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/queries-on-number-of-points-inside-a-circle/discuss/1353057/One-liner-78-speed | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
return [sum(pow(x - xc, 2) + pow(y - yc, 2) - r * r <= 0
for x, y in points) for xc, yc, r in queries] | queries-on-number-of-points-inside-a-circle | One liner, 78% speed | EvgenySH | 1 | 193 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,100 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/2835422/python-code | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
m=len(points)
n=len(queries)
ans=[]
for i in range(n):
count=0
for j in range(m):
if (queries[i][0]-points[j][0])**2 + (queries[i][1]-points[j][1])**2 <= queries[i][2]**2:
count+=1
ans.append(count)
return ans | queries-on-number-of-points-inside-a-circle | python code | ayushigupta2409 | 0 | 5 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,101 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/2799913/Python-beats-80-(Clean-solution) | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
innerPointsCount = []
for i, q in enumerate(queries):
count = 0
for j, p in enumerate(points):
if ((p[0] - q[0]) ** 2) + ((p[1] - q[1]) ** 2) <= q[2] ** 2: count += 1
innerPointsCount.append(count)
return innerPointsCount | queries-on-number-of-points-inside-a-circle | Python beats 80% (Clean solution) | farruhzokirov00 | 0 | 4 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,102 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/2692623/Python3-Simple-Solution | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
res = []
for Cx, Cy, r in queries:
count = 0
for x, y in points:
if (Cx - x)**2 + (Cy - y)**2 <= r**2:
count += 1
res.append(count)
return res | queries-on-number-of-points-inside-a-circle | Python3 Simple Solution | mediocre-coder | 0 | 5 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,103 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/2663463/Python-Easy-understanding | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
result = []
for i in range(len(queries)):
cx = queries[i][0]
cy = queries[i][1]
r = queries[i][2]
total = 0
for j in range(len(points)):
x = points[j][0]
y = points[j][1]
distance = (cx - x)**2 + (cy-y)**2
distance = distance**(1/2)
if distance <= r:
total += 1
result.append(total)
return result | queries-on-number-of-points-inside-a-circle | Python Easy understanding | Noisy47 | 0 | 12 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,104 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/2657456/Python3-Keeping-It-Integer | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
c = [0 for i in range(len(queries))]
for i in range(len(queries)):
q = queries[i]
r2 = q[-1]*q[-1]
for p in points:
d2 = (q[0]-p[0])**2 + (q[1]-p[1])**2
if d2 <= r2:
c[i] += 1
return c | queries-on-number-of-points-inside-a-circle | Python3 Keeping It Integer | godshiva | 0 | 2 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,105 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/2615317/Python-Basic-Math-(Iterative-solution) | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
l=[]
for circle in range(len(queries)):
count=0
for point in range(len(points)):
if (queries[circle][0]-points[point][0])**2 + (queries[circle][1]-points[point][1])**2 <= queries[circle][2]**2:
count+=1
l.append(count)
return l | queries-on-number-of-points-inside-a-circle | Python - Basic Math (Iterative solution) | utsa_gupta | 0 | 38 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,106 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/2583760/Python-One-Liner-for-fun | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
return [sum(((px-rx)**2 + (py-ry)**2 <= r**2) for px, py in points) for rx, ry, r in queries] | queries-on-number-of-points-inside-a-circle | [Python] - One-Liner for fun | Lucew | 0 | 29 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,107 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/2491260/Simple-python-solution | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
res = []
for circle in queries:
x,y,r = circle
count = 0
for point in points:
xi,yi = point
distance = sqrt((xi - x)**2 + (yi - y)**2)
if distance <=r:
count += 1
res.append(count)
return res | queries-on-number-of-points-inside-a-circle | Simple python solution | aruj900 | 0 | 50 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,108 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/2227446/python-solution-oror-one-liner-oror-My-first-one-liner | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
return [sum([math.dist(point, [query[0], query[1]]) <= query[2] for point in points]) for query in queries] | queries-on-number-of-points-inside-a-circle | python solution || one liner || My first one-liner | lamricky11 | 0 | 44 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,109 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/2212612/Python-Euclidean-distance-solution | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
# Euclidean distance should be less than or equal to radius
res = []
for query in queries:
count = 0
for point in points:
