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/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2848431/Easiest-solution-in-Python | class Solution:
def subtractProductAndSum(self, n: int) -> int:
su=0
pro=1
while(n>0):
a=n%10
n=int(n/10)
su=su+a
pro=pro*a
return pro-su | subtract-the-product-and-sum-of-digits-of-an-integer | Easiest solution in Python | ArijitDeveloper | 0 | 1 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,000 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2847130/python-solution-time-O(logn)-space-O(1) | class Solution:
def subtractProductAndSum(self, n: int) -> int:
sum_val=0
pro_val=1
while(n!=0):
num=n%10
sum_val+=num
pro_val*=num
n//=10
return pro_val-sum_val | subtract-the-product-and-sum-of-digits-of-an-integer | python solution time O(logn) space O(1) | sintin1310 | 0 | 1 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,001 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2846338/97-Python-(list-comprehension-%2B-reduce) | class Solution:
def subtractProductAndSum(self, n: int) -> int:
S = [int(x) for x in str(n)]
return reduce(lambda a, b: a * b, S) - reduce(lambda a, b: a + b, S) | subtract-the-product-and-sum-of-digits-of-an-integer | 97% Python (list comprehension + reduce) ✔ | amarildoaliaj | 0 | 1 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,002 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2843572/Python-Solution-faster-than-90-of-solutions | class Solution:
def subtractProductAndSum(self, n: int):
product = 1
total = 0
arr = []
arr = [int(i) for i in str(n)]
for i in range(len(arr)):
total += arr[i]
product = product * arr[i]
return product - total | subtract-the-product-and-sum-of-digits-of-an-integer | Python Solution - faster than 90% of solutions | heli_kolambekar | 0 | 1 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,003 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2837492/Python3-Straightforward-solution | class Solution:
def subtractProductAndSum(self, n: int) -> int:
m = [int(x) for x in str(n)]
j = 1
for i in m:
j *= i
return j - sum(m) | subtract-the-product-and-sum-of-digits-of-an-integer | [Python3] Straightforward solution | Peanut_in_Motion | 0 | 1 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,004 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2836681/The-simple-and-understandable-way | class Solution:
def subtractProductAndSum(self, n: int) -> int:
m = 1
a = 0
for i in str(n):
m *= int(i)
a += int(i)
return m-a | subtract-the-product-and-sum-of-digits-of-an-integer | The simple and understandable way | Abraha111 | 0 | 1 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,005 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2801526/1281.-Subtract-the-Product-and-Sum-of-Digits-of-an-Integer | class Solution:
def subtractProductAndSum(self, n: int) -> int:
l = [int(x) for x in str(n)]
mul = 1
for _ in l:
mul *= _
sum = 0
for _ in l:
sum += _
return (mul - sum) | subtract-the-product-and-sum-of-digits-of-an-integer | 1281. Subtract the Product and Sum of Digits of an Integer | jmukesh99 | 0 | 2 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,006 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2789719/Subtract-the-Product-and-sum-in-python-very-easy-solution | class Solution:
def subtractProductAndSum(self, n: int) -> int:
sub = 0
mul = 1
while n > 0:
r = n % 10
sub += r
mul *= r
n = n // 10
return mul - sub | subtract-the-product-and-sum-of-digits-of-an-integer | Subtract the Product and sum in python very easy solution | jashii96 | 0 | 2 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,007 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2789718/Subtract-the-Product-and-sum-in-python-very-easy-solution | class Solution:
def subtractProductAndSum(self, n: int) -> int:
sub = 0
mul = 1
while n > 0:
r = n % 10
sub += r
mul *= r
n = n // 10
return mul - sub | subtract-the-product-and-sum-of-digits-of-an-integer | Subtract the Product and sum in python very easy solution | jashii96 | 0 | 0 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,008 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2780115/Python-Easy-Solution-(Simple-Approach) | class Solution:
def subtractProductAndSum(self, n: int) -> int:
sum = 0
prod = 1
while n > 0 :
rem = n % 10
sum = sum + rem
prod = prod * rem
n = n//10
return (prod - sum) | subtract-the-product-and-sum-of-digits-of-an-integer | Python Easy Solution (Simple Approach) | Asmogaur | 0 | 4 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,009 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2774195/python-easy-solution | class Solution:
def subtractProductAndSum(self, n: int) -> int:
s=0
p=1
while n:
r=n%10
s=s+r
p=p*r
n=n//10
return p-s | subtract-the-product-and-sum-of-digits-of-an-integer | python easy solution | dummu_chandini | 0 | 2 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,010 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2757002/Python3-26ms-13.8MB-Solution | class Solution:
def subtractProductAndSum(self, n: int) -> int:
str_n = str(n)
product = 1
sum_n = 0
for char in str_n:
product = product * int(char)
sum_n = sum_n + int(char)
return product - sum_n | subtract-the-product-and-sum-of-digits-of-an-integer | Python3 - 26ms 13.8MB Solution | NikhileshNanduri | 0 | 2 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,011 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2752773/Simple-python-code-with-explanation | class Solution:
def subtractProductAndSum(self, n):
#cat(sum) is a variable to store the sum of digits in integer
cat = 0
#rat(product) is a variable to store the product of digits in integer
rat = 1
#while n is not 0 this while loop will work
while n != 0:
#add the last digit to cat
cat += (n%10)
#multiply the last digit with rat
rat *= (n%10)
#remove the last digit
n = n//10
#return the difference of rat(product) - cat(sum)
return (rat - cat) | subtract-the-product-and-sum-of-digits-of-an-integer | Simple python code with explanation | thomanani | 0 | 2 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,012 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2719521/simple-python-solution-%3A-) | class Solution:
def subtractProductAndSum(self, n: int) -> int:
a = str(n)
d = []
for i in a:
d.append(int(i))
z = 1
for i in d:
z = z * i
return z - sum(d) | subtract-the-product-and-sum-of-digits-of-an-integer | simple python solution :-) | ft3793 | 0 | 4 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,013 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2714615/Python-3-or-O(n)-or-Two-Approaches-or-One-Liner | class Solution:
def subtractProductAndSum(self, n: int) -> int:
nums = []
while(n>0): #This loop will continue looping until all the digits in n are removed
nums.append(n % 10) # This appends the last digit of n
n //= 10 # This removes the last digit of n
