
currentOne = temp = list1
currentTwo = list2
# find node before where A starts
for _ in range(indexA-1):
currentOne = currentOne.next
# find node after where B ends
for _ in range(indexB+1):
temp = temp.next
# find end of list2
while currentTwo.next is not None:
currentTwo = currentTwo.next
currentOne.next = list2 # connect list1 to the head of list2 where before A starts
currentTwo.next = temp # connect the end of list2 to the rest of list1 where after B ends
return list1
``` | merge-in-between-linked-lists | Simple [Python] Solution, Easy to Understand! | ufultra | 0 | 27 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,137 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/1999818/python-3-oror-simple-O(b-%2B-m)-solution | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
prevA = dummy1 = ListNode(0, list1)
for _ in range(a):
prevA = prevA.next
nextB = prevA
for _ in range(b - a + 2):
nextB = nextB.next
end2 = list2
while end2.next:
end2 = end2.next
prevA.next = list2
end2.next = nextB
return dummy1.next | merge-in-between-linked-lists | python 3 || simple O(b + m) solution | dereky4 | 0 | 23 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,138 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/1093472/Python3-2-pass | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
node = list1
for k in range(b+1):
if k == a-1: start = node
node = node.next
end = node
start.next = node = list2
while node.next: node = node.next
node.next = end
return list1 | merge-in-between-linked-lists | [Python3] 2-pass | ye15 | 0 | 24 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,139 |
https://leetcode.com/problems/merge-in-between-linked-lists/discuss/1071317/python-3-solution-beat-74 | class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
head=list1
c=0
while(c<=b):
if c==a-1:
prev=list1
list1=list1.next
prev.next=list2
else:
list1=list1.next
c+=1
while(list2.next):
list2=list2.next
list2.next=list1
return head | merge-in-between-linked-lists | python 3 solution beat 74% | AchalGupta | 0 | 58 | merge in between linked lists | 1,669 | 0.745 | Medium | 24,140 |
https://leetcode.com/problems/minimum-number-of-removals-to-make-mountain-array/discuss/1543614/Python-oror-Easy-Solution | class Solution:
def minimumMountainRemovals(self, lst: List[int]) -> int:
l = len(lst)
dp = [0] * l
dp1 = [0] * l
for i in range(l): # for increasing subsequence
maxi = 0
for j in range(i):
if lst[i] > lst[j]:
if dp[j] > maxi:
maxi = dp[j]
dp[i] = maxi + 1
for i in range(l - 1, -1, -1): # for decreasing subsequence
maxi1 = 0
for j in range(l - 1, i, -1):
if lst[i] > lst[j]:
if dp1[j] > maxi1:
maxi1 = dp1[j]
dp1[i] = maxi1 + 1
ans = 0
for i in range(l):
if dp[i] > 1 and dp1[i] > 1:
temp = dp[i] + dp1[i] - 1
if temp > ans:
ans = temp
return l - ans | minimum-number-of-removals-to-make-mountain-array | Python || Easy Solution | naveenrathore | 2 | 209 | minimum number of removals to make mountain array | 1,671 | 0.425 | Hard | 24,141 |
https://leetcode.com/problems/minimum-number-of-removals-to-make-mountain-array/discuss/1722955/python3%3A-TIme-Complexity-O(n2) | class Solution:
def minimumMountainRemovals(self, nums: List[int]) -> int:
n = len(nums)
left = [0 for i in range(n)]
right = [0 for i in range(n)]
for i in range(1, n):
val = 0
for j in range(i):
if nums[j] < nums[i]:
val = max(val, left[j] + 1)
left[i] = val
for i in range(n - 2, -1, -1):
val = 0
for j in range(i + 1, n):
if nums[j] < nums[i]:
val = max(val, right[j] + 1)
right[i] = val
ans = inf
for i in range(1, n - 1):
if left[i] > 0 and right[i] > 0:
ans = min(ans, n - 1 - left[i] - right[i])
return ans | minimum-number-of-removals-to-make-mountain-array | python3: TIme Complexity O(n^2) | reynaldocv | 1 | 64 | minimum number of removals to make mountain array | 1,671 | 0.425 | Hard | 24,142 |
https://leetcode.com/problems/minimum-number-of-removals-to-make-mountain-array/discuss/1093648/Python3-linearithmic-solution | class Solution:
def minimumMountainRemovals(self, nums: List[int]) -> int:
def fn(nums):
"""Return length of LIS (excluding x) ending at x."""