if sqrt(pow((query[0] - point[0]), 2) + pow((query[1] - point[1]), 2)) <= query[2]:
count += 1
res.append(count)
return res | queries-on-number-of-points-inside-a-circle | Python Euclidean distance solution | Gp05 | 0 | 30 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,110 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/1769788/Queries-on-Number-of-Points-Inside-a-Circle-python-solution | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
fans = []
count = 0
for i in range(len(queries)):
for j in range(len(points)):
if (points[j][0]-queries[i][0])**2 + (points[j][1]-queries[i][1])**2 <= (queries[i][2])**2:
count+=1
fans.append(count)
answer = [fans[0]]
for x in range(1,len(fans)):
answer.append(fans[x] - fans[x-1])
return answer | queries-on-number-of-points-inside-a-circle | Queries on Number of Points Inside a Circle python solution | seabreeze | 0 | 78 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,111 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/1309676/Python-Simple-Mathor-Memory-lesser-than-90 | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
#(x-x1)**2+(y-y1)**2<=r**2 for point to lie inside a circle
res=[0]*(len(queries))
i=0
for x,y,r in queries:
for x1,y1 in points:
if (x-x1)**2+(y-y1)**2<=r**2:
res[i]+=1
i+=1
return res | queries-on-number-of-points-inside-a-circle | Python Simple Math| Memory lesser than 90% | ana_2kacer | 0 | 105 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,112 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/1275658/Python3-simple-%22one-liner%22-solution | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
return [sum([1 for j in points if (i[0]-j[0])**2+(i[1]-j[1])**2-i[2]**2 <= 0]) for i in queries] | queries-on-number-of-points-inside-a-circle | Python3 simple "one-liner" solution | EklavyaJoshi | 0 | 86 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,113 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/1163145/Python3-brute-force | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
ans = []
for x, y, r in queries:
val = 0
for xx, yy in points:
if (x-xx)**2 + (y-yy)**2 <= r**2:
val += 1
ans.append(val)
return ans | queries-on-number-of-points-inside-a-circle | [Python3] brute-force | ye15 | 0 | 39 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,114 |
https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/discuss/1163125/Python-Solution | class Solution:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
res = []
for x,y,r in queries:
rsq = pow(r,2)
count = 0
for px,py in points:
ans = pow(px-x,2)+pow(py-y,2)
if ans<= rsq:
count += 1
res.append(count)
return res | queries-on-number-of-points-inside-a-circle | Python Solution | SaSha59 | 0 | 104 | queries on number of points inside a circle | 1,828 | 0.864 | Medium | 26,115 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1281679/Python3-solution-using-single-for-loop | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
res = []
for i in range(1,len(nums)):
res.append(2**maximumBit - 1 - nums[i-1])
nums[i] = nums[i-1]^nums[i]
res.append(2**maximumBit - 1 - nums[-1])
return res[::-1] | maximum-xor-for-each-query | Python3 solution using single for loop | EklavyaJoshi | 3 | 111 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,116 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1167144/Python-3-One-liner-faster-than-100 | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
return list(accumulate([nums[0] ^ 2 ** maximumBit - 1] + nums[1:], ixor))[::-1] | maximum-xor-for-each-query | [Python 3] One-liner faster than 100% | kingjonathan310 | 3 | 160 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,117 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1167144/Python-3-One-liner-faster-than-100 | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
return list(accumulate([nums[0] ^ (1 << maximumBit) - 1] + nums[1:], ixor))[::-1] | maximum-xor-for-each-query | [Python 3] One-liner faster than 100% | kingjonathan310 | 3 | 160 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,118 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/2771136/Simple-solution-beats-99 | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
ans = [0] * len(nums)
x = (2**maximumBit-1)
for i, n in enumerate(nums):
x = x ^ n
ans[-1-i] = x
return ans | maximum-xor-for-each-query | Simple solution beats 99% | Mencibi | 1 | 23 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,119 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/2845941/Python-or-2-line-or-o(n)-or-well-commented-or-prefix-sum-or-one-pass-or-bit-manipulation | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
m,prefix_xor = 2**maximumBit - 1 , list(accumulate(nums,xor))[::-1]
return (i^m for i in prefix_xor)
"""
2 3 4 7 ( Given )
010 011 100 111 (given in binary)
010 001 101 010 (since constrain is high && we have to repeatively calculate xor
we should use prefix-xor)
^ k k k k (k'th val to xor to get all set bit i.e 111)
111 111 111 111 ( this is result and we have to find val of k for each query)
NOTE: i ^ k = res, then i ^ res = k
So take the prefix-xor and xor every ele with all set bit.
