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https://leetcode.com/problems/minimum-total-distance-traveled/discuss/2783378/O(n3)-solution-sort-%2B-shift-the-minimum-subarray-of-assignments | class Solution:
def minimumTotalDistance(self, robot: List[int], factory: List[List[int]]) -> int:
robot.sort()
factory.sort()
cap = []
for x, limit in factory:
cap.extend([x] * limit)
m = len(robot)
n = len(cap)
indices = list(range(m))
ans = sum(abs(x - y) for x, y in zip(robot, cap))
def increment(i):
diff = 0
while i < m:
if indices[i] + 1 < n:
diff -= abs(robot[i] - cap[indices[i]])
diff += abs(robot[i] - cap[indices[i] + 1])
else:
return math.inf, i + 1
if i + 1 < m and indices[i] + 1 == indices[i + 1]:
i += 1
else:
return diff, i + 1
for i in reversed(range(m)):
while True:
diff, j = increment(i)
if diff <= 0:
ans += diff
for x in range(i, j):
indices[x] += 1
else:
break
return ans | minimum-total-distance-traveled | O(n^3) solution, sort + shift the minimum subarray of assignments | chuan-chih | 1 | 50 | minimum total distance traveled | 2,463 | 0.397 | Hard | 33,800 |
https://leetcode.com/problems/minimum-total-distance-traveled/discuss/2787810/My-Python3-solution | class Solution:
def minimumTotalDistance(self, A: List[int], B: List[List[int]]) -> int:
n, m = len(A), len(B)
dp = [inf] * (n + 1)
dp[n] = 0
A.sort()
B.sort()
for j in range(m-1,-1,-1):
for i in range(n):
cur = 0
for k in range(1, min(B[j][1], n - i) + 1):
cur += abs(A[i + k - 1] - B[j][0])
dp[i] = min(dp[i], dp[i + k] + cur)
return dp[0] | minimum-total-distance-traveled | My Python3 solution | Chiki1601 | 0 | 8 | minimum total distance traveled | 2,463 | 0.397 | Hard | 33,801 |
https://leetcode.com/problems/minimum-total-distance-traveled/discuss/2783549/python3-O(N-3)-dp-solution | class Solution:
def minimumTotalDistance(self, robot: List[int], factory: List[List[int]]) -> int:
robot.sort()
factory.sort()
m, n = len(robot), len(factory)
@lru_cache(None)
def dp(i, j, k):
if i >= m: # all robots fixed
return 0
if j >= n: # not all robots get fixed but run out of factories.
return math.inf
if k <= 0: # the factory uses up repair limit
if j + 1 < n: # use next factory if it's valid
return dp(i, j + 1, factory[j + 1][1])
else: # no more factory to use, return inf
return math.inf
dist = abs(robot[i] - factory[j][0]) # cost to use current factory
# Use current factory or skip it
return min(dist + dp(i + 1, j, k - 1), dp(i, j, -1))
return dp(0, 0, factory[0][1]) | minimum-total-distance-traveled | [python3] O(N ^ 3) dp solution | caijun | 0 | 20 | minimum total distance traveled | 2,463 | 0.397 | Hard | 33,802 |
https://leetcode.com/problems/minimum-total-distance-traveled/discuss/2783490/Python-DP-Solution-with-explanation | class Solution:
def minimumTotalDistance(self, robot: List[int], factory: List[List[int]]) -> int:
m = len(robot)
n = len(factory)
factory.sort()
robot.sort()
opt = [[float('inf') for j in range(n + 1)] for i in range(m + 1)]
for j in range(n + 1):
opt[0][j] = 0
for i in range(1, m + 1):
for j in range(1, n + 1):
cur = 0
for x in range(min(factory[j - 1][1], i) + 1):
opt[i][j] = min(opt[i][j], opt[i-x][j-1] + cur)
cur += abs(factory[j-1][0] - robot[i-1-x])
return opt[-1][-1] | minimum-total-distance-traveled | [Python] DP Solution with explanation | codenoob3 | 0 | 22 | minimum total distance traveled | 2,463 | 0.397 | Hard | 33,803 |
https://leetcode.com/problems/minimum-total-distance-traveled/discuss/2783254/Memorized-DFS | class Solution:
def minimumTotalDistance(self, robot: List[int], factory: List[List[int]]) -> int:
@cache
def dfs(i, j, k):
if j == 0: #no robot available
return 0
if i == 0: #no factory available
return float('inf')
if k == 0: #the ith factory has 0 spot to use
if i == 1: #if there'ls only one factory left
return float('inf')
return dfs(i - 1, j, factory[i - 2][1]) #since the ith factory has no spot, we have to not use it
result1 = dfs(i - 1, j, factory[i - 2][1]) if i >= 2 else float('inf') #condition 1: don't use the ith factory
result2 = dfs(i, j - 1, k - 1) + abs(robot[j - 1] - factory[i - 1][0])#condition 2: use the ith factory
result = min(result1, result2)
return result
m, n = len(robot), len(factory)
robot.sort()
factory.sort()
result = dfs(n, m, factory[n - 1][1])
return result | minimum-total-distance-traveled | Memorized DFS | zhanghaotian19 | 0 | 30 | minimum total distance traveled | 2,463 | 0.397 | Hard | 33,804 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2817811/Easy-Python-Solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
av=[]
nums.sort()
while nums:
av.append((nums[-1]+nums[0])/2)
nums.pop(-1)
nums.pop(0)
return len(set(av)) | number-of-distinct-averages | Easy Python Solution | Vistrit | 2 | 75 | number of distinct averages | 2,465 | 0.588 | Easy | 33,805 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2831078/Python3-set | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
seen = set()
for i in range(len(nums)//2):
seen.add((nums[i] + nums[~i])/2)
return len(seen) | number-of-distinct-averages | [Python3] set | ye15 | 1 | 4 | number of distinct averages | 2,465 | 0.588 | Easy | 33,806 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2818793/Python-oror-Easy-oror-Sorting-oror-O(nlogn)-Solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
n=len(nums)
i=0
j=n-1
s=set()
nums.sort()
while i<=j:
s.add((nums[i]+nums[j])/2)
i+=1
j-=1
return len(s) | number-of-distinct-averages | Python || Easy || Sorting || O(nlogn) Solution | DareDevil_007 | 1 | 13 | number of distinct averages | 2,465 | 0.588 | Easy | 33,807 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2808058/Python-Simple-Python-Solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
a=[]
for i in range(len(nums)//2):
a.append((max(nums)+min(nums))/2)
nums.remove(max(nums))
nums.remove(min(nums))
b=set(a)
print(a)
print(b)
return len(b) | number-of-distinct-averages | ✅✅[ Python ] 🐍🐍 Simple Python Solution ✅✅ | sourav638 | 1 | 26 | number of distinct averages | 2,465 | 0.588 | Easy | 33,808 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807918/Python-Answer-Heap-and-Set-Solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
h1 = nums.copy()
h2 = [-i for i in nums.copy()]
ans = set()
heapify(h1)
heapify(h2)
while h1 and h2:
n1 = heappop(h1)
n2= - heappop(h2)
ans.add(mean([n2,n1]))
return len(ans) | number-of-distinct-averages | [Python Answer🤫🐍🐍🐍] Heap and Set Solution | xmky | 1 | 13 | number of distinct averages | 2,465 | 0.588 | Easy | 33,809 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807189/easy-python-solutionoror-easy-to-understand | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
lis = []
while nums!=[]:
if (min(nums)+max(nums))/2 not in lis:
lis.append((min(nums)+max(nums))/2)
nums.remove(max(nums))
nums.remove(min(nums))
return len(lis) | number-of-distinct-averages | ✅✅easy python solution|| easy to understand | chessman_1 | 1 | 30 | number of distinct averages | 2,465 | 0.588 | Easy | 33,810 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2849863/90-speed-python-solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
ans = set()
while nums:
ans.add((nums.pop(nums.index(min(nums))) + nums.pop(nums.index(max(nums))))/2)
return len(ans) | number-of-distinct-averages | 90% speed python solution | StikS32 | 0 | 1 | number of distinct averages | 2,465 | 0.588 | Easy | 33,811 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2849228/easy-python-sort.-Detailed-explanation. | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
avg_list=[]
while nums:
_len = len(nums)
temp = [nums[0],nums[-1]]
nums = nums[1:_len-1]
avg_list.append(sum(temp)/2)
return len(set(avg_list)) | number-of-distinct-averages | easy python - sort. Detailed explanation. | ATHBuys | 0 | 1 | number of distinct averages | 2,465 | 0.588 | Easy | 33,812 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2843315/Python-Two-Pointers-Solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