return prod(nums)-sum(nums) | subtract-the-product-and-sum-of-digits-of-an-integer | ✔️Python 3 | 🕝 O(n) | Two Approaches | One-Liner | keioon | 0 | 4 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,014 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2693581/Python-solution-faster-than-92-and-less-memory-than-95-or-Without-converting-to-string | class Solution:
def subtractProductAndSum(self, n: int) -> int:
product = 1
sum = 0
while n > 0:
digit = n % 10
product = product * digit
sum = sum + digit
n = n // 10
return product - sum | subtract-the-product-and-sum-of-digits-of-an-integer | 🐍 Python solution faster than 92% & less memory than 95% | Without converting to string | anandanshul001 | 0 | 4 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,015 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2687773/Python%2BNumPy | class Solution:
def subtractProductAndSum(self, n: int) -> int:
from numpy import prod,sum
N=[int(x) for x in str(n)]
return prod(N)-sum(N) | subtract-the-product-and-sum-of-digits-of-an-integer | Python+NumPy | Leox2022 | 0 | 1 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,016 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2684120/Python-solution | class Solution:
def subtractProductAndSum(self, n: int) -> int:
product = 1
sum_digits = 0
for num in str(n):
product *= int(num)
sum_digits += int(num)
return product - sum_digits | subtract-the-product-and-sum-of-digits-of-an-integer | Python solution | samanehghafouri | 0 | 16 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,017 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2678670/Easy-solution-or-faster-than-99.7 | class Solution:
def subtractProductAndSum(self, n: int) -> int:
def prod(x):
p = 1
while x>0:
p *= x%10
x//=10
return p
def add(x):
s = 0
while x>0:
s += x%10
x//=10
return s
return prod(n)-add(n) | subtract-the-product-and-sum-of-digits-of-an-integer | Easy solution | faster than 99.7% | MockingJay37 | 0 | 20 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,018 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2668005/Easy-python-solution | class Solution:
def subtractProductAndSum(self, n: int) -> int:
a=1
b=0
res = [int(a) for a in str(n)]
for i in res:
a*=i
b+=i
return a-b | subtract-the-product-and-sum-of-digits-of-an-integer | Easy python solution | Navaneeth7 | 0 | 3 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,019 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2664645/Python3-Solutions | class Solution:
def subtractProductAndSum(self, n: int) -> int:
p = 1 # product
s = 0 # sum
for i in str(n):
p *= int(i)
s += int(i)
return p - s | subtract-the-product-and-sum-of-digits-of-an-integer | Python3 Solutions | AnzheYuan1217 | 0 | 20 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,020 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2664645/Python3-Solutions | class Solution:
def subtractProductAndSum(self, n: int) -> int:
p = 1
s = 0
for i in range(len(str(n))):
digit = n % 10
n //= 10
p *= digit
s += digit
return p - s | subtract-the-product-and-sum-of-digits-of-an-integer | Python3 Solutions | AnzheYuan1217 | 0 | 20 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,021 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2664645/Python3-Solutions | class Solution:
def subtractProductAndSum(self, n: int) -> int:
p = 1
s = 0
while n != 0:
digit = n % 10
n //= 10
p *= digit
s += digit
return p - s | subtract-the-product-and-sum-of-digits-of-an-integer | Python3 Solutions | AnzheYuan1217 | 0 | 20 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,022 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2664645/Python3-Solutions | class Solution:
def subtractProductAndSum(self, n: int) -> int:
p = 1
s = 0
length = len(str(n)) - 1
for i in range(len(str(n))):
digit = n // (10 ** length)
n = n % (10 ** length)
length -= 1
p *= digit
s += digit
return p - s | subtract-the-product-and-sum-of-digits-of-an-integer | Python3 Solutions | AnzheYuan1217 | 0 | 20 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,023 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2658825/Python-or-Easy-solution | class Solution:
def subtractProductAndSum(self, n: int) -> int:
s, p = 0, 1
while n:
mod = n % 10
s += mod
p *= mod
n //= 10
return p - s | subtract-the-product-and-sum-of-digits-of-an-integer | Python | Easy solution | LordVader1 | 0 | 24 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,024 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2645586/easy-approach! | class Solution:
def subtractProductAndSum(self, n: int) -> int:
sum,pro = 0,1
for i in str(n): sum += int(i) ; pro *= int(i)
return pro - sum | subtract-the-product-and-sum-of-digits-of-an-integer | easy approach! | sanjeevpathak | 0 | 2 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,025 |
https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/discuss/2622443/Python-Simple-Solution | class Solution:
def subtractProductAndSum(self, n: int) -> int:
digit_prod = 1
digit_sum = 0
for i in range(len(str(n))):
digit_prod *= int(str(n)[i])
digit_sum += int(str(n)[i])
return digit_prod - digit_sum | subtract-the-product-and-sum-of-digits-of-an-integer | Python Simple Solution | sharondev | 0 | 23 | subtract the product and sum of digits of an integer | 1,281 | 0.867 | Easy | 19,026 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/712693/Python-O(n)-Easy-to-Understand | class Solution:
# Time: O(n)
# Space: O(n)
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
res, dic = [], {}
for idx, group in enumerate(groupSizes):
if group not in dic:
dic[group] = [idx]
else:
dic[group].append(idx)
if len(dic[group]) == group:
res.append(dic[group])
del dic[group]
return res | group-the-people-given-the-group-size-they-belong-to | Python O(n) Easy to Understand | whissely | 5 | 214 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,027 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1942719/Python-Hashmap-Simple-Approach | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
# Step 1 : Categorise using hashmap
h = {}
for index,value in enumerate(groupSizes):
h[value] = h.get(value,[]) + [index]
ans = []
# Step 2 : Prepare the groups
for size in h.keys():
num_grps = len(h[size])//size # // means integer division
for x in range(num_grps):
temp = []
for j in range(size):
temp.append(h[size].pop(0))
ans.append(temp)
# Return the answer
return ans | group-the-people-given-the-group-size-they-belong-to | Python Hashmap Simple Approach | BeardedOwl1357 | 1 | 40 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,028 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1777376/Python-3-HashMap-solution | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
res = []
groups = collections.defaultdict(list)
for i, size in enumerate(groupSizes):
groups[size].append(i)
if len(groups[size]) == size:
res.append(groups[size])
groups[size] = []