ans, vals = [], []
for i, x in enumerate(nums):
k = bisect_left(vals, x)
if k == len(vals): vals.append(x)
else: vals[k] = x
ans.append(k)
return ans
left, right = fn(nums), fn(nums[::-1])[::-1]
ans = inf
for i in range(1, len(nums)-1):
if left[i] and right[i]:
ans = min(ans, len(nums) - left[i] - right[i] - 1)
return ans | minimum-number-of-removals-to-make-mountain-array | [Python3] linearithmic solution | ye15 | 0 | 93 | minimum number of removals to make mountain array | 1,671 | 0.425 | Hard | 24,143 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2675823/Python-or-1-liner-simple-solution | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max([sum(acc) for acc in accounts]) | richest-customer-wealth | Python | 1-liner simple solution | LordVader1 | 36 | 2,200 | richest customer wealth | 1,672 | 0.882 | Easy | 24,144 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1077433/faster-than-98.4-submissions....python-basics | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
res=[]
for i in accounts:
res.append(sum(i))
return max(res)
```easy solution using python basics | richest-customer-wealth | faster than 98.4% submissions....python basics | yashwanthreddz | 5 | 795 | richest customer wealth | 1,672 | 0.882 | Easy | 24,145 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2231650/Beginner-Friendly-Solution-oror-Fast!!-oror-Python | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
# Use the variable below to remember the highest sum.
wealth = 0
for customer in accounts:
# The sum() function below can be replaced with another for-loop in order to traverse the nested lists.
# In this case, however, the function approach makes it simpler.
s = sum(customer)
if wealth < s:
wealth = s
return wealth | richest-customer-wealth | Beginner Friendly Solution || Fast!! || Python | cool-huip | 4 | 114 | richest customer wealth | 1,672 | 0.882 | Easy | 24,146 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2071570/Python-Easy-Solution-oror-One-Liner | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max([sum(i) for i in accounts]) | richest-customer-wealth | Python Easy Solution || One Liner | Shivam_Raj_Sharma | 4 | 138 | richest customer wealth | 1,672 | 0.882 | Easy | 24,147 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2071570/Python-Easy-Solution-oror-One-Liner | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
arr = []
for i in range(len(accounts)):
wealth = 0
for j in accounts[i]:
wealth += j
arr.append(wealth)
return max(arr) | richest-customer-wealth | Python Easy Solution || One Liner | Shivam_Raj_Sharma | 4 | 138 | richest customer wealth | 1,672 | 0.882 | Easy | 24,148 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1655228/Python-Easy-Solution-or-One-liner | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max(map(sum, accounts)) | richest-customer-wealth | Python Easy Solution | One-liner ✔ | leet_satyam | 4 | 179 | richest customer wealth | 1,672 | 0.882 | Easy | 24,149 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1733401/Python-3-(40ms)-or-2-Pythonic-Solutions-or-One-Liner-Easy-to-Understand | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
m=0
for i in accounts:
m = max(m,sum(i))
return m | richest-customer-wealth | Python 3 (40ms) | 2 Pythonic Solutions | One-Liner Easy to Understand | MrShobhit | 3 | 161 | richest customer wealth | 1,672 | 0.882 | Easy | 24,150 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1733401/Python-3-(40ms)-or-2-Pythonic-Solutions-or-One-Liner-Easy-to-Understand | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max(map(sum,accounts)) | richest-customer-wealth | Python 3 (40ms) | 2 Pythonic Solutions | One-Liner Easy to Understand | MrShobhit | 3 | 161 | richest customer wealth | 1,672 | 0.882 | Easy | 24,151 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1697292/Python-code%3A-Richest-Customer-Wealth | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
arr=[]
for i in accounts:
arr.append(sum(i))
return max(arr) | richest-customer-wealth | Python code: Richest Customer Wealth | Anilchouhan181 | 3 | 160 | richest customer wealth | 1,672 | 0.882 | Easy | 24,152 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1154494/Python-using-sum-and-max | class Solution(object):
def maximumWealth(self, accounts):