print(nums)
left, right = 0, len(nums) - 1
visited = set()
while left < right:
avg = (nums[left] + nums[right]) / 2
if avg not in visited:
visited.add(avg)
left += 1
right -= 1
return len(visited) | number-of-distinct-averages | Python Two Pointers Solution | Vayne1994 | 0 | 1 | number of distinct averages | 2,465 | 0.588 | Easy | 33,813 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2839775/Python-Solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
s=set()
l=nums
n=len(l)
while(n>0):
s.add((max(l)+min(l))/2)
l.remove(max(l))
l.remove(min(l))
n=len(l)
return len(s) | number-of-distinct-averages | Python Solution | CEOSRICHARAN | 0 | 5 | number of distinct averages | 2,465 | 0.588 | Easy | 33,814 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2839542/Python3-Unreadable-One-Liner-with-Sorting-1-Line | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
return len({(num+nums[idx])/2 for idx, num in enumerate(reversed(nums[len(nums)//2:]))}) if nums.sort() is None else 0 | number-of-distinct-averages | [Python3] - Unreadable One-Liner with Sorting - 1 Line | Lucew | 0 | 3 | number of distinct averages | 2,465 | 0.588 | Easy | 33,815 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2834349/Python-(Simple-Maths) | class Solution:
def dfs(self, arr):
n = len(arr)
n1, n2 = min(arr), max(arr)
arr.remove(n1)
arr.remove(n2)
return (n1+n2)/2
def distinctAverages(self, nums):
n, ans = len(nums), set()
while n:
ans.add(self.dfs(nums))
n -= 2
return len(ans) | number-of-distinct-averages | Python (Simple Maths) | rnotappl | 0 | 5 | number of distinct averages | 2,465 | 0.588 | Easy | 33,816 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2831884/Python-Solution-Clean-Code-oror-set-min-max-oror-contest-question | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
n=len(nums)
l=[]
while n!=0:
mi=min(nums)
ma=max(nums)
l.append((mi+ma)/2)
if mi==ma:
nums.remove(mi)
else:
nums.remove(mi)
nums.remove(ma)
n=len(nums)
return len(set(l)) | number-of-distinct-averages | Python Solution - Clean Code || set , min , max || contest question | T1n1_B0x1 | 0 | 4 | number of distinct averages | 2,465 | 0.588 | Easy | 33,817 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2824119/Easy-understanding-beginner-friendly | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
d = set()
left = 0
right = len(nums)-1
nums = sorted(nums)
while left < right:
num = (nums[left] + nums[right])/2
d.add(num)
left += 1
right -= 1
print(num)
#print(d)
return len(d)
#count = 0
#for i in d:
#count += 1
#return count | number-of-distinct-averages | Easy understanding - beginner friendly | Asselina94 | 0 | 2 | number of distinct averages | 2,465 | 0.588 | Easy | 33,818 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2816758/Python-code | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
reslist=[]
zero_list = [x for x in nums if x==0]
if zero_list == nums:
return 1
while nums:
print("Nums:",nums)
if(nums[0] and nums[-1]):
avg = (nums[0]+nums[-1])/2
del nums[0]
del nums[-1]
reslist.append(avg)
elif nums[0]:
avg = nums[0]/2
del nums[0]
del nums[-1]
reslist.append(avg)
elif nums[-1]:
avg = nums[-1]/2
del nums[-1]
del nums[0]
reslist.append(avg)
else:
break
reslist=set(reslist)
print(reslist)
return (len(reslist)) | number-of-distinct-averages | Python code | user1079z | 0 | 4 | number of distinct averages | 2,465 | 0.588 | Easy | 33,819 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2816420/Number-of-Distinct-Averages(Solution) | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
x,l = [],len(nums)
for i in range(l//2) : x.append((nums[i]+nums[l-i-1])/2)
return len(set(x)) | number-of-distinct-averages | Number of Distinct Averages(Solution) | NIKHILTEJA | 0 | 1 | number of distinct averages | 2,465 | 0.588 | Easy | 33,820 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2812981/Python-solution | class Solution(object):
def distinctAverages(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
s=set()
nums.sort()
i=0
j=len(nums)-1
while i<j:
avg=(float(nums[i])+float(nums[j]))/2
s.add(avg)
i+=1
j-=1
return len(s) | number-of-distinct-averages | Python solution | indrajitruidascse | 0 | 3 | number of distinct averages | 2,465 | 0.588 | Easy | 33,821 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2811254/Sort-and-pick-up-pairs-100-speed | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
len_nums = len(nums)
len_nums1 = len_nums - 1
return len(set(nums[i] + nums[len_nums1 - i]
for i in range(len_nums // 2))) | number-of-distinct-averages | Sort and pick up pairs, 100% speed | EvgenySH | 0 | 7 | number of distinct averages | 2,465 | 0.588 | Easy | 33,822 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2810106/Python3-easy-to-understand | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
i, j = 0, len(nums) - 1
res = set()
while i < j:
res.add((nums[i] + nums[j]) / 2)
i += 1
j -= 1
return len(res) | number-of-distinct-averages | Python3 - easy to understand | mediocre-coder | 0 | 5 | number of distinct averages | 2,465 | 0.588 | Easy | 33,823 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2810096/Python%3A-sort%2Bzip-average-is-not-important | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
return len(set([x+y for x,y in zip(nums,nums[::-1])])) | number-of-distinct-averages | Python: sort+zip - average is not important | Leox2022 | 0 | 2 | number of distinct averages | 2,465 | 0.588 | Easy | 33,824 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2809774/Two-pointers-Python3 | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
i,j=0,len(nums)-1
vi=set()
while i<j:
vi.add((nums[i]+nums[j])/2)
i+=1
j-=1
return len(vi) | number-of-distinct-averages | Two pointers Python3 ✅ | Xhubham | 0 | 5 | number of distinct averages | 2,465 | 0.588 | Easy | 33,825 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2809183/Efficient-solution-using-sorting-and-two-pointers | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
hash_dict = {}
i = 0
j = len(nums) - 1
while i < j:
min_element = nums[i]
max_element = nums[j]
avg = (min_element + max_element)/2
hash_dict[avg] = 1
i += 1
j -= 1
return len(hash_dict) | number-of-distinct-averages | Efficient solution using sorting and two-pointers | akashp2001 | 0 | 4 | number of distinct averages | 2,465 | 0.588 | Easy | 33,826 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2809172/python-easy-solution-using-sorting | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
hash_dict = {}
while len(nums) > 0:
min_element = nums[0]
max_element = nums[len(nums) - 1]
avg = (min_element + max_element)/2
hash_dict[avg] = 1
nums.pop(0)
nums.pop()
return len(hash_dict) | number-of-distinct-averages | python easy solution using sorting | akashp2001 | 0 | 1 | number of distinct averages | 2,465 | 0.588 | Easy | 33,827 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2809074/Python-Using-set-beats-100.00 | class Solution:
def distinctAverages(self, nums):
avgs = set()
while len(nums):
a, b = max(nums), min(nums)
avgs.add(a+b) # sum is sufficient, dividing by 2 not necessary
nums.remove(a)
nums.remove(b)
return len(avgs) | number-of-distinct-averages | [Python] Using set, beats 100.00% | U753L | 0 | 7 | number of distinct averages | 2,465 | 0.588 | Easy | 33,828 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2808885/Python-solution-using-Queues | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
memo = set()
nums.sort()
queue = deque(nums)
while queue:
average = (queue[-1] + queue[0])/2
if average not in memo:
memo.add(average)
queue.pop()
queue.popleft()
return len(memo) | number-of-distinct-averages | Python solution using Queues | dee7 | 0 | 4 | number of distinct averages | 2,465 | 0.588 | Easy | 33,829 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2808331/Number-of-Distinct-Averages-Python-Easy-Approach!!!! | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums=sorted(nums)