return res | group-the-people-given-the-group-size-they-belong-to | Python 3, HashMap solution | dereky4 | 1 | 149 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,029 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1479892/Python3-solution-or-faster-than-76-or-Comments | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
d = {}
for i in range(len(groupSizes)): # creating a dictionary to group all persons that have the same groupSize
if groupSizes[i] not in d:
d[groupSizes[i]] = []
d[groupSizes[i]].append(i)
else:
d[groupSizes[i]].append(i)
res = []
k = 0 # index of the list we return
for i, ele in enumerate(d): # we iterate through the dictionary and every list
count = 0
res.append([])
for j in range(len(d[ele])):
res[k].append(d[ele][j])
count += 1 # we count how many persons we add -> if there are more persons than the number of groupSize we append another empty list
if count == ele:
if j < len(d[ele])-1:
res.append([])
k += 1
count = 0
k += 1
return res | group-the-people-given-the-group-size-they-belong-to | Python3 solution | faster than 76% | Comments | FlorinnC1 | 1 | 117 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,030 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/446452/Python3-group-ids-according-to-size | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
ids = dict() #mapping from size to ids
for i, size in enumerate(groupSizes):
ids.setdefault(size, []).append(i)
ans = []
for size, ids in ids.items():
ans.extend([ids[i:i+size] for i in range(0, len(ids), size)]) #split list into list of list
return ans | group-the-people-given-the-group-size-they-belong-to | [Python3] group ids according to size | ye15 | 1 | 55 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,031 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2831803/Python-Solution-Hashmap-approach-oror-Explained | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
d={}
for i in range(len(groupSizes)):
#putting array of indeces into the values
#of keys (which are the same numbers present in groupSizes)
#d={key:[values]}
if groupSizes[i] in d:
d[groupSizes[i]].append(i)
else:
d[groupSizes[i]]=[i]
ans=[]
for i in d.keys():
j=0
l=[]
while j<len(d[i]): #from every key storing indices array as value
l.append(d[i][j]) #make subarrays same to the size of i
if len(l)==i: #if one subarray reach to the length of i
#and still numbers remaining in d[i] then make another subarray
ans.append(l) #answer array
l=[]
j+=1 #traversing in d[i] values
if len(l)!=0: #if values finished in d[i] then l will be having few values so append to ans
ans.append(l)
return ans | group-the-people-given-the-group-size-they-belong-to | Python Solution - Hashmap approach || Explained✔ | T1n1_B0x1 | 0 | 3 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,032 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2825501/Python-3-HashTable-O(n) | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
lookup = defaultdict(list)
result = []
# Build lookup dictionary key: group size, value: list of ids that are in that group size
for id, size in enumerate(groupSizes):
lookup[size].append(id)
# Check for too many ids for group size and break up into groups of 'key' size
for key, value in lookup.items():
left = 0
right = key
while right <= len(value):
result.append(value[left:right])
left += key
right += key
return result | group-the-people-given-the-group-size-they-belong-to | Python 3 HashTable O(n) | bettend | 0 | 2 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,033 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2771714/python-solution-Group-the-People-Given-the-Group-Size-They-Belong | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
counter = {}
main_list = []
for i in range(len(groupSizes)):
if groupSizes[i] not in counter:
counter[groupSizes[i]] = [i]
else:
counter[groupSizes[i]].append(i)
if len(counter[groupSizes[i]]) == groupSizes[i]:
main_list.append(counter[groupSizes[i]])
counter[groupSizes[i]] = []
return (main_list) | group-the-people-given-the-group-size-they-belong-to | python solution Group the People Given the Group Size They Belong | sarthakchawande14 | 0 | 3 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,034 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2702608/Simple-and-Easy-to-Understand-or-Beginner's-Friendly-or-Python | class Solution(object):
def groupThePeople(self, g):
hashT = {}
ans = []
for i in range(len(g)):
if g[i] not in hashT: hashT[g[i]] = [i]
else:
if len(hashT[g[i]]) == g[i]:
ans.append(hashT[g[i]])
hashT[g[i]] = [i]
else: hashT[g[i]].append(i)
for n in hashT: ans.append(hashT[n])
return ans | group-the-people-given-the-group-size-they-belong-to | Simple and Easy to Understand | Beginner's Friendly | Python | its_krish_here | 0 | 14 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,035 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2695449/Python3-Simple-Solution | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
d = defaultdict(list)
for i, size in enumerate(groupSizes):
d[size].append(i)
res = []
for k, v in d.items():
for i in range(0, len(v) - k + 1, k):
res.append(d[k][i: i + k])
return res | group-the-people-given-the-group-size-they-belong-to | Python3 Simple Solution | mediocre-coder | 0 | 6 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,036 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2478766/A-straight-forward-Python-Solution | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
lookup_table = {}
output = []
for person, groupSize in enumerate(groupSizes):
if groupSize in lookup_table:
lookup_table[groupSize].append(person)
continue
lookup_table[groupSize] = [person]
for key in lookup_table.keys():
if key == len(lookup_table[key]):
output.append(lookup_table[key])
continue
group_stub = [0] * key
for idx in range(len(lookup_table[key]) // key):
output.append(lookup_table[key][idx * key:(idx*key) + key])
return output | group-the-people-given-the-group-size-they-belong-to | A straight forward Python Solution | cengleby86 | 0 | 28 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,037 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2475318/Easiest-Solution-using-Hash-Map | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
#Step 1: Store people who have group with the same size
group_size_people = {}
for i, size in enumerate(groupSizes):
if size not in group_size_people:
group_size_people[size]=[i]
else:
group_size_people[size].append(i)
#Step 2: Put them in to group
res = []
for size, people in group_size_people.items():
if size < len(people):
for i in range(len(people)):
if (i+1)%size ==0:
res.append(people[i+1-size:i+1])
else:
res.append(people)
return res | group-the-people-given-the-group-size-they-belong-to | Easiest Solution using Hash Map | williamhuybui | 0 | 10 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,038 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2473110/Python-Solution-or-defaultdict-or-easy-understanding | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