all set-bit = pow(2,maximumBit) - 1
prefix-xor = accumulate(nums,xor)
make sure to use prefix-xor in reverse order because
in question we have asked xor of complete list in first query
""" | maximum-xor-for-each-query | Python | 2-line | o(n) | well-commented | prefix-sum | one-pass | bit-manipulation | YaBhiThikHai | 0 | 1 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,120 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/2646520/Python3-or-Prefix-Sum-With-Bit-Manip.-Fact-Max-XOR-2maximumBit-1 | class Solution:
#Let n = len(nums)!
#Time-Complexity: O(n + n) -> O(N)
#Space-Complexity: O(N)
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
#Approach:
#1. First, initialize prefix sum array of exclusive or results up to each index position!
prefix_sum = [nums[0]]
for a in range(1, len(nums)):
prefix_sum.append(prefix_sum[len(prefix_sum)-1] ^ nums[a])
#2.For each query, take the prefix sum of current index starting from the end and moving left, and
#utilize binary search on search space of possible values for non-negative integer k, that is
#maximal and satisfies the inequality!
comparison = 2 ** maximumBit
ans = []
#for each query
for i in range(len(prefix_sum)-1, -1, -1):
#get the res so far!
current_res = prefix_sum[i]
#largest XOR is going to be 2^maximumBit - (1)!
#Thus, knowing current_res prefix result and largest possible XOR,
#if we exclusive or btw the two, we will be able to find
#non-negative k that maximizes XOR for current query!
res = current_res ^ (comparison - 1)
ans.append(res)
return ans | maximum-xor-for-each-query | Python3 | Prefix Sum With Bit Manip. Fact Max XOR = 2^maximumBit - 1 | JOON1234 | 0 | 21 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,121 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1571069/Python-solution | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
prefix = [nums[0]]
for idx in range(1, len(nums)):
prefix.append(prefix[idx-1] ^ nums[idx])
print(prefix)
ans = []
maxValue = 2 ** maximumBit
for idx in range(len(nums)-1, -1, -1):
k=prefix[idx] ^ (maxValue-1)
ans.append(k)
return ans | maximum-xor-for-each-query | Python solution | akshaykumar19002 | 0 | 62 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,122 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1527716/Python-or-O(n)-Solution-or-Prefix-array-or-Simple-approach | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
pre=[]
x=0
ans=[]
for i in range(len(nums)):
x=x^nums[i]
pre.append(x)
bit=2**maximumBit
for i in range(len(nums)-1,-1,-1):
k=pre[i]^(bit-1)
ans.append(k)
return ans | maximum-xor-for-each-query | Python | O(n) Solution | Prefix array | Simple approach | meghanakolluri | 0 | 108 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,123 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1234115/O(N)-solution-WIth-Concept-of-XORING-Explained | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
list_arr,xor = [],0
for i in range(0,len(nums)) :
xor ^= nums[i]
list_arr.append(xor ^ ((1 << maximumBit)-1))
return list_arr[::-1] | maximum-xor-for-each-query | O(N) solution WIth Concept of XORING Explained | lankesh | 0 | 76 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,124 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1165473/Python3-O(n)-Solution-with-Explanation-and-Comments-100-fast-and-100-memory-efficient | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
'''
1. The maximum value achieved from a number of maximumBit is 2^(maximumBit)-1
2. Need to find the d in the equation a XOR b XOR c XOR d = x
3. If a XOR b XOR c XOR d = x, then a XOR b XOR c XOR x = d