lst=[]
if(len(nums)==2):
return 1
while(len(nums)!=0):
x=nums[0]+nums[-1]
lst.append(x)
del nums[0]
del nums[-1]
s=set(lst)
return len(s) | number-of-distinct-averages | Number of Distinct Averages Python Easy Approach!!!! | knock_knock47 | 0 | 3 | number of distinct averages | 2,465 | 0.588 | Easy | 33,830 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2808294/Python-3-solution%3A | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
i=0
l=[]
while(i<len(nums)):
avg=(max(nums)+min(nums))/2
if avg not in l:
l.append(avg)
nums.pop(nums.index(max(nums)))
nums.pop(nums.index(min(nums)))
return len(l) | number-of-distinct-averages | Python 3 solution: | CharuArora_ | 0 | 3 | number of distinct averages | 2,465 | 0.588 | Easy | 33,831 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2808179/Python-sort-%2B-set | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
return len(set((nums[i] + nums[-i-1])/2 for i in range(len(nums)//2))) | number-of-distinct-averages | Python, sort + set | blue_sky5 | 0 | 6 | number of distinct averages | 2,465 | 0.588 | Easy | 33,832 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2808062/Easy-Python-Solution-explained | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
d={}
nums.sort()
for i in range(len(nums)//2):
# set.add(nums[i]+nums[len(nums)-i-1])
d[nums[i]+nums[len(nums)-i-1]]=1//checks uniqueness
return len(d) | number-of-distinct-averages | Easy Python Solution explained | yashleetcode123 | 0 | 2 | number of distinct averages | 2,465 | 0.588 | Easy | 33,833 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807762/Faster-than-100-Simple-Python3-Solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
averages = set()
while len(nums) != 0:
min_num = min(nums)
nums.remove(min_num)
max_num = max(nums)
nums.remove(max_num)
averages.add((max_num + min_num) / 2)
return len(averages) | number-of-distinct-averages | Faster than 100% Simple Python3 Solution | y-arjun-y | 0 | 5 | number of distinct averages | 2,465 | 0.588 | Easy | 33,834 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807521/Python-5-lines-use-fast-bitwise-operators | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
avgs = set()
nums.sort()
for i in range(len(nums) >> 1):
avgs.add((nums[i] + nums[~i]) / 2)
return len(avgs) | number-of-distinct-averages | Python 5 lines - use fast bitwise operators | SmittyWerbenjagermanjensen | 0 | 3 | number of distinct averages | 2,465 | 0.588 | Easy | 33,835 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807480/Python-or-JavaScript-2-Pointers-with-sort | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
result = set()
sortedNums = sorted(nums)
left = 0
right = len(sortedNums) - 1
while left < right:
result.add((sortedNums[left] + sortedNums[right]) / 2)
left += 1
right -= 1
return len(result) | number-of-distinct-averages | [Python | JavaScript] 2 Pointers with sort | ChaseDho | 0 | 13 | number of distinct averages | 2,465 | 0.588 | Easy | 33,836 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807426/pythonjava-easy-to-understand | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
seen = set()
for i in range(len(nums) // 2):
seen.add((nums[i] + nums[len(nums) - 1 - i]) / 2)
print(seen)
return len(seen)
class Solution {
public int distinctAverages(int[] nums) {
Arrays.sort(nums);
Set<Float> seen = new HashSet<>();
for (int i = 0; i < nums.length / 2; i ++) {
seen.add((float)(nums[i] + nums[nums.length - 1 - i]) / 2);
System.out.println(seen);
}
return seen.size();
}
} | number-of-distinct-averages | python/java easy to understand | akaghosting | 0 | 5 | number of distinct averages | 2,465 | 0.588 | Easy | 33,837 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807335/Python-Progress-from-O(NlogN)-to-O(N)-solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
avgs = set()
sorted_nums = collections.deque(sorted(nums))
while sorted_nums:
lo, hi = sorted_nums.popleft(), sorted_nums.pop()
avgs.add((lo+hi)/2)
return len(avgs) | number-of-distinct-averages | [Python] Progress from O(NlogN) to O(N) solution | sc1233 | 0 | 10 | number of distinct averages | 2,465 | 0.588 | Easy | 33,838 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807335/Python-Progress-from-O(NlogN)-to-O(N)-solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
freq_by_num = collections.Counter(nums)
lo, hi = min(nums), max(nums)
avgs = set()
while len(freq_by_num) > 0:
if lo not in freq_by_num:
lo += 1
continue
if hi not in freq_by_num:
hi -= 1
continue
avgs.add((lo+hi)/2)
freq_by_num[lo] -= 1
freq_by_num[hi] -= 1
if freq_by_num[lo] == 0:
del freq_by_num[lo]
if freq_by_num[hi] == 0:
del freq_by_num[hi]
return len(avgs) | number-of-distinct-averages | [Python] Progress from O(NlogN) to O(N) solution | sc1233 | 0 | 10 | number of distinct averages | 2,465 | 0.588 | Easy | 33,839 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807308/Easy-Python-Solution-oror-100 | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
avgList = []
nums.sort()
while nums:
maxNumber = nums.pop(0)
minNumber = nums.pop(-1)
avgList.append((maxNumber + minNumber) / 2)
avgList = set(avgList)
return len(avgList) | number-of-distinct-averages | Easy Python Solution || 100% | arefinsalman86 | 0 | 8 | number of distinct averages | 2,465 | 0.588 | Easy | 33,840 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807285/Easy-Solution-O(nlog(n))-and-O(1) | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
q=collections.deque(nums)
s=set()
while len(q)>0:
s.add(q.popleft()+q.pop())
return len(s) | number-of-distinct-averages | Easy Solution O(nlog(n)) and O(1) | Motaharozzaman1996 | 0 | 5 | number of distinct averages | 2,465 | 0.588 | Easy | 33,841 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807278/Python3-Two-Lines | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
return len({(nums[i]+nums[-i-1])/2 for i in range(len(nums)//2)}) | number-of-distinct-averages | Python3, Two Lines | Silvia42 | 0 | 8 | number of distinct averages | 2,465 | 0.588 | Easy | 33,842 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807278/Python3-Two-Lines | class Solution02:
def distinctAverages(self, nums: List[int]) -> int:
s=set()
nums.sort()
for i in range(len(nums)//2):
x = (nums[i]+nums[-i-1])/2
s.add(x)
return len(s) | number-of-distinct-averages | Python3, Two Lines | Silvia42 | 0 | 8 | number of distinct averages | 2,465 | 0.588 | Easy | 33,843 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807271/Python-Solution | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
avgs = []
for i in range(len(nums)//2):
avgs.append((nums[i]+nums[-i-1])/2)
return len(set(avgs)) | number-of-distinct-averages | Python Solution | saivamshi0103 | 0 | 6 | number of distinct averages | 2,465 | 0.588 | Easy | 33,844 |
https://leetcode.com/problems/number-of-distinct-averages/discuss/2807235/Simple-Python%3A-Queue-%2B-Set | class Solution:
def distinctAverages(self, nums: List[int]) -> int:
ans = set()
nums.sort()
queue = deque(nums)
while queue:
low = queue.popleft()
high = queue.pop()
ans.add((low+high)/2)
return len(ans) | number-of-distinct-averages | Simple Python: Queue + Set | pokerandcode | 0 | 15 | number of distinct averages | 2,465 | 0.588 | Easy | 33,845 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2813134/Easy-solution-from-Recursion-to-memoization-oror-3-approaches-oror-PYTHON3 | class Solution:
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
def recursive(s,ans):
if len(s)>=low and len(s)<=high:
ans+=[s]
if len(s)>high:
return
recursive(s+"0"*zero,ans)
recursive(s+"1"*one,ans)
return
ans=[]
recursive("",ans)