ans = []
d = defaultdict(list)
for i, g in enumerate(groupSizes):
d[g].append(i)
if len(d[g]) == g:
ans.append(d[g])
d[g] = []
return ans | group-the-people-given-the-group-size-they-belong-to | Python Solution | defaultdict | easy understanding | imneeraj_kumar | 0 | 33 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,039 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2412445/Hashtable-Python3-Statistic | class Solution:
def groupThePeople(self, g: List[int]) -> List[List[int]]:
N = len(g)
count = {}
res = []
for i in range(N):
if g[i] in count:
l = count[g[i]]
if len(l) < g[i] - 1:
count[g[i]] = count[g[i]] + [i]
else:
res.append(count[g[i]] + [i])
count.pop(g[i])
else:
if g[i] > 1: count[g[i]] = [i]
else: res.append([i])
return res | group-the-people-given-the-group-size-they-belong-to | Hashtable Python3 Statistic | jdai1234 | 0 | 25 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,040 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2275059/No-additional-collections-just-sorting-with-indices | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
indexed = sorted([(i, v) for i, v in enumerate(groupSizes)], key=lambda t: t[1])
answer = []
i = 0
while i < len(groupSizes):
answer.append([v[0] for v in indexed[i: i + indexed[i][1]]])
i += indexed[i][1]
return answer | group-the-people-given-the-group-size-they-belong-to | No additional collections, just sorting with indices | amaargiru | 0 | 25 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,041 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2238575/Python3-intuitive-solution-w-zipping | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
pos = [i for i in range(len(groupSizes))]
zipped = zip(pos, groupSizes)
A = sorted(zipped, key = lambda x:x[1])
i, ans = 0, []
while i < len(groupSizes):
chunk = A[i][1]
tmp = []
for j in range(i, i+ chunk):
tmp.append(A[j][0])
i += chunk
ans.append(tmp)
return(ans) | group-the-people-given-the-group-size-they-belong-to | Python3 intuitive solution w/ zipping | TheKivs | 0 | 35 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,042 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2153851/Python-Solution-Using-Dictionaries-(BASIC)-O(n2)%3ATime-Complexity | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
d={}
for i in set(groupSizes):
d[i]=[]
for i in range(len(groupSizes)):
d[groupSizes[i]]=d.get(groupSizes[i])+[i]
res=[]
for i in list(d.keys()):
for j in range(0,len(d[i]),i):
res.append(d[i][j:j+i])
return res | group-the-people-given-the-group-size-they-belong-to | Python Solution Using Dictionaries (BASIC) O(n^2):Time Complexity | Kunalbmd | 0 | 28 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,043 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/2120706/Python-Dictionary-Implementation | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
mapp = {}
for i in range(len(groupSizes)):
if groupSizes[i] in mapp:
mapp[groupSizes[i]].append(i)
else:
mapp[groupSizes[i]] = [i]
ans = []
for key,value in mapp.items():
temp = []
j =0
for i in range(len(value)):
temp.append(value[i])
j+=1
if j == key:
ans.append(temp)
temp = []
j = 0
return ans | group-the-people-given-the-group-size-they-belong-to | Python Dictionary Implementation | somendrashekhar2199 | 0 | 59 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,044 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1832191/Personal-python-solution | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
# define dict to sort groups
buckets = {}
#define ans list to return later
ans = []
#iterate through list with enumerate
for index,val in enumerate(groupSizes):
#if val in buckets append index of person to group bucket
if val in buckets:
temp = buckets[val]
temp.append(index)
buckets[val] = temp
else:
#if val not in buckets add val as with list of index of first person in group
buckets[val] = [index]
#iterate through dict keys (groups)
for val in buckets:
#define pointers to slice the groups for ans
pointer1 = 0
pointer2 = val
#define list of curr group indexes to be sliced
bucket_list = buckets[val]
#iterate through list with pointers and slice it into groups until out of vals
while pointer2 <= len(bucket_list):
group_slice = bucket_list[pointer1:pointer2]
ans.append(group_slice)
pointer1 += val
pointer2 += val
return(ans) | group-the-people-given-the-group-size-they-belong-to | Personal python solution | LegendaryBeagle | 0 | 46 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,045 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1832191/Personal-python-solution | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
buckets = {}
ans = []
for index,val in enumerate(groupSizes):
if val in buckets:
temp = buckets[val]
temp.append(index)
buckets[val] = temp
else:
buckets[val] = [index]
pointer1 = 0
for val in buckets:
pointer1 = 0
pointer2 = val
bucket_list = buckets[val]
while pointer2 <= len(bucket_list):
group_slice = bucket_list[pointer1:pointer2]
ans.append(group_slice)
pointer1 += val
pointer2 += val
return(ans) | group-the-people-given-the-group-size-they-belong-to | Personal python solution | LegendaryBeagle | 0 | 46 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,046 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1546657/Python3-Solution-with-using-map | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
d = {}
res = []
for idx, gs in enumerate(groupSizes):
if gs not in d:
d[gs] = []
d[gs].append(idx)
if len(d[gs]) == gs:
res.append(d[gs])
d[gs] = []
return res | group-the-people-given-the-group-size-they-belong-to | [Python3] Solution with using map | maosipov11 | 0 | 49 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,047 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1425156/Python-3 | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
dictn = {}
for i, v in enumerate(groupSizes):
dictn.setdefault(v, []).append(i)
res = []
for x, y in dictn.items():
p = x
j = 0
if(len(y) < x):
res.append(y)
else:
for _ in range(0, int(len(y)/x)):
res.append(y[j:p])
j = p
p += x
res.sort(key=len)
return res | group-the-people-given-the-group-size-they-belong-to | Python 3 | amestri890 | 0 | 59 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,048 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1416688/Python3-simple-solution-99.41-faster | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
dict = {}
outList = []
for i in range(0,len(groupSizes)):
if groupSizes[i] not in dict:
dict[groupSizes[i]] = [i]
else:
dict[groupSizes[i]] += [i]
for k,v in dict.items():
for i in range(0,len(v),k):
outList.append(v[i:i+k])
return outList | group-the-people-given-the-group-size-they-belong-to | Python3 simple solution 99.41% faster | sunnysharma03 | 0 | 99 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,049 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1297655/python-easy-solution-beats-~91 | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