"""
:type accounts: List[List[int]]
:rtype: int
"""
max_wealth = 0
for account in accounts:
wealth = sum(account)
max_wealth = max(wealth, max_wealth)
return max_wealth | richest-customer-wealth | Python using sum and max | 5tigerjelly | 3 | 237 | richest customer wealth | 1,672 | 0.882 | Easy | 24,153 |
https://leetcode.com/problems/richest-customer-wealth/discuss/952803/Python3-1-line | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max(map(sum, accounts)) | richest-customer-wealth | [Python3] 1-line | ye15 | 3 | 121 | richest customer wealth | 1,672 | 0.882 | Easy | 24,154 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2361059/one-line-or-Success-Details-faster-than-100.00-or-Python3 | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max([sum(i) for i in accounts]) | richest-customer-wealth | one line | Success Details faster than 100.00% | Python3 | vimla_kushwaha | 2 | 94 | richest customer wealth | 1,672 | 0.882 | Easy | 24,155 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2234469/Python3-simple-solution | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
result = 0
for i in range(len(accounts)):
result = max(result, sum(accounts[i]))
return result | richest-customer-wealth | 📌 Python3 simple solution | Dark_wolf_jss | 2 | 69 | richest customer wealth | 1,672 | 0.882 | Easy | 24,156 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2145674/Python-Easy-and-Simple-Solution | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
maxwealth=0
for i in range(len(accounts)):
maxwealth=max(sum(accounts[i]), maxwealth)
return maxwealth | richest-customer-wealth | Python Easy and Simple Solution | pruthashouche | 2 | 67 | richest customer wealth | 1,672 | 0.882 | Easy | 24,157 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1312458/WEEB-DOES-PYTHON | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
cur_max = 0
for money in accounts:
if sum(money) > cur_max:
cur_max = sum(money)
return cur_max | richest-customer-wealth | WEEB DOES PYTHON | Skywalker5423 | 2 | 118 | richest customer wealth | 1,672 | 0.882 | Easy | 24,158 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1151108/Python3-Simple-Solution | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
ans = 0
for i in accounts:
ans = max(ans , sum(i))
return ans | richest-customer-wealth | [Python3] Simple Solution | Lolopola | 2 | 78 | richest customer wealth | 1,672 | 0.882 | Easy | 24,159 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2348319/Riches-man's-wealth-Python-simple-one | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
# Initialize the maximum wealth seen so far to 0 (the minimum wealth possible)
max_wealth_so_far = 0
# Iterate over accounts
for account in accounts:
# Add the money in each bank
curr_customer_wealth = sum(account)
# Update the maximum wealth seen so far if the current wealth is greater
# If it is less than the current sum
max_wealth_so_far = max(max_wealth_so_far, curr_customer_wealth)
# Return the maximum wealth
return max_wealth_so_far | richest-customer-wealth | Riches man's wealth Python simple one | RohanRob | 1 | 96 | richest customer wealth | 1,672 | 0.882 | Easy | 24,160 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2328819/Python3-or-Easy-Solution-or-3-line-soution | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
res = 0
temp = list(map(lambda x: sum(x), accounts))
return max(temp) | richest-customer-wealth | Python3 | Easy Solution | 3 line soution | irapandey | 1 | 86 | richest customer wealth | 1,672 | 0.882 | Easy | 24,161 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2250216/Python3-Runtime%3A-60ms-oror-Memory%3A-13.9mb-32.16 | class Solution:
# O(rc) || O(n)
# Runtime: 60ms || Memory: 13.9mb 32.16%
def maximumWealth(self, accounts: List[List[int]]) -> int:
maxWealth = float('-inf')
for row in range(len(accounts)):
runningSum = 0
for col in range(len(accounts[row])):
runningSum += accounts[row][col]
maxWealth = max(maxWealth, runningSum)
return maxWealth | richest-customer-wealth | Python3 Runtime: 60ms || Memory: 13.9mb 32.16% | arshergon | 1 | 30 | richest customer wealth | 1,672 | 0.882 | Easy | 24,162 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1972374/maximum-sum-easy-python3-solution | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