return len(ans) | count-ways-to-build-good-strings | Easy solution from Recursion to memoization || 3 approaches || PYTHON3 | dabbiruhaneesh | 1 | 32 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,846 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2813134/Easy-solution-from-Recursion-to-memoization-oror-3-approaches-oror-PYTHON3 | class Solution:
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
ans=[0]
def recursive(s):
if s>high:
return 0
p=recursive(s+zero)+recursive(s+one)
if s>=low and s<=high:
p+=1
return p
return recursive(0)%(10**9+7) | count-ways-to-build-good-strings | Easy solution from Recursion to memoization || 3 approaches || PYTHON3 | dabbiruhaneesh | 1 | 32 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,847 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2813134/Easy-solution-from-Recursion-to-memoization-oror-3-approaches-oror-PYTHON3 | class Solution:
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
ans=[0]
mod=10**9+7
def recursive(s):
if s>high:
return 0
if dp[s]!=-1:
return dp[s]
p=recursive(s+zero)+recursive(s+one)
if s>=low and s<=high:
p+=1
dp[s]=p%mod
return dp[s]
dp=[-1]*(high+1)
return recursive(0)%mod | count-ways-to-build-good-strings | Easy solution from Recursion to memoization || 3 approaches || PYTHON3 | dabbiruhaneesh | 1 | 32 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,848 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2812003/FASTEST-PYTHON-SOLUTION-USING-DYNAMIC-PROGRAMMING | class Solution:
def countGoodStrings(self, low: int, high: int, a: int, b: int) -> int:
ct=0
lst=[-1]*(high+1)
def sol(a,b,lst,i):
if i==0:
return 1
if i<0:
return 0
if lst[i]!=-1:
return lst[i]
lst[i]=sol(a,b,lst,i-a)+sol(a,b,lst,i-b)
return lst[i]
for i in range(low,high+1):
ct+=sol(a,b,lst,i)
return ct%(10**9+7) | count-ways-to-build-good-strings | FASTEST PYTHON SOLUTION USING DYNAMIC PROGRAMMING | beneath_ocean | 1 | 12 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,849 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2808241/Python-DP-tabulation-approach | class Solution:
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
MOD = 1_000_000_007
limit = high + 1 - min(zero, one) # last dp array index to be processed
dp_size = limit + max(zero, one) # dp array size
dp: List[int] = [0] * dp_size
dp[0] = 1
for i in range(limit):
if dp[i]:
dp[i+zero] += dp[i] % MOD
dp[i+one] += dp[i] % MOD
return sum(dp[low:high+1]) % MOD | count-ways-to-build-good-strings | [Python] DP tabulation approach | olzh06 | 1 | 19 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,850 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2829014/Python3-DP | class Solution:
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
dp = [0] * (high+1)
for i in range(high, -1, -1):
if low <= i: dp[i] = 1
if i+zero <= high: dp[i] += dp[i+zero]
if i+one <= high: dp[i] += dp[i+one]
dp[i] %= 1_000_000_007
return dp[0] | count-ways-to-build-good-strings | [Python3] DP | ye15 | 0 | 6 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,851 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2825194/Python-(Simple-DP) | class Solution:
def countGoodStrings(self, low, high, zero, one):
dp = [0]*(high+1)
dp[0] = 1
for i in range(1,max(one,zero)+1):
if i-zero >= 0 and i-one >= 0:
dp[i] = dp[i-zero] + dp[i-one]
elif i-zero >= 0:
dp[i] = dp[i-zero]
elif i-one >= 0:
dp[i] = dp[i-one]
for i in range(max(one,zero)+1,high+1):
dp[i] = dp[i-zero] + dp[i-one]
return sum([dp[i] for i in range(low,high+1)])%(10**9+7) | count-ways-to-build-good-strings | Python (Simple DP) | rnotappl | 0 | 3 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,852 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2810983/Python-or-EnumerationRecurrence-Problem | class Solution:
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
p = 10**9 + 7
dp = Counter({0:1})
for i in range(1,high+1):
dp[i] = (dp[i - zero] + dp[i - one]) % p
return reduce(lambda x, y : (x + y) % p, (dp[i] for i in range(low,high+1)), 0) | count-ways-to-build-good-strings | Python | Enumeration/Recurrence Problem | on_danse_encore_on_rit_encore | 0 | 5 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,853 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2808895/python3-dp-solution-for-reference. | class Solution:
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
dp = [0]*(high)
dp[zero-1] = 1
dp[one-1] += 1
for i in range(high):
if i-zero >= 0:
dp[i] += 1+dp[i-zero]
if i-one >= 0:
dp[i] += 1+dp[i-one]
return (dp[high-1] - (dp[low-2] if low-2>=0 else 0))%(10**9+7) | count-ways-to-build-good-strings | [python3] dp solution for reference. | vadhri_venkat | 0 | 9 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,854 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2807924/Python-Answer-DP-Solution | class Solution:
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
mod = 10**9 + 7
@cache
def calc(i):
res = 0
ro = 0
rz = 0
if i == zero:
res += 1
if i == one:
res += 1
if i > zero:
rz = calc(i-zero)%mod
if i > one:
ro = calc(i-one)%mod
return (res + rz + ro)%mod
mod = 10**9 + 7
res = 0
for i in range(low, high+1):
val = calc(i)
res = (res+ val )%mod
return res | count-ways-to-build-good-strings | [Python Answer🤫🐍🐍🐍] DP Solution | xmky | 0 | 8 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,855 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2807475/python-Good-old-DP | class Solution:
def __init__(self):
self.dp = {}
def OPT(self, i):
if i <= 0:
return 0
return self.dp[i]
def countGoodStrings(self, low, high, zero, one):
mod = 1e9 + 7
self.dp[zero] = 1
if one in self.dp:
self.dp[one] += 1
else:
self.dp[one] = 1
for i in range(1, high + 1):
if i not in self.dp:
self.dp[i] = (self.OPT(i - one) + self.OPT(i - zero)) % mod
else:
self.dp[i] += (self.OPT(i - one) + self.OPT(i - zero)) % mod
count = 0
for i in range(low, high + 1):
count += self.dp[i]
return int(count % mod) | count-ways-to-build-good-strings | [python] Good old DP | uzairfassi | 0 | 11 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,856 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2807305/DP-Python-solution | class Solution:
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
dp = [0] * (high+1)
dp[0] = 1
mod = 10**9 + 7
for i in range(1, high+1):
dp[i] = (dp[i-zero] + dp[i-one])
return sum(dp[i] for i in range(low, high+1)) % mod | count-ways-to-build-good-strings | DP Python solution | qcx60990 | 0 | 19 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,857 |
https://leetcode.com/problems/count-ways-to-build-good-strings/discuss/2807274/Simple-Python3-solution. | class Solution:
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
dp = [0]*(high+1)
dp[zero] += 1
dp[one] += 1
mod = (10**9)+7
for i in range(min(zero,one)+1,high+1):
if i-zero>=0:
dp[i]+=dp[i-zero]
if i-one>=0:
dp[i]+=dp[i-one]
s = 0
for i in range(low,high+1):
s+=dp[i]
return s%mod | count-ways-to-build-good-strings | Simple Python3 solution. | kamal0308 | 0 | 23 | count ways to build good strings | 2,466 | 0.419 | Medium | 33,858 |
https://leetcode.com/problems/most-profitable-path-in-a-tree/discuss/2838006/BFS-%2B-Graph-%2B-level-O(n) | class Solution:
def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
path_b = set([bob])
lvl_b = {bob:0}
g = defaultdict(list)
for u, v in edges:
g[u].append(v)
g[v].append(u)
n = len(amount)
node_lvl = [0] * n
q = deque([0])
lvl = 0
seen = set([0])
while q:
size = len(q)
for _ in range(size):
u = q.popleft()
node_lvl[u] = lvl
for v in g[u]:
if v in seen:
continue
q.append(v)
seen.add(v)
lvl += 1
b = bob
lvl = 1
while b != 0:
for v in g[b]:
if node_lvl[v] > node_lvl[b]:
continue
b = v
cost = amount[b]
path_b.add(b)
lvl_b[b] = lvl
break
lvl += 1
# print(f"lvl_b {lvl_b} path_b {path_b} ")
cost_a = []
q = deque([(0, amount[0])])
seen = set([0])
lvl = 1
while q:
size = len(q)
for _ in range(size):
u, pre_cost = q.popleft()
child_cnt = 0
for v in g[u]:
if v in seen:
continue
seen.add(v)
child_cnt += 1
cost = pre_cost
inc = amount[v]
if v in path_b:
if lvl_b[v] == lvl:
cost += inc//2
elif lvl_b[v] > lvl:
cost += inc
else:
cost += 0
else:
cost += amount[v]
q.append((v, cost))
if child_cnt == 0:
cost_a.append(pre_cost)
lvl += 1
ans = max(cost_a)
return ans | most-profitable-path-in-a-tree | BFS + Graph + level, O(n) | goodgoodwish | 0 | 2 | most profitable path in a tree | 2,467 | 0.464 | Medium | 33,859 |
https://leetcode.com/problems/most-profitable-path-in-a-tree/discuss/2831080/Python3-dfs | class Solution:
def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
n = 1 + len(edges)
tree = [[] for _ in range(n)]
for u, v in edges:
tree[u].append(v)
tree[v].append(u)
seen = [False] * n
def dfs(u, d):
"""Return """
seen[u] = True
ans = -inf
dd = 0 if u == bob else n
for v in tree[u]:
if not seen[v]:
x, y = dfs(v, d+1)
ans = max(ans, x)
dd = min(dd, y)
if ans == -inf: ans = 0
if d == dd: ans += amount[u]//2
if d < dd: ans += amount[u]
return ans, dd+1
return dfs(0, 0)[0] | most-profitable-path-in-a-tree | [Python3] dfs | ye15 | 0 | 3 | most profitable path in a tree | 2,467 | 0.464 | Medium | 33,860 |
https://leetcode.com/problems/most-profitable-path-in-a-tree/discuss/2814835/Python3-or-Two-DFS-Traversal-or-Explanation | class Solution:
def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
n = len(amount)
parent = [-1 for i in range(n)]
dist = [0 for i in range(n)]
graph = [[] for i in range(n)]
for u,v in edges:
graph[u].append(v)
graph[v].append(u)
def findDistance(curr,currParent):
for it in graph[curr]:
if it!=currParent:
dist[it] = dist[curr] + 1
findDistance(it,curr)
parent[it] = curr
return
def modifyCost(currBob,currDist):
while currBob!=0:
if currDist < dist[currBob]:
amount[currBob] = 0
elif currDist == dist[currBob]:
amount[currBob] = amount[currBob]//2
currBob = parent[currBob]
currDist+=1
return
def findMaxProfitPath(curr,currParent):
if len(graph[curr]) == 1 and curr > 0:
return amount[curr]
val = -math.inf
for it in graph[curr]:
if it!=currParent:
val = max(val,findMaxProfitPath(it,curr))
return val + amount[curr]
findDistance(0,-1)
modifyCost(bob,0)
return findMaxProfitPath(0,-1) | most-profitable-path-in-a-tree | [Python3] | Two DFS Traversal | Explanation | swapnilsingh421 | 0 | 14 | most profitable path in a tree | 2,467 | 0.464 | Medium | 33,861 |
https://leetcode.com/problems/most-profitable-path-in-a-tree/discuss/2812390/python | class Solution:
def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
adj = defaultdict(set)
for u, v in edges:
adj[u].add(v)
adj[v].add(u)
path = [[bob, 0]]
bobSteps = dict()
aliceMaxScore = float('-inf')
def dfsBob(parent, root, steps):
if root == 0:
for node, step in path:
bobSteps[node] = step
return
for child in adj[root]:
if child == parent:
continue
path.append([child, steps + 1])
dfsBob(root, child, steps + 1)
path.pop()
def dfsAlice(parent, root, steps, score):
nonlocal aliceMaxScore
if (root not in bobSteps) or (steps < bobSteps[root]):
score += amount[root]
elif steps > bobSteps[root]:
score += 0
elif steps == bobSteps[root]:
score += amount[root] // 2
if (len(adj[root]) == 1) and (root != 0):
aliceMaxScore = max(aliceMaxScore, score)
return
for neighbor in adj[root]:
if neighbor == parent:
continue
dfsAlice(root, neighbor, steps + 1, score)
dfsBob(-1, bob, 0)
dfsAlice(-1, 0, 0, 0)
return aliceMaxScore | most-profitable-path-in-a-tree | python | yzhao156 | 0 | 10 | most profitable path in a tree | 2,467 | 0.464 | Medium | 33,862 |
https://leetcode.com/problems/most-profitable-path-in-a-tree/discuss/2811225/Python-3-DFS-Backtracking-with-Explanation | class Solution:
def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
n = len(amount) # number of nodes
graph = [set() for _ in range(n)] # adjacency list presentation of the tree
for i,j in edges:
graph[i].add(j)
graph[j].add(i)
bobpath = dict() # use hashmap to record nodes on the path from bob to root 0
# as well as time: node = key, time = value
self.stop = False # True when Bob reaches root 0, this helps stop the DFS
visited = [False]*n # True if node is visited, initialized as False for all nodes
def backtrackbob(node,time): # the first backtracking, with time at each node
bobpath[node] = time # add node:time to the hashmap
visited[node] = True # mark node as visited
if node==0: # root 0 is reached
self.stop = True # this helps stop the DFS
return None
count = 0 # this helps determine if a node is leaf node
for nei in graph[node]:
if not visited[nei]:
count += 1
break
if count==0: # node is leaf node if all neighbors are already visited
del bobpath[node] # delete leaf node from hashmap before retreating from leaf node
return None
for nei in graph[node]: # explore unvisited neighbors of node
if self.stop: return None # if root 0 is already reached, stop the DFS
if not visited[nei]:
backtrackbob(nei,time+1)
if not self.stop: # if root 0 is reached, keep node in the hashmap when backtracking
del bobpath[node] # otherwise, delete node before retreating
return None
backtrackbob(bob,0) # execute the first backtracking, time at bob is initialized = 0
self.ans = float(-inf) # answer of the problem
self.income = 0 # income of Alice when travelling to leaf nodes, initialized = 0
visited = [False]*n # True if node is visited, initialized as False for all nodes
def backtrackalice(node,time): # second backtracking, with time at each node
visited[node] = True
if node in bobpath: # if the node Alice visits is on Bob's path, there are 3 cases
if time == bobpath[node]: # Alice and Bob reach the node at the same time
reward = amount[node]//2
elif time<bobpath[node]: # Alice reachs node before Bob
reward = amount[node]
else: # Alice reachs node after Bob
reward = 0
else: # if the node Alice visits is not on Bob's path
reward = amount[node]
self.income += reward # add the reward (price) at this node to the income
count = 0 # this helps determine if a node is leaf node
for nei in graph[node]:
if not visited[nei]:
count += 1
break
if count==0: # node is leaf node if all neighbors are already visited
self.ans = max(self.ans,self.income) # update the answer
self.income -= reward # remove the reward (price) at leaf node before backtracking
return None
for nei in graph[node]: # explore unvisited neighbors of node
if not visited[nei]:
backtrackalice(nei,time+1)
self.income -= reward # remove the reward (price) at this node before retreating
return None
backtrackalice(0,0) # execute the second backtracking, time at root 0 is initialized = 0
return self.ans | most-profitable-path-in-a-tree | Python 3, DFS Backtracking with Explanation | cuongmng | 0 | 14 | most profitable path in a tree | 2,467 | 0.464 | Medium | 33,863 |
https://leetcode.com/problems/most-profitable-path-in-a-tree/discuss/2809797/Python3-BFS%2BDFS-step-by-step | class Solution:
def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
class Node:
def __init__(self, val, gate, parent, steps_from_zero):
self.val = val
self.gate = gate
self.parent = parent
self.steps_from_zero = steps_from_zero
self.steps_from_bob = len(amount)
self.children = set()
# build a graph from input
graph = defaultdict(list)
for e in edges:
graph[e[0]].append(e[1])
graph[e[1]].append(e[0])
seen = set([0])
root = Node(val = 0,
gate = amount[0],
parent = None,
steps_from_zero = 0)
# BFS to construct the tree
queue = deque([root])
level = 1
while queue:
curr_level_length = len(queue)
for _ in range(curr_level_length):
curr_node = queue.popleft()
for i in graph[curr_node.val]:
if i not in seen:
seen.add(i)
new_node = Node(val = i,
gate = amount[i],
parent = curr_node,
steps_from_zero = level)
if i == bob:
bob_node = new_node
curr_node.children.add(new_node)
queue.append(new_node)
level += 1
# iterate bob path and fill in steps_from_bob
tmp_node = bob_node
i = 0
while tmp_node:
tmp_node.steps_from_bob = i
i += 1
tmp_node = tmp_node.parent
# DFS to find max net_income
def dfs(curr_alice, net_income):
nonlocal ans