h={}
for person,size in enumerate(groupSizes):
if size not in h:
h[size]=[person]
else:
h[size].append(person)
ans=[]
for i in h:
if len(h[i])!=i:
for end in range(0,len(h[i]),i):
ans.append(h[i][end:end+i])
else:
ans.append(h[i])
return (ans) | group-the-people-given-the-group-size-they-belong-to | python easy solution beats ~91% | prakharjagnani | 0 | 73 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,050 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1273671/Simple-and-Clean-Python-solution | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
dic=defaultdict(list)
#Store in a dictionary[i] the list of indices of input array groupSizes that belong to a group size i
for index,ele in enumerate(groupSizes):
dic[ele].append(index)
answer=[]
for key,value in dic.items():
for idx in range(0,len(value),key):
answer.append(value[idx:idx + key])
return answer | group-the-people-given-the-group-size-they-belong-to | Simple and Clean Python solution | jaipoo | 0 | 150 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,051 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1265774/Python-O(N)-solution | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
groups = {}
ans = []
for i, size in enumerate(groupSizes):
if not groups.get(size, None):
groups[size] = []
groups[size].append(i) # O(1)
if len(groups[size]) == size:
ans.append(groups[size])
groups[size] = []
return ans | group-the-people-given-the-group-size-they-belong-to | Python O(N) solution | user5573CJ | 0 | 62 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,052 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1239434/Python-3-Solution-using-dictionary | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
d = {}
final = []
# Create a dic with number as key and its positions as value
for i, x in enumerate(groupSizes):
if x not in d:
d[x] = [i]
else:
d[x].append(i)
tmp = dict(sorted(d.items()))
for x,y in tmp.items():
if len(y) == x:
final.append(y)
else:
for i in range(0,len(y),x):
final.append(y[i:i+x])
return(final) | group-the-people-given-the-group-size-they-belong-to | [Python 3] Solution using dictionary | SushilG96 | 0 | 43 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,053 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1102734/WEEB-EXPLAINS-SIMPLE-PYTHON-CODE | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
adjList = defaultdict(list)
for idx, value in enumerate(groupSizes):
adjList[value].append(idx) # construct adjacency list
result = []
for i in adjList:
level = [] # for each group we create a new level
for j in adjList[i]: # for the people in that particular group
level += [j] # add to group
if len(level) == i: # if group is full
result.append(level) # put in result
level = [] # clean the group cuz if not then it will overflow
return result | group-the-people-given-the-group-size-they-belong-to | WEEB EXPLAINS SIMPLE PYTHON CODE | Skywalker5423 | 0 | 103 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,054 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/1078169/Python-Queue | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
graph = collections.defaultdict(collections.deque)
for idx, i in enumerate(groupSizes):
graph[i].append(idx)
result = []
for key, val in graph.items():
while val:
tmp = []
for i in range(key):
tmp += [val.popleft()]
result.append(tmp)
return result | group-the-people-given-the-group-size-they-belong-to | Python Queue | dev-josh | 0 | 83 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,055 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/996513/Simple-solution-in-python3-handling-case-of-groupSizei-1-seperately-using-hashmap | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
hashmap = dict()
for i in range(len(groupSizes)):
if groupSizes[i] != 1:
if groupSizes[i] not in hashmap:
hashmap[groupSizes[i]] = [i]
else:
hashmap[groupSizes[i]].append(i)
ans = []
for k in hashmap:
v = hashmap[k]
number_groups = len(v) // k
start = 0
for i in range(number_groups):
ans.append(v[start: start + k])
start = start + k
for i in range(len(groupSizes)):
if groupSizes[i] == 1:
ans.append([i])
return ans | group-the-people-given-the-group-size-they-belong-to | Simple solution in python3 handling case of groupSize[i] == 1 seperately using hashmap | amoghrajesh1999 | 0 | 53 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,056 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/765755/Simple-python-solution-using-dictionary-and-list. | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
sol = {}
for i in range(len(groupSizes)):
if groupSizes[i] not in sol:
sol[groupSizes[i]] = [i]
else:
sol[groupSizes[i]].append(i)
res = []
for key in sol:
val = sol[key]
temp = []
for i in range(len(val)):
temp.append(val[i])
if (i + 1) % key == 0:
res.append(temp)
temp = []
return res | group-the-people-given-the-group-size-they-belong-to | Simple python solution using dictionary and list. | C3Wizard | 0 | 57 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,057 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/755013/Simple-Python3-List-Solution-Beats-92 | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
# Maximum possible number of groups is the same as number of people
groups = [[] for _ in range(len(groupSizes)+1)]
# Add invididuals to groups without subdivisions
for i in range(len(groupSizes)):
groups[groupSizes[i]].append(i)
res = []
# Split the groups into subgroups if necessary
for i in range(len(groups)):
if len(groups[i]) == 0: continue
ind = 0
for p in range(0, len(groups[i]), i):
res.append(groups[i][i*ind:i*(ind+1)])
ind += 1
return res | group-the-people-given-the-group-size-they-belong-to | Simple Python3 List Solution Beats 92% | jacksilver | 0 | 122 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,058 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/639225/Python-easy-to-understand-two-step-solution-using-list | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
groups_of_same_size = [[i for i, e in enumerate(groupSizes) if e == s] for s in set(groupSizes)]
result = []
for groups in groups_of_same_size:
size = groupSizes[groups[0]]
count = len(groups) // size
result.extend(groups[i*size:(i+1)*size] for i in range(count))
return result | group-the-people-given-the-group-size-they-belong-to | Python easy to understand two-step solution using list | usualwitch | 0 | 92 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,059 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/550393/Noob-Python3-Answer-(First-Attempt)-Improvement-Suggestions-Welcome | class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
#We are wanting to return a list of groups, where each group is a list that contains IDs of persons. A person's
#ID is equal to their index in the groupSizes list (which is the input parameter to this function).