c=0 #Intialising a counter variable for comparison
for i in range(len(accounts)): #looping through the entire list of lists
temp_list = accounts[i] #storing the list
sum=0
for j in range(len(temp_list)): #looping through the particular list
sum=sum+temp_list[j] #adding all the elements in the list
if sum>c: #for getting the maximum value
c=sum
sum=0
return c | richest-customer-wealth | maximum sum easy python3 solution | vishwahiren16 | 1 | 72 | richest customer wealth | 1,672 | 0.882 | Easy | 24,163 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1954990/Richest-customer-wealth | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
max_num = max(sum(i) for i in accounts)
return max_num | richest-customer-wealth | Richest customer wealth | vidamaleki | 1 | 29 | richest customer wealth | 1,672 | 0.882 | Easy | 24,164 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1936445/Python-3-One-Liner-FASTER-THAN-988 | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max(map(sum, accounts)) | richest-customer-wealth | [Python 3] One-Liner, FASTER THAN 98,8% | tender777 | 1 | 50 | richest customer wealth | 1,672 | 0.882 | Easy | 24,165 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1733409/Python-Easy-To-Understand-Approach-With-Explanation | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
maxWealth = 0
for money in accounts:
# Finding the max value between our current max and previous max
maxWealth = max(sum(money), maxWealth)
return maxWealth | richest-customer-wealth | Python Easy To Understand Approach With Explanation | pniraj657 | 1 | 41 | richest customer wealth | 1,672 | 0.882 | Easy | 24,166 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1717329/1672-richest-customer-wealth-or-greater-97 | class Solution(object):
def maximumWealth(self, accounts):
"""
:type accounts: List[List[int]]
:rtype: int
"""
max_money = 0
for sbi in accounts:
money = sum(sbi)
if money > max_money:
max_money = money
return max_money | richest-customer-wealth | 1672 - richest customer wealth | > 97% | ankit61d | 1 | 51 | richest customer wealth | 1,672 | 0.882 | Easy | 24,167 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1715043/1672.-Richest-Customer-Wealth-easy-python-solution | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
lis=[]
for x in accounts:
lis.append(sum(x))
return max(lis) | richest-customer-wealth | 1672. Richest Customer Wealth easy python solution | apoorv11jain | 1 | 76 | richest customer wealth | 1,672 | 0.882 | Easy | 24,168 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1450079/One-line-simple-solution-faster-than-98.42 | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max([sum(acc) for acc in accounts ]) | richest-customer-wealth | One line simple solution faster than 98.42% | 1_d99 | 1 | 232 | richest customer wealth | 1,672 | 0.882 | Easy | 24,169 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1374464/Python3-One-Liner-or-Using-map-max-sum | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max(map(sum, accounts)) | richest-customer-wealth | [Python3] One Liner | Using map, max, sum | say4n | 1 | 55 | richest customer wealth | 1,672 | 0.882 | Easy | 24,170 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1172860/Simple-Python-Solutions%3A-Multiple-Approaches_O(n*m) | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
max_wealth = 0
for c in accounts:
c_wealth = sum(c)
if c_wealth > max_wealth:
max_wealth = c_wealth
return max_wealth | richest-customer-wealth | Simple Python Solutions: Multiple Approaches_O(n*m) | smaranjitghose | 1 | 96 | richest customer wealth | 1,672 | 0.882 | Easy | 24,171 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1172860/Simple-Python-Solutions%3A-Multiple-Approaches_O(n*m) | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max(map(sum,accounts)) | richest-customer-wealth | Simple Python Solutions: Multiple Approaches_O(n*m) | smaranjitghose | 1 | 96 | richest customer wealth | 1,672 | 0.882 | Easy | 24,172 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1157010/Python3-one-liner-easy-and-understandable | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max([sum(i) for i in accounts]) | richest-customer-wealth | Python3 one-liner easy and understandable | adarsh__kn | 1 | 89 | richest customer wealth | 1,672 | 0.882 | Easy | 24,173 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1142266/Python-99.77-faster | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