# Bob and Alice are in the same node
if curr_alice.steps_from_zero == curr_alice.steps_from_bob:
net_income += curr_alice.gate / 2
# Alice reaches the node before Bob therefore gain/pay the gate fee
elif curr_alice.steps_from_zero < curr_alice.steps_from_bob:
net_income += curr_alice.gate
# Alice reaches leaf node
if len(curr_alice.children) == 0:
ans = max(ans, net_income)
return
for child_node in curr_alice.children:
dfs(child_node, net_income)
ans = float('-inf')
dfs(root, 0)
return int(ans) | most-profitable-path-in-a-tree | [Python3] BFS+DFS step-by-step | user1793l | 0 | 9 | most profitable path in a tree | 2,467 | 0.464 | Medium | 33,864 |
https://leetcode.com/problems/most-profitable-path-in-a-tree/discuss/2807551/Python-Easiest-DFS-Solution-100-Faster | class Solution:
def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
graph = defaultdict(list)
for i,j in edges:
graph[i].append(j)
graph[j].append(i)
parent=[-1 for _ in range(len(amount)+1)]
time = [-1 for _ in range(len(amount)+1)]
def dfs(node,par,cnt):
parent[node]=par
time[node]=cnt
for nei in graph[node]:
if nei!=par:
dfs(nei,node,cnt+1)
dfs(0,-1,0)
curr = bob
t=0
while curr!=0:
if t<time[curr]:
amount[curr]=0
elif t==time[curr]:
amount[curr]//=2
curr = parent[curr]
t+=1
best = -sys.maxsize
def dfs2(node,par,curr):
down =0
for v in graph[node]:
if v!=par:
dfs2(v,node,curr+amount[v])
down+=1
if down==0:
nonlocal best
best = max(best,curr)
dfs2(0,-1,amount[0])
return best | most-profitable-path-in-a-tree | Python Easiest DFS Solution 100% Faster | prateekgoel7248 | 0 | 35 | most profitable path in a tree | 2,467 | 0.464 | Medium | 33,865 |
https://leetcode.com/problems/most-profitable-path-in-a-tree/discuss/2807500/Easy-solution-with-python100-speed | class Solution:
def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
graph = defaultdict(list)
for a, b in edges:
graph[a].append(b)
graph[b].append(a)
arr = []
def dfs(node, path, parent = -1):
if node == 0:
arr.extend(list(path))
return
for nbr in graph[node]:
if nbr == parent: continue
path.append(nbr)
dfs(nbr, path, node)
path.pop()
dfs(bob, [bob])
for i in range(len(arr)//2):
amount[arr[i]] = 0
if len(arr)%2:
amount[arr[i+1]] //= 2
visited = set()
def dfs2(node, parent = -1):
if len(graph[node]) == 1 and graph[node][0] == parent:
return amount[node]
cur = -inf
for nbr in graph[node]:
if parent == nbr: continue
cur = max(cur, dfs2(nbr, node))
return amount[node] + cur
return dfs2(0) | most-profitable-path-in-a-tree | Easy solution with python100% speed | Samuelabatneh | 0 | 11 | most profitable path in a tree | 2,467 | 0.464 | Medium | 33,866 |
https://leetcode.com/problems/most-profitable-path-in-a-tree/discuss/2807464/Python-Solution-Explained | class Solution:
def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
graph = defaultdict(list)
for edge in edges:
graph[edge[0]].append(edge[1])
graph[edge[1]].append(edge[0])
def bob_dfs(curr_node, prev_node, count):
if curr_node == 0:
return count
if prev_node != None and len(graph[curr_node]) == 1 and curr_node != 0:
return -1
for node in graph[curr_node]:
if node == prev_node:
continue
res = bob_dfs(node, curr_node, count + 1)
if res != -1:
if count < (res + 1) // 2:
amount[curr_node] = 0
if res % 2 == 0 and res // 2 == count:
amount[curr_node] /= 2
return res
return -1
bob_dfs(bob, None, 0)
leafs_res = []
def sum_dfs(curr_node, curr_sum, prev_node):
if len(graph[curr_node]) == 1 and curr_node != 0:
leafs_res.append(curr_sum + amount[curr_node])
for node in graph[curr_node]:
if node == prev_node:
continue
sum_dfs(node, curr_sum + amount[curr_node], curr_node)
sum_dfs(0, 0, None)
res = max(leafs_res)
if res % 1 == 0:
return int(res)
else:
return res | most-profitable-path-in-a-tree | Python Solution - Explained | eden33335 | 0 | 13 | most profitable path in a tree | 2,467 | 0.464 | Medium | 33,867 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2807491/Python-Simple-dumb-O(N)-solution | class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
def splitable_within(parts_limit):
# check the message length achievable with <parts_limit> parts
length = sum(limit - len(str(i)) - len(str(parts_limit)) - 3 for i in range(1, parts_limit + 1))
return length >= len(message)
parts_limit = 9
if not splitable_within(parts_limit):
parts_limit = 99
if not splitable_within(parts_limit):
parts_limit = 999
if not splitable_within(parts_limit):
parts_limit = 9999
if not splitable_within(parts_limit):
return []
# generate the actual message parts
parts = []
m_index = 0 # message index
for part_index in range(1, parts_limit + 1):
if m_index >= len(message): break
length = limit - len(str(part_index)) - len(str(parts_limit)) - 3
parts.append(message[m_index:m_index + length])
m_index += length
return [f'{part}<{i + 1}/{len(parts)}>' for i, part in enumerate(parts)] | split-message-based-on-limit | [Python] Simple dumb O(N) solution | vudinhhung942k | 2 | 62 | split message based on limit | 2,468 | 0.483 | Hard | 33,868 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2843586/Binary-Search-solution-with-digit-count-check(It-passes-%22abbababbbaaa-aabaa-a%22-test-case-) | class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
if limit <= 5: return []
# Important: Determine number of digit of b. BS monotone holds for same digit-count
# This will fix the testcase like: message = "abbababbbaaa aabaa a", limit = 8
nd = -1
for dc in range(1, 7):
remain = len(message)
for d in range(1, dc+1):
if limit - 3 - d - dc < 0: break
remain -= (limit - 3 - d - dc)*9*10**(d-1)
if remain <= 0:
nd = dc
break
if nd == -1: return []
# binary search of b
start, end = 10**(nd - 1), 10 ** nd
while start != end:
mid = (start + end) // 2
dc = len(str(mid))
remain = len(message)
for d in range(1, dc):
remain -= (limit - 3 - d - dc)*9*10**(d-1)
for i in range(10**(dc-1), mid + 1):
remain -= limit - 3 - 2*dc
if remain < 0: break
# print(start, end, mid, remain)
if remain > 0: start = mid + 1
else: end = mid
# construct answer
ans = []
dc = len(str(start))
cur = 0
for d in range(1, dc + 1):
for i in range(10**(d-1), 10**d):
nxt = min(cur + limit - 3 - d - dc, len(message))
ans.append(message[cur:nxt] + '<' + str(i) + '/' + str(start) + '>')
cur = nxt
if nxt >= len(message): break
if cur < len(message): return []
return ans | split-message-based-on-limit | Binary Search solution with digit-count check(It passes "abbababbbaaa aabaa a" test case ) | qian48 | 0 | 4 | split message based on limit | 2,468 | 0.483 | Hard | 33,869 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2831094/Python3-linear-search | class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
prefix = b = 0
while 3 + len(str(b))*2 < limit and len(message) + prefix + (3+len(str(b)))*b > limit * b:
b += 1
prefix += len(str(b))
ans = []
if 3 + len(str(b))*2 < limit:
i = 0
for a in range(1, b+1):
step = limit - (len(str(a)) + len(str(b)) + 3)
ans.append(f"{message[i:i+step]}<{a}/{b}>")
i += step
return ans | split-message-based-on-limit | [Python3] linear search | ye15 | 0 | 9 | split message based on limit | 2,468 | 0.483 | Hard | 33,870 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2823276/Python-Binary-Search-On-range(1-9)-(10-99)... | class Solution:
def splitMessage(self, s: str, limit: int) -> List[str]:
if limit <= 4: return []
n = len(s)
ans, step = float('inf'), 1
while True:
if 2 * len(str(step)) + 3 > limit: break
lo, hi, suffix = step, min(n, step * 10 - 1), len(str(step)) + 3
while lo <= hi:
mid = (lo + hi) // 2
idx, i = 1, 0
while i < n:
i += limit - (suffix + len(str(idx)))
idx += 1
if idx - 1 > mid: break
idx -= 1
if idx <= mid:
ans = min(ans, mid)
hi = mid - 1
else: lo = mid + 1
if ans != float('inf'): break
step *= 10
if ans == float('inf'): return []
ans_split = []
idx, i, suffix = 1, 0, len(str(ans)) + 3
while i < n:
ans_split.append(s[i:i + limit - (suffix + len(str(idx)))] + '<' + str(idx) + '/' + str(ans) + '>')