groupsOfPeople = list()
#Dictionary to track the occurrences for each group size
groupSizeToOccurrences = dict()
#Count the occurrences and populate the dictionary
for i in range(len(groupSizes)):
groupSize = groupSizes[i]
personId = i
if groupSize in groupSizeToOccurrences:
groupSizeToOccurrences[groupSize] += 1
else:
groupSizeToOccurrences[groupSize] = 1
#Make the groups
for groupSize in groupSizeToOccurrences:
groupSizeOccurrences = groupSizeToOccurrences[groupSize]
numGroupsToCreate = groupSizeOccurrences//groupSize
groupsList = self.getListOfGroups(groupSize, numGroupsToCreate, groupSizes)
groupsOfPeople += groupsList
return groupsOfPeople
def getListOfGroups(self, groupSize: int, numGroupsToCreate: int, groupSizes: List[int]) -> List[List[int]]:
listOfGroups = self.createEmptyListOfGroups(numGroupsToCreate)
for i in range(len(groupSizes)):
#Only look at the groupSize that matches the input parameter to this function
if(groupSizes[i] == groupSize):
#Get the index (personId) and put it into a list
personId = i
nextGroup = self.getNextGroup(listOfGroups, groupSize)
nextGroup.append(personId)
return listOfGroups
def getNextGroup(self, listOfGroups: List[List[int]], groupSize):
#Get the next group that has not yet reached it's expected size and return it
for i in range(len(listOfGroups)):
hasReachedGroupSize = (len(listOfGroups[i]) == groupSize)
if not hasReachedGroupSize:
return listOfGroups[i]
def createEmptyListOfGroups(self, numGroupsToCreate: int):
emptyGroups = list()
for i in range(numGroupsToCreate):
emptyGroups.append(list())
return emptyGroups | group-the-people-given-the-group-size-they-belong-to | Noob Python3 Answer (First Attempt) - Improvement Suggestions Welcome | jleikam | 0 | 42 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,060 |
https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/discuss/446439/Python-3-(three-lines)-(beats-100) | class Solution:
def groupThePeople(self, G: List[int]) -> List[List[int]]:
S, D = set(G), collections.defaultdict(list)
for i, g in enumerate(G): D[g].append(i)
return [D[i][i*j:i*(j+1)] for i in S for j in range(G.count(i)//i)]
- Junaid Mansuri
- Chicago, IL | group-the-people-given-the-group-size-they-belong-to | Python 3 (three lines) (beats 100%) | junaidmansuri | 0 | 286 | group the people given the group size they belong to | 1,282 | 0.857 | Medium | 19,061 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/863333/Python3-Binary-search-with-explanation | class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
left, right = 1, max(nums)
while left + 1 < right:
mid = (left + right) // 2
div_sum = self.get_sum(mid, nums)
if div_sum > threshold:
left = mid
else:
right = mid
div_sum = self.get_sum(left, nums)
if div_sum <= threshold:
return left
return right
def get_sum(self, divisor, nums):
res = 0
for n in nums:
tmp = n // divisor
if tmp * divisor < n:
tmp += 1
res += tmp
return res | find-the-smallest-divisor-given-a-threshold | Python3 Binary search with explanation | ethuoaiesec | 3 | 313 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,062 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/840531/Python-3-or-Binary-Search-or-Explanations | class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
def ok(mid):
ans = 0
for num in nums:
ans += math.ceil(num / mid)
if ans > threshold: return False
return True
l, r = 1, int(1e6)
while l <= r:
mid = (l+r) // 2
if ok(mid): r = mid - 1
else: l = mid + 1
return l | find-the-smallest-divisor-given-a-threshold | Python 3 | Binary Search | Explanations | idontknoooo | 3 | 668 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,063 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/2761376/Python-Binary-Serach | class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
left, right = 1, max(nums)
while left < right:
mid = (left + right )//2
count = 0
for num in nums:
count += math.ceil(num/mid)
if count > threshold:
left = mid + 1
else:
right = mid
return left | find-the-smallest-divisor-given-a-threshold | Python Binary Serach | vijay_2022 | 0 | 2 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,064 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/2597711/Python3-or-Solved-Using-Binary-Search-on-Divisor-Range-of-Values | class Solution:
#Let n = len(nums)!
#Time-Complexity: O(log(max(nums)) * n)
#Space-Complexity: O(1)
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
#Approach: Define the search space for divisors, ranging from 1 to
#maximum element in input array nums!
#Perform binary search on such search space and update minimal divisor
#s.t. it meets the condition as we go!
#Once binary search stops, the answer variable should store value
#which is the minimal divisor to not exceed threshold!
#define a helper function to find sum of elements after dividing each
#and every element in nums by divisor!
def helper(divisor):
nonlocal nums
accumulator = 0
for e in nums:
divided = e / divisor
#if it's integer, leave it!
if(divided.is_integer()):
accumulator += int(divided)
continue
else:
divided = int(divided)
#round up
divided += 1
accumulator += divided
return accumulator
L, H = 1, max(nums)
ans = None
while L <= H:
cur = (L + H) // 2
total_sum = helper(cur)
#check against the threshold!
if(total_sum <= threshold):
ans = cur
H = cur - 1
continue
else:
L = cur + 1
continue
return ans | find-the-smallest-divisor-given-a-threshold | Python3 | Solved Using Binary Search on Divisor Range of Values | JOON1234 | 0 | 7 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,065 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/2398462/Python-Binary-Search | class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
high = max(nums)
low = 1
while (low < high):
mid = (low + high)//2
if self.helper(nums, mid, threshold):
high = mid
else:
low = mid + 1
return low
def helper(self, nums, thres, threshold):
divSum = 0
for i in range(len(nums)):
divSum += math.ceil(nums[i]/thres)
return (divSum <= threshold) | find-the-smallest-divisor-given-a-threshold | Python Binary Search | logeshsrinivasans | 0 | 31 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,066 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/2381064/python3-binary-search-solution | class Solution:
# O(nlogn) time,
# O(1) space,
# Approach: binary search,
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
def evalFunction(div):
tot = 0
for num in nums:
qutnt = num//div
tot +=qutnt
if num % div != 0:
tot +=1
return tot
def binarySearch(start, end):
mid = (start + end) // 2
qutnt_sum = evalFunction(mid)
if qutnt_sum <= threshold and (end-start < 2):
return mid
elif start >= end-1:
return end
elif qutnt_sum <= threshold:
return binarySearch(start, mid)
else:
return binarySearch(mid, end)
return binarySearch(1, max(nums)) | find-the-smallest-divisor-given-a-threshold | python3 binary search solution | destifo | 0 | 11 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,067 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/2094606/Python-Solution-neat-and-clean-easy-to-understand | class Solution:
def helper(self,nums,m):
Sum = 0
for n in nums:
Sum += math.ceil(n/m)
return Sum
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
l,r = 1, max(nums)
while l < r:
mid = (l+r)//2
Sum = self.helper(nums,mid)
if Sum > threshold:
l = mid + 1
else:
r = mid
return r | find-the-smallest-divisor-given-a-threshold | Python Solution neat and clean easy to understand | __Asrar | 0 | 78 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,068 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/1933364/6-Lines-Python-Solution-oror-90-Faster-oror-Memory-less-than-80 | class Solution:
def smallestDivisor(self, nums: List[int], t: int) -> int:
lo=1 ; hi=max(nums)
while lo<=hi:
mid=(lo+hi)//2
if sum(ceil(num/mid) for num in nums)<=t: hi=mid-1
else: lo=mid+1
return lo | find-the-smallest-divisor-given-a-threshold | 6-Lines Python Solution || 90% Faster || Memory less than 80% | Taha-C | 0 | 62 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,069 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/1915781/Python-or-Binary-Search-or-Easy | class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
# print(((1-1)//5)+1)
def helper(x):
ans = 0
for i in range(len(nums)):
ans+=(((nums[i]-1)//x)+1)
return ans<=threshold
low = 1
high = max(nums)
while low<high:
mid = low+(high-low)//2
if helper(mid):
high = mid
else:
low=mid+1
return low | find-the-smallest-divisor-given-a-threshold | Python | Binary Search | Easy | Brillianttyagi | 0 | 36 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,070 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/1078615/Python3-binary-search-beats-99 | class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
compute_sum = lambda x : sum([ceil(n / x) for n in nums])
real_sum = sum(nums)
if real_sum < threshold:
return 1
if threshold == len(nums):
return max(nums)
left = ceil(real_sum / threshold)
right = ceil(real_sum / (threshold - len(nums)))
if compute_sum(left) <= threshold:
return left
while left < right - 1:
mid = (right - left) // 2 + left
if compute_sum(mid) > threshold:
left = mid
else:
right = mid
return right | find-the-smallest-divisor-given-a-threshold | Python3 binary search beats 99% | fengzhixiwu | 0 | 95 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,071 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/718528/Python3-bisect_left-Find-the-Smallest-Divisor-Given-a-Threshold | class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
class Wrapper:
def __getitem__(self, x):
return sum(-n//x for n in nums)
return bisect.bisect_left(Wrapper(), -threshold, 1, max(nums)) | find-the-smallest-divisor-given-a-threshold | Python3 bisect_left - Find the Smallest Divisor Given a Threshold | r0bertz | 0 | 113 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,072 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/446492/Python3-Binary-search | class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
# True if sum of numbers divided by d <= threshold
fn = lambda d: sum(ceil(x/d) for x in nums) <= threshold
# "first true" binary search
lo, hi = 1, max(nums)
while lo < hi:
mid = lo + hi >> 1
if fn(mid): hi = mid
else: lo = mid + 1
return lo | find-the-smallest-divisor-given-a-threshold | [Python3] Binary search | ye15 | 0 | 93 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,073 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/446492/Python3-Binary-search | class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
lo, hi = 1, 10**6
while lo < hi:
mid = lo + hi >> 1
if sum(ceil(x/mid) for x in nums) <= threshold: hi = mid
else: lo = mid + 1
return lo | find-the-smallest-divisor-given-a-threshold | [Python3] Binary search | ye15 | 0 | 93 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,074 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/446396/Python-3-(beats-100)-(Optimized-w-Proof)-(Bisection-Search)-(six-lines) | class Solution:
def smallestDivisor(self, N: List[int], t: int) -> int:
N.sort(); S, L, ceil = sum(N), len(N), math.ceil; a, b = ceil(S/t), min(ceil(S/(t - L)),N[-1])
while a < b:
m = (a+b)//2
if sum(ceil(n/m) for n in N) > t: a = m + 1
else: b = m
return a | find-the-smallest-divisor-given-a-threshold | Python 3 (beats 100%) (Optimized w/ Proof) (Bisection Search) (six lines) | junaidmansuri | 0 | 300 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,075 |
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/discuss/446396/Python-3-(beats-100)-(Optimized-w-Proof)-(Bisection-Search)-(six-lines) | class Solution:
def smallestDivisor(self, N: List[int], t: int) -> int:
N.sort(); a, b, ceil = 1, N[-1], math.ceil
while a < b:
m = (a+b)//2
if sum(ceil(n/m) for n in N) > t: a = m + 1
else: b = m
return a | find-the-smallest-divisor-given-a-threshold | Python 3 (beats 100%) (Optimized w/ Proof) (Bisection Search) (six lines) | junaidmansuri | 0 | 300 | find the smallest divisor given a threshold | 1,283 | 0.554 | Medium | 19,076 |
https://leetcode.com/problems/minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix/discuss/446552/Python-3-(ten-lines)-(Check-All-Permutations) | class Solution:
def minFlips(self, G: List[List[int]]) -> int:
M, N = len(G), len(G[0])
P = [(i,j) for i,j in itertools.product(range(M),range(N))]
for n in range(M*N+1):
for p in itertools.permutations(P,n):
H = list(map(list,G))
for (x,y) in p:
for (i,j) in (x,y-1),(x,y),(x,y+1),(x-1,y),(x+1,y):
if 0 <= i < M and 0 <= j < N: H[i][j] = 1 - H[i][j]
if max(max(H)) == 0: return n
return -1
- Junaid Mansuri
- Chicago, IL | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | Python 3 (ten lines) (Check All Permutations) | junaidmansuri | 3 | 441 | minimum number of flips to convert binary matrix to zero matrix | 1,284 | 0.72 | Hard | 19,077 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/452166/Python-3-(four-different-one-line-solutions)-(beats-100) | class Solution:
def findSpecialInteger(self, A: List[int]) -> int:
return collections.Counter(A).most_common(1)[0][0]
from statistics import mode
class Solution:
def findSpecialInteger(self, A: List[int]) -> int:
return mode(A)
class Solution:
def findSpecialInteger(self, A: List[int]) -> int:
return max(set(A), key = A.count)
class Solution:
def findSpecialInteger(self, A: List[int]) -> int:
return (lambda C: max(C.keys(), key = lambda x: C[x]))(collections.Counter(A))
- Junaid Mansuri
- Chicago, IL | element-appearing-more-than-25-in-sorted-array | Python 3 (four different one-line solutions) (beats 100%) | junaidmansuri | 18 | 1,800 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,078 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/2056417/Python-easy-to-understand-solution | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
per = len(arr)//4
for i in arr:
occ = arr.count(i)
if occ > per:
return i | element-appearing-more-than-25-in-sorted-array | Python easy to understand solution | Shivam_Raj_Sharma | 3 | 121 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,079 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/1103979/Python3-simple-solution-using-four-approaches | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
n = len(arr)
for i in arr:
if arr.count(i) > n/4:
return i | element-appearing-more-than-25-in-sorted-array | Python3 simple solution using four approaches | EklavyaJoshi | 2 | 122 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,080 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/1103979/Python3-simple-solution-using-four-approaches | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
return sorted(arr, key = lambda x: arr.count(x), reverse = True)[0] | element-appearing-more-than-25-in-sorted-array | Python3 simple solution using four approaches | EklavyaJoshi | 2 | 122 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,081 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/1103979/Python3-simple-solution-using-four-approaches | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
a = 0
count = 0
n = len(arr)
for i in arr:
if i == a:
count += 1
else:
a = i
count = 1
if count > n/4:
return a | element-appearing-more-than-25-in-sorted-array | Python3 simple solution using four approaches | EklavyaJoshi | 2 | 122 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,082 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/1332139/WEEB-DOES-PYTHON-(ONE-LINER) | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