newX = 0
for x in accounts:
if newX < sum(x):
newX = sum(x)
return newX | richest-customer-wealth | Python 99.77% faster | Valarane | 1 | 101 | richest customer wealth | 1,672 | 0.882 | Easy | 24,174 |
https://leetcode.com/problems/richest-customer-wealth/discuss/1137766/Python3-32ms-(with-comments) | class Solution(object):
def maximumWealth(self, accounts):
"""
:type accounts: List[List[int]]
:rtype: int
"""
#declaring the sum of 1st list in the accounts as the biggest value
result = sum(accounts[0])
#looping through each list in the accounts list
for account in accounts:
#finding the sum of each list
maximum = sum(account)
#checking if this list is the largest ever encountered
if maximum > result:
#If yes, then declaring it as the largest in the result variable
result = maximum
return result | richest-customer-wealth | Python3 32ms (with comments) | Akshar-code | 1 | 112 | richest customer wealth | 1,672 | 0.882 | Easy | 24,175 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2838948/Python-Fast-and-Simple-Solution | class Solution:
def maximumWealth(self, accounts):
richest = 0
for i in range(len(accounts)):
localTotal = 0
for j in range((len(accounts[i]))):
localTotal += accounts[i][j]
if localTotal > richest:
richest = localTotal
return richest | richest-customer-wealth | Python - Fast & Simple Solution | BladeStew | 0 | 1 | richest customer wealth | 1,672 | 0.882 | Easy | 24,176 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2838422/Solution-(Beginner) | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
y =[]
for x in accounts:
t = sum(x)
y.append(t)
return max(y) | richest-customer-wealth | Solution (Beginner) | arob3303 | 0 | 1 | richest customer wealth | 1,672 | 0.882 | Easy | 24,177 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2828803/Python-3-lines-oror-Faster-than-97.95-and-Memory-Beats-98.38!! | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
m = 0
for a in accounts: m = max(sum(a), m)
return m | richest-customer-wealth | Python 3-lines ✅✅✅ || Faster than 97.95% and Memory Beats 98.38%!! | qiy2019 | 0 | 3 | richest customer wealth | 1,672 | 0.882 | Easy | 24,178 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2826236/Fast-and-Simple-Python-Solution | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
reference = 0
for i in accounts:
wealth = 0
for j in i:
wealth += j
if wealth > reference:
reference = wealth
return reference | richest-customer-wealth | Fast and Simple Python Solution | PranavBhatt | 0 | 2 | richest customer wealth | 1,672 | 0.882 | Easy | 24,179 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2821549/One-Liner-approach | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max(sum(i) for i in accounts) | richest-customer-wealth | One Liner approach | prathambhatiapersonal | 0 | 3 | richest customer wealth | 1,672 | 0.882 | Easy | 24,180 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2818359/The-simplest | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
result = []
for i in range(len(accounts)):
result.append(sum(accounts[i]))
return max(result) | richest-customer-wealth | The simplest | pkozhem | 0 | 2 | richest customer wealth | 1,672 | 0.882 | Easy | 24,181 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2810584/Easily-steps-Python3 | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
wealth = 0
for i in range(len(accounts)):
wealth = max(sum(accounts[i]), wealth)
return wealth | richest-customer-wealth | Easily steps Python3 | Ar_elazazi007 | 0 | 3 | richest customer wealth | 1,672 | 0.882 | Easy | 24,182 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2805396/Pure-Algorithmic-Solution | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
max = 0 # initialize the max value
for i in range(len(accounts)): # loop through all of the rows
sum = 0 # define and set a local running sum
for j in range(len(accounts[i])): # loop through all of the columns
sum += accounts[i][j] # add up the values in the columns
if sum > max: # check to see if the total of this column is the largest
max = sum # if it is, set it to the max.