i += limit - (suffix + len(str(idx)))
idx += 1
return ans_split | split-message-based-on-limit | [Python] Binary Search On range(1, 9), (10, 99)... | cava | 0 | 8 | split message based on limit | 2,468 | 0.483 | Hard | 33,871 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2818925/Faster-than-95 | class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
l=len(message)
if limit<=5:
return []
min_size=None
min_ct=0
for ct in range(1, min(((limit-4)>>1)+1, 6)):
total=0
for i in range(1, ct):
total+=(9*10**(i-1))*(limit-i-ct-3)
left=l-total
if left<0:
continue
left_size=(left-1)//(limit-2*ct-3)+1
if left_size>9*10**(ct-1):
continue
total_size=left_size+10**(ct-1)-1
if (min_size==None or total_size<min_size):
min_size=total_size
min_ct=ct
if min_size==None:
return []
else:
str_total_size=str(min_size)
ans=[]
curs=0
for i in range(min_size-1):
new_curs=curs+limit-(int(log10(i+1))+1)-min_ct-3
# print(limit, (int(log10(i+1))+1), min_ct)
ans.append(message[curs:new_curs]+'<'+str(i+1)+'/'+str_total_size+'>')
curs=new_curs
ans.append(message[curs:]+'<'+str_total_size+'/'+str_total_size+'>')
return ans | split-message-based-on-limit | Faster than 95% | mbeceanu | 0 | 4 | split message based on limit | 2,468 | 0.483 | Hard | 33,872 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2807907/Python-Answer-Greedy-Buffer-and-Special-Character | class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
def buildnum(start,total):
return f'<{start}/{total}>'
output = []
if limit-5 <= 0:
return []
else:
total = ceil(len(message) / (limit-3-2))
total = '&'* len(str(total))
buff = []
count = 1
for i,c in enumerate(message):
buff.append(c)
if len(buff) + len(buildnum(count,total)) == limit:
output.append( ''.join(buff) + buildnum(count,total) )
buff = []
count += 1
elif len(buff) + len(buildnum(count,total)) > limit :
return []
if len(buff) > 0:
output.append( ''.join(buff) + buildnum(count,total) )
count += 1
count -= 1
output = [s.replace(total,str(count)) for s in output]
return output | split-message-based-on-limit | [Python Answer🤫🐍🐍🐍] Greedy Buffer and Special Character | xmky | 0 | 11 | split message based on limit | 2,468 | 0.483 | Hard | 33,873 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2807809/Python-3Binary-Search | class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
n = len(message)
l, h = 1, n
def bs(x):
dl = 3 + len(str(x))
n = len(message)
for i in range(1, x+1):
k = limit - dl - len(str(i))
if k <= 0: return False
n -= k
if n <= 0: return True
return False
while l < h:
mid = l + (h - l) // 2
if bs(mid):
h = mid
else:
l = mid + 1
dl = 3 + len(str(l))
n = len(message)
j = 0
ans = []
for i in range(1, l+1):
k = limit - dl - len(str(i))
if k <= 0: return []
tmp = message[j:j+k] + f"<{i}/{l}>"
ans.append(tmp)
j += k
return ans | split-message-based-on-limit | [Python 3]Binary Search | chestnut890123 | 0 | 24 | split message based on limit | 2,468 | 0.483 | Hard | 33,874 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2807488/Python-Solution-Explained | class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
res = []
curr_start = 0
if (limit - 5) * 9 > len(message):
calc = ceil(len(message) / (limit - 5))
for i in range(1, calc+1):
res.append(f"{message[curr_start:curr_start+(limit-5)]}<{i}/{calc}>")
curr_start += (limit - 5)
return res
elif (limit - 6) * 9 + (limit - 7) * 90 > len(message):
calc = ceil((len(message) - (limit - 6) * 9) / (limit - 7)) + 9
for i in range(1, 10):
res.append(f"{message[curr_start:curr_start+(limit-6)]}<{i}/{calc}>")
curr_start += (limit - 6)
for i in range(10, calc+1):
res.append(f"{message[curr_start:curr_start+(limit-7)]}<{i}/{calc}>")
curr_start += (limit - 7)
return res
elif (limit - 7) * 9 + (limit - 8) * 90 + (limit - 9) * 900 > len(message):
calc = ceil((len(message) - (limit - 7) * 9 - (limit - 8) * 90) / (limit - 9)) + 99
for i in range(1, 10):
res.append(f"{message[curr_start:curr_start+(limit-7)]}<{i}/{calc}>")
curr_start += (limit - 7)
for i in range(10, 100):
res.append(f"{message[curr_start:curr_start+(limit-8)]}<{i}/{calc}>")
curr_start += (limit - 8)
for i in range(100, calc+1):
res.append(f"{message[curr_start:curr_start+(limit-9)]}<{i}/{calc}>")
curr_start += (limit - 9)
return res
elif (limit - 8) * 9 + (limit - 9) * 90 + (limit - 10) * 900 + (limit - 11) * 9000 > len(message):
calc = ceil((len(message) - (limit - 8) * 9 - (limit - 9) * 90 - (limit - 10) * 900) / (limit - 11)) + 999
for i in range(1, 10):
res.append(f"{message[curr_start:curr_start+(limit-8)]}<{i}/{calc}>")
curr_start += (limit - 8)
for i in range(10, 100):
res.append(f"{message[curr_start:curr_start+(limit-9)]}<{i}/{calc}>")
curr_start += (limit - 9)
for i in range(100, 1000):
res.append(f"{message[curr_start:curr_start+(limit-10)]}<{i}/{calc}>")
curr_start += (limit - 10)
for i in range(1000, calc+1):
res.append(f"{message[curr_start:curr_start+(limit-11)]}<{i}/{calc}>")
curr_start += (limit - 11)
return res
else:
return [] | split-message-based-on-limit | Python Solution - Explained | eden33335 | 0 | 7 | split message based on limit | 2,468 | 0.483 | Hard | 33,875 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2807409/Fast-Tricky-Python3-Explained | class Solution:
def split(self, message, limit, parts):
i, res = 0, []
for part in range(1, parts+1):
suffix = '<' + str(part) + '/' + str(parts) + '>'
prefix_len = limit - len(suffix)
res.append(message[i:i+prefix_len] + suffix)
i += prefix_len
return res
def splitMessage(self, message: str, limit: int) -> List[str]:
if limit < 6:
return []
n = len(message)
for parts in range(1, n+1):
tot_parts = parts
cur_parts = 9
suffix = len(str(parts)) + 4
prefix = limit - suffix
mess = n
while tot_parts > 0 and mess > 0 and prefix > 0:
max_take = cur_parts * prefix
if mess > max_take:
tot_parts -= cur_parts
mess -= max_take
else:
full_parts = mess // prefix
tot_parts -= full_parts
mess -= full_parts * prefix
if mess > 0:
tot_parts -= 1
mess = 0
cur_parts *= 10
prefix -= 1
suffix += 1
if tot_parts == mess == 0:
return self.split(message, limit, parts)
return [] | split-message-based-on-limit | Fast, Tricky Python3 Explained | ryangrayson | 0 | 21 | split message based on limit | 2,468 | 0.483 | Hard | 33,876 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2807297/O(N)-Simple-Straightforward-Approach-Python | class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
symbol_count = 3
page_count = 0
N = len(message)
ans_page = -1
for i in range(1,len(message)+1):
page_count += len(str(i))
tot_count = (len(str(i))+3)*i
suffix_count = page_count + tot_count
x , y = divmod(suffix_count+N,i)
if (x == limit and y == 0) or (x < limit):
ans_page = i
break
if ans_page == -1:
return []
curr_page = 1
suff_start = '<'
suff_end = '/'+str(ans_page)+'>'
i = 0
res = []
while curr_page <= ans_page:
if i >= N:
return []
suffix = suff_start + str(curr_page) + suff_end
j = i + limit - len(suffix)
res.append(message[i:j]+suffix)
i = j
curr_page += 1
return res | split-message-based-on-limit | O(N) - Simple Straightforward Approach - Python | ananth_19 | 0 | 28 | split message based on limit | 2,468 | 0.483 | Hard | 33,877 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2807296/Binary-Search-%2B-Construction-Python-Solution | class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
l, r = 0, len(message)
def is_possible(n):
rem = len(message)
for i in range(1,n+1):
used = 3 + len(str(i)) + len(str(n))
taken = limit - used
rem -= taken
return rem <= 0
while l < r:
mid = (l+r) // 2
if is_possible(mid):
r = mid
else:
l = mid + 1
if not is_possible(l):
return []
ret = []
mptr = 0
for i in range(1,l+1):
used = 3 + len(str(i)) + len(str(l))
curr = message[mptr:mptr+(limit-used)]
mptr += (limit-used)
ret.append(curr + "<{0}/{1}>".format(str(i), str(l)))
return ret | split-message-based-on-limit | Binary Search + Construction Python Solution | seanjaeger | 0 | 17 | split message based on limit | 2,468 | 0.483 | Hard | 33,878 |