return Counter(arr).most_common(1)[0][0] | element-appearing-more-than-25-in-sorted-array | WEEB DOES PYTHON (ONE-LINER) | Skywalker5423 | 1 | 102 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,083 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/451307/Python3-one-pass | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
for i in range(len(arr)):
if arr[i] == arr[i+len(arr)//4]: return arr[i] | element-appearing-more-than-25-in-sorted-array | [Python3] one pass | ye15 | 1 | 71 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,084 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/2781682/python-easy-(2-lines) | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
a =len(arr) * (25/100);
for i in arr :
if arr.count(i) > a:
return i | element-appearing-more-than-25-in-sorted-array | python easy (2 lines) | seifsoliman | 0 | 2 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,085 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/2687124/Python3-Solution-oror-O(N)-Time-and-O(1)-Space-Complexity | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
n=len(arr)
element=arr[0]
count=1
for i in arr[1:]:
if count>n//4:
break
if element==i:
count+=1
else:
element=i
count=1
return element | element-appearing-more-than-25-in-sorted-array | Python3 Solution || O(N) Time & O(1) Space Complexity | akshatkhanna37 | 0 | 3 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,086 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/2391977/Very-simple | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
l=len(arr)
c=(l//4)+1
d={}
for i in arr:
if i in d:
d[i]+=1
else:
d[i]=1
if d[i]>=c:
return i | element-appearing-more-than-25-in-sorted-array | Very simple | sunakshi132 | 0 | 26 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,087 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/2308037/Using-HashMap-or-faster-from-67 | class Solution(object):
def findSpecialInteger(self, arr):
"""
:type arr: List[int]
:rtype: int
"""
n=len(arr)
record={}
pcnt=n/4
for i in arr:
try:
record[i]+=1
if record[i]>pcnt:
return i
except:
record[i]=1
return record.keys()[0] | element-appearing-more-than-25-in-sorted-array | Using HashMap | faster from 67 | abdulhaseeb13 | 0 | 25 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,088 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/2298962/Two-Python-1-line-solutions | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
return collections.Counter(arr).most_common(1)[0][0] | element-appearing-more-than-25-in-sorted-array | Two Python 1 line solutions | zip_demons | 0 | 18 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,089 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/2047203/Simple-python-solution | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
return max([x for x in arr if arr.count(x) > len(arr)//4]) | element-appearing-more-than-25-in-sorted-array | Simple python solution | StikS32 | 0 | 38 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,090 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/1921797/Python-two-beginner-friendly-methods | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
return Counter(arr).most_common()[0][0] | element-appearing-more-than-25-in-sorted-array | Python two beginner friendly methods | alishak1999 | 0 | 50 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,091 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/1921797/Python-two-beginner-friendly-methods | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
for i in set(arr):
if arr.count(i) > (0.25 * len(arr)):
return i | element-appearing-more-than-25-in-sorted-array | Python two beginner friendly methods | alishak1999 | 0 | 50 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,092 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/1853075/Python-(Simple-Approach-and-Beginner-friendly) | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
dict = {}
number = int(0.25*len(arr))
for i in arr:
if i not in dict:
dict[i] = 1
else:
dict[i]+=1
for i,n in dict.items():
if n>number:
return i | element-appearing-more-than-25-in-sorted-array | Python (Simple Approach and Beginner-friendly) | vishvavariya | 0 | 26 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,093 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/1848778/1-Line-Python-Solution-oror-90-Faster-oror-Memory-less-than-40 | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
return Counter(arr).most_common(1)[0][0] | element-appearing-more-than-25-in-sorted-array | 1-Line Python Solution || 90% Faster || Memory less than 40% | Taha-C | 0 | 45 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,094 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/1848778/1-Line-Python-Solution-oror-90-Faster-oror-Memory-less-than-40 | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
for a in set(arr):
if arr.count(a)>len(arr)//4: return a | element-appearing-more-than-25-in-sorted-array | 1-Line Python Solution || 90% Faster || Memory less than 40% | Taha-C | 0 | 45 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,095 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/1311342/pyhton-3-%3A-super-simple-solution-using-dict | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
dic = {}
for i in arr :
if i in dic :
dic[i] += 1
else :
dic[i] = 1
return max(dic,key=dic.get) | element-appearing-more-than-25-in-sorted-array | pyhton 3 : super simple solution using dict | rohitkhairnar | 0 | 75 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,096 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/924684/Python%3A-O(N)-Time-%2B-O(1)-Space-(No-CounterZip) | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
# Setting equal to the first element
# since length of the array can be 1
maxInt = arr[0]
maxCount = 1
currInt = arr[0]
currCount = 1
# O(N) linear scan
for i in range(1, len(arr)):
# print (f"arr[i]: {arr[i]}, currInt: {currInt}, currCount: {currCount}, maxInt: {maxInt}, maxCount: {maxCount}")
if arr[i] == currInt:
currCount += 1
else:
if currCount > maxCount:
maxCount = currCount
maxInt = currInt
else:
# refreshing out curr values and setting them
# equal to the new value we just encountered
currCount = 1
currInt = arr[i]
# For the edge case where the solution
# happens to be towards the end
if currCount > maxCount:
return currInt
else:
return maxInt | element-appearing-more-than-25-in-sorted-array | Python: O(N) Time + O(1) Space (No Counter/Zip) | JuanTheDoggo | 0 | 62 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,097 |
https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/discuss/501594/Runtime%3A-88-ms-faster-than-92.56-of-Python3-online-submissions | class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
larr = len(arr)
if larr == 0:
return None
v = len(arr)*.25
prev = None
ct = 0
for e in arr:
if e == prev:
ct += 1
else:
ct = 1
prev = e
if ct > v:
return e
return None | element-appearing-more-than-25-in-sorted-array | Runtime: 88 ms, faster than 92.56% of Python3 online submissions | jcravener | 0 | 123 | element appearing more than 25 percent in sorted array | 1,287 | 0.595 | Easy | 19,098 |
https://leetcode.com/problems/remove-covered-intervals/discuss/1784520/Python3-SORTING-Explained | class Solution:
def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:
res, longest = len(intervals), 0
srtd = sorted(intervals, key = lambda i: (i[0], -i[1]))
for _, end in srtd:
if end <= longest:
res -= 1
else:
longest = end
return res | remove-covered-intervals | ✔️ [Python3] SORTING 👀, Explained | artod | 46 | 1,600 | remove covered intervals | 1,288 | 0.572 | Medium | 19,099 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.