return max | richest-customer-wealth | Pure Algorithmic Solution | cyber-patriot517 | 0 | 3 | richest customer wealth | 1,672 | 0.882 | Easy | 24,183 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2803202/Simple-and-easy-way | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
a=[]
for i in accounts:
a.append(sum(i))
return max(a) | richest-customer-wealth | Simple and easy way | chetan21 | 0 | 1 | richest customer wealth | 1,672 | 0.882 | Easy | 24,184 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2799357/simple-solution-with-python-list-functions | class Solution:
def maximumWealth(self, accounts: List[List[int]]):
listout = []
for li in accounts:
listout.append(sum(li))
return max(listout) | richest-customer-wealth | simple solution with python list functions | peralaarun91 | 0 | 2 | richest customer wealth | 1,672 | 0.882 | Easy | 24,185 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2799356/simple-solution-with-python-list-functions | class Solution:
def maximumWealth(self, accounts: List[List[int]]):
listout = []
for li in accounts:
listout.append(sum(li))
return max(listout) | richest-customer-wealth | simple solution with python list functions | peralaarun91 | 0 | 1 | richest customer wealth | 1,672 | 0.882 | Easy | 24,186 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2788547/Python-easy-solution | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
richest = 0
for li in accounts:
totalAmount = 0
for num in li:
totalAmount += num
if totalAmount > richest:
richest = totalAmount
return richest | richest-customer-wealth | Python easy solution | kamman626 | 0 | 1 | richest customer wealth | 1,672 | 0.882 | Easy | 24,187 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2787219/python3-%3A-T-O(n)-S-O(1) | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
richest=0
print(len(accounts))
for i in accounts:
if sum(i)>richest:
richest=sum(i)
return richest | richest-customer-wealth | python3 : T-O(n) , S-O(1) | fancyuserid | 0 | 2 | richest customer wealth | 1,672 | 0.882 | Easy | 24,188 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2786424/Easy-with-Python-for-the-Richest-Customer-Wealth-Task | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max(sum(customer_account) for customer_account in accounts) | richest-customer-wealth | Easy with Python for the Richest Customer Wealth Task | Bassel_Alf | 0 | 2 | richest customer wealth | 1,672 | 0.882 | Easy | 24,189 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2783447/Simple-Python-Solution-oror-Easy-and-Explained | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
max1 = 0
sum1 = 0
for i in accounts:
sum1 = sum(i)
if sum1>max1:
max1 = sum1
return max1 | richest-customer-wealth | Simple Python Solution || Easy & Explained | dnvavinash | 0 | 1 | richest customer wealth | 1,672 | 0.882 | Easy | 24,190 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2781059/Python3-or-Faster-than-97-or-Simple-w-less15-lines | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
most = 0
for e in accounts:
current = 0
current += sum(e)
if current > most:
most = current
else:
continue
return most | richest-customer-wealth | [Python3] | Faster than 97% | Simple w/ <15 lines | xzvg | 0 | 4 | richest customer wealth | 1,672 | 0.882 | Easy | 24,191 |
https://leetcode.com/problems/richest-customer-wealth/discuss/2772551/Python-Brute-Force-simple-solution-for-beginners | class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
maxwealth=0
for i in range(len(accounts)):
maxwealth=max(maxwealth,sum(accounts[i]))
return maxwealth | richest-customer-wealth | Python Brute Force simple solution for beginners | mritunjayyy | 0 | 2 | richest customer wealth | 1,672 | 0.882 | Easy | 24,192 |
https://leetcode.com/problems/find-the-most-competitive-subsequence/discuss/953711/Python3-greedy-O(N) | class Solution:
def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