https://leetcode.com/problems/split-message-based-on-limit/discuss/2807269/Nothing-fancy%3A-check-potential-of-parts-one-by-one-O(nlogn) | class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
n = len(message)
best = math.inf
def check(x):
avail = 0
start = 1
while start <= x:
next_start = start * 10
suffix = f"<{start}/{x}>"
if len(suffix) > limit:
return False
avail += (min(next_start - 1, x) - start + 1) * (limit - len(suffix))
start = next_start
last_suffix = f"<{x}/{x}>"
return avail >= n >= avail - (limit - len(last_suffix))
for n_part in range(1, n + 1):
if check(n_part):
break
else:
return []
ans = []
start = 0
for i in range(1, n_part + 1):
suffix = f"<{i}/{n_part}>"
next_start = start + limit - len(suffix)
ans.append(message[start:next_start] + suffix)
start = next_start
return ans | split-message-based-on-limit | Nothing fancy: check potential # of parts one by one, O(nlogn) | chuan-chih | 0 | 26 | split message based on limit | 2,468 | 0.483 | Hard | 33,879 |
https://leetcode.com/problems/convert-the-temperature/discuss/2839560/Python-or-Easy-Solution | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [(celsius + 273.15),(celsius * 1.80 + 32.00)] | convert-the-temperature | Python | Easy Solution✔ | manayathgeorgejames | 4 | 127 | convert the temperature | 2,469 | 0.909 | Easy | 33,880 |
https://leetcode.com/problems/convert-the-temperature/discuss/2810126/Very-difficult-%3A( | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [celsius + 273.15, celsius * 1.8 + 32] | convert-the-temperature | Very difficult :( | mediocre-coder | 4 | 610 | convert the temperature | 2,469 | 0.909 | Easy | 33,881 |
https://leetcode.com/problems/convert-the-temperature/discuss/2817777/Python-1-liner | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [celsius + 273.15, celsius * 1.80 + 32.00] | convert-the-temperature | Python 1-liner | Vistrit | 3 | 137 | convert the temperature | 2,469 | 0.909 | Easy | 33,882 |
https://leetcode.com/problems/convert-the-temperature/discuss/2818442/PYTHON-Simple-or-One-liner | class Solution:
def convertTemperature(self, c: float) -> List[float]:
return [c+273.15, c*1.80+32] | convert-the-temperature | [PYTHON] Simple | One-liner | omkarxpatel | 2 | 8 | convert the temperature | 2,469 | 0.909 | Easy | 33,883 |
https://leetcode.com/problems/convert-the-temperature/discuss/2818760/Python-Easy-Solution-in-One-Line | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [celsius+273.15,celsius*1.8+32] | convert-the-temperature | Python Easy Solution in One Line | DareDevil_007 | 1 | 41 | convert the temperature | 2,469 | 0.909 | Easy | 33,884 |
https://leetcode.com/problems/convert-the-temperature/discuss/2808822/Python-solution | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
ans=[]
Kelvin = celsius + 273.15
Fahrenheit = celsius * (9/5) + 32
ans.append(Kelvin)
ans.append(Fahrenheit)
return ans | convert-the-temperature | Python solution | shashank_2000 | 1 | 50 | convert the temperature | 2,469 | 0.909 | Easy | 33,885 |
https://leetcode.com/problems/convert-the-temperature/discuss/2849858/Python-oneliner | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [celsius+273.15, celsius*1.80+32] | convert-the-temperature | Python oneliner | StikS32 | 0 | 2 | convert the temperature | 2,469 | 0.909 | Easy | 33,886 |
https://leetcode.com/problems/convert-the-temperature/discuss/2849468/Python3-solution | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [round(celsius + 273.15, 5), round((celsius *1.80) + 32.00, 5)] | convert-the-temperature | Python3 solution | SupriyaArali | 0 | 1 | convert the temperature | 2,469 | 0.909 | Easy | 33,887 |
https://leetcode.com/problems/convert-the-temperature/discuss/2845500/1-line-Python | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [celsius + 273.15, celsius * 1.80 + 32.00] | convert-the-temperature | 1 line, Python | Tushar_051 | 0 | 2 | convert the temperature | 2,469 | 0.909 | Easy | 33,888 |
https://leetcode.com/problems/convert-the-temperature/discuss/2844004/Simple-Python-Solution-%3A) | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [
celsius + 273.15, # kelvin
celsius * 1.80 + 32.00 # farenheit
] | convert-the-temperature | Simple Python Solution :) | EddSH1994 | 0 | 1 | convert the temperature | 2,469 | 0.909 | Easy | 33,889 |
https://leetcode.com/problems/convert-the-temperature/discuss/2841244/one-line-solution-python | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [celsius + 273.15,celsius * (9/5) + 32.00] | convert-the-temperature | one line solution python | ft3793 | 0 | 1 | convert the temperature | 2,469 | 0.909 | Easy | 33,890 |
https://leetcode.com/problems/convert-the-temperature/discuss/2838076/Kid-Problem-Eaiest-One-Line-Code | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [celsius+273.15,celsius*1.80 + 32.00] | convert-the-temperature | Kid Problem Eaiest One Line Code | khanismail_1 | 0 | 3 | convert the temperature | 2,469 | 0.909 | Easy | 33,891 |
https://leetcode.com/problems/convert-the-temperature/discuss/2831897/Python-Solution-Math-oror-Contest-Question | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
k=celsius + 273.15
f=celsius * 1.80 + 32.00
ans=[round(k, 5),round(f, 5)]
return ans | convert-the-temperature | Python Solution - Math || Contest Question | T1n1_B0x1 | 0 | 3 | convert the temperature | 2,469 | 0.909 | Easy | 33,892 |
https://leetcode.com/problems/convert-the-temperature/discuss/2822760/hello-world | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [(celsius+273.15),((celsius*1.8)+32.0)] | convert-the-temperature | hello world | ATHBuys | 0 | 5 | convert the temperature | 2,469 | 0.909 | Easy | 33,893 |
https://leetcode.com/problems/convert-the-temperature/discuss/2822121/Python3-one-line-solution | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [round(celsius + 273.15, 5), round(celsius * 1.8 + 32,5)] | convert-the-temperature | Python3 one-line solution | sipi09 | 0 | 4 | convert the temperature | 2,469 | 0.909 | Easy | 33,894 |
https://leetcode.com/problems/convert-the-temperature/discuss/2821695/Python3-Solution-with-using-simple-conversion | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [celsius + 273.15, celsius * 1.8 + 32.0] | convert-the-temperature | [Python3] Solution with using simple conversion | maosipov11 | 0 | 4 | convert the temperature | 2,469 | 0.909 | Easy | 33,895 |
https://leetcode.com/problems/convert-the-temperature/discuss/2820055/Python-1-Line-Code | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [celsius + 273.15, celsius * 1.80 + 32.00] | convert-the-temperature | Python 1 - Line Code | Asmogaur | 0 | 5 | convert the temperature | 2,469 | 0.909 | Easy | 33,896 |
https://leetcode.com/problems/convert-the-temperature/discuss/2818001/Easy-1-line-python-code | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return(celsius + 273.15, celsius * 1.80 + 32.00) | convert-the-temperature | Easy 1-line python code | Xand0r | 0 | 1 | convert the temperature | 2,469 | 0.909 | Easy | 33,897 |
https://leetcode.com/problems/convert-the-temperature/discuss/2816282/Convert-the-temperature(Solution) | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
return [celsius+273.15,celsius*1.80+32.00] | convert-the-temperature | Convert the temperature(Solution) | NIKHILTEJA | 0 | 1 | convert the temperature | 2,469 | 0.909 | Easy | 33,898 |
https://leetcode.com/problems/convert-the-temperature/discuss/2815507/python-solution-no-need-for-10**-5 | class Solution:
def convertTemperature(self, celsius: float) -> List[float]:
kelvin = celsius + 273.15
fahrenheit = (celsius * 1.80) + 32.00
return [kelvin, fahrenheit] | convert-the-temperature | python solution no need for 10**-5 | samanehghafouri | 0 | 2 | convert the temperature | 2,469 | 0.909 | Easy | 33,899 |