stack = [] # (increasing) mono-stack
for i, x in enumerate(nums):
while stack and stack[-1] > x and len(stack) + len(nums) - i > k: stack.pop()
if len(stack) < k: stack.append(x)
return stack | find-the-most-competitive-subsequence | [Python3] greedy O(N) | ye15 | 6 | 342 | find the most competitive subsequence | 1,673 | 0.493 | Medium | 24,193 |
https://leetcode.com/problems/find-the-most-competitive-subsequence/discuss/953711/Python3-greedy-O(N) | class Solution:
def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
ans, pq = [], []
prev = -inf
for i, x in enumerate(nums):
heappush(pq, (x, i))
if i+k >= len(nums):
while pq and pq[0][1] < prev: heappop(pq)
x, prev = heappop(pq)
ans.append(x)
return ans | find-the-most-competitive-subsequence | [Python3] greedy O(N) | ye15 | 6 | 342 | find the most competitive subsequence | 1,673 | 0.493 | Medium | 24,194 |
https://leetcode.com/problems/find-the-most-competitive-subsequence/discuss/1027570/O(n)-time-and-O(n)-space-python3-solution-accepted-(STACK) | class Solution:
def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
i=0
n=len(nums)
res_len=0
stack=[]
while(i<n):
while(stack and (n-i)>k-res_len and stack[-1]>nums[i]):
stack.pop()
res_len-=1
else:
if(res_len<k):
stack.append(nums[i])
res_len+=1
i+=1
return stack | find-the-most-competitive-subsequence | O(n) time and O(n) space python3 solution accepted (STACK) | _Rehan12 | 2 | 54 | find the most competitive subsequence | 1,673 | 0.493 | Medium | 24,195 |
https://leetcode.com/problems/find-the-most-competitive-subsequence/discuss/2790817/Monotonic-Stack-Solution | class Solution:
def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
stack,N=[],len(nums)
for i,x in enumerate(nums):
while(stack and stack[-1] >x and len(stack) + N-i >k ): ##put some condition on i also so that relevant length subs is formed
stack.pop()
stack.append(x)
return stack[:k] | find-the-most-competitive-subsequence | Monotonic Stack Solution | shadowfax_1024 | 0 | 3 | find the most competitive subsequence | 1,673 | 0.493 | Medium | 24,196 |
https://leetcode.com/problems/find-the-most-competitive-subsequence/discuss/2333991/Python-monotonic-stack.-O(N)O(k) | class Solution:
def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
stack = []
for i, n in enumerate(nums):
while stack and stack[-1] > n and len(stack) + len(nums) - i > k:
stack.pop()
if len(stack) < k:
stack.append(n)
return stack | find-the-most-competitive-subsequence | Python, monotonic stack. O(N)/O(k) | blue_sky5 | 0 | 33 | find the most competitive subsequence | 1,673 | 0.493 | Medium | 24,197 |
https://leetcode.com/problems/find-the-most-competitive-subsequence/discuss/1736203/Python3-or-Monotonic-Stack | class Solution:
def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
n=len(nums)
stack=[]
top=0
for i in range(n):
diff=n-i
while stack and top+diff>k and stack[top-1]>nums[i]:
stack.pop()
top-=1
if top<k:
stack.append(nums[i])
top+=1
return stack | find-the-most-competitive-subsequence | [Python3] | Monotonic Stack | swapnilsingh421 | 0 | 34 | find the most competitive subsequence | 1,673 | 0.493 | Medium | 24,198 |
https://leetcode.com/problems/find-the-most-competitive-subsequence/discuss/953091/Python3-Solution-using-stack | class Solution:
def __init__(self):
self.stack = []
self.toRemove = 0
def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
self.stack.append(nums[0])
self.toRemove = len(nums) - k
for i in range(1,len(nums)):
while(self.stack and self.toRemove>0 and self.stack[-1]>nums[i]):
self.toRemove -= 1
self.stack.pop()
self.stack.append(nums[i])
while(self.toRemove>0 or len(self.stack)>k):
self.toRemove -= 1
self.stack.pop()
return self.stack | find-the-most-competitive-subsequence | Python3 Solution using stack | swap2001 | 0 | 53 | find the most competitive subsequence | 1,673 | 0.493 | Medium | 24,199 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.