4. Starting from index 0, XOR nums[i] with 2^(maximumBit)-1 and store in the list
5. Up until the last element, XOR nums[i] with the previous XOR result and append to the list
6. Return the list in reverse order
'''
maxm = 2**maximumBit - 1 #the maximum value of maxiumBit count
res = [nums[0]^maxm] #store the initial xor result in res list
XOR = res[0]
if len(nums)>=2:
for num in nums[1:]:
XOR^=num #bitwise XOR of previous xor result with nums[i]
res.append(XOR)
return res[::-1] #return the reverse list | maximum-xor-for-each-query | Python3 O(n) Solution with Explanation and Comments, 100% fast and 100% memory efficient | bPapan | 0 | 44 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,125 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1164591/100-faster-100-less-time-python-solution | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
res = reduce(operator.xor, nums) # taking the xor of 1st n elements
lim = (1<<maximumBit)-1 #
store = []
for val in nums[::-1]:
store.append(res^lim)
res = res^val
return store | maximum-xor-for-each-query | 100% faster, 100% less time python solution | deleted_user | 0 | 41 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,126 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1163923/Simple-Python3-solution-with-O(n)-DP-one-time-list-walkthrough | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
# Compute XOR for each query
# e.g., [0, 1, 1, 3]
# first query k: k XOR (0 XOR 1 XOR 1 XOR 3) = 2**(maximumBit)-1
# => k = 2**(maxiumBit)-1 XOR (0 XOR 1 XOR 1 XOR 3)
# second query k: k XOR (0 XOR 1 XOR 1) = 2**(maximumBit)-1
# => k = 2**(maxiumBit)-1 XOR (0 XOR 1 XOR 1)
for i in range(len(nums)):
if i == 0:
nums[i] = nums[i] ^ (2**(maximumBit)-1)
continue
nums[i] = nums[i] ^ nums[i-1]
return nums[::-1] | maximum-xor-for-each-query | Simple Python3 solution with O(n) DP one-time list walkthrough | tkuo-tkuo | 0 | 34 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,127 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1163431/Python3-Solution | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
maxiNum = 2**maximumBit
ans = []
xor = 0
for i in nums:
xor ^= i
while len(nums) != 0:
num,sum_ = 0,0
for i in range(32):
num = 1<<i
if (num<=(maxiNum-1)):
if not xor&num:
sum_ += num
else:
break
ans.append(sum_)
last = nums.pop()
xor ^= last
return ans | maximum-xor-for-each-query | Python3 Solution | swap2001 | 0 | 34 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,128 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1163217/Python3-1-pass | class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
ans = [0]*len(nums)
prefix = 0
for i, x in enumerate(nums):
prefix ^= x
ans[~i] = prefix ^ ((1 << maximumBit) -1)
return ans | maximum-xor-for-each-query | [Python3] 1-pass | ye15 | 0 | 38 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,129 |
https://leetcode.com/problems/maximum-xor-for-each-query/discuss/1163115/Python-Solution | class Solution:
'''
Step 1: Building a xor array.
eg :[0,1,1,3]
xor = [0]
now 0 xor 1 [0,1]
now 1 xor 1 [0,1,0]
finally 0 xor 3 [0,1,0,3]
step 2:
Now to find the value of k at each query we just want to take the len(2^maxbit-1 )so here its 2 digit(maxBit =2 so 2^2 = 4 ) .
no.of bits in 3 are 2
value of k will be inverse of xor value[i]
eg :
when i = 0 xor[i] = 0 what will be the value of k in this case when bits are 2
00 xor inverse(00)
inverse(00) = 11
hence k = 3
'''
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
xor = []
maxb = len(bin(pow(2,maximumBit)-1)[2:])
n = len(nums)
xor.append(nums[0])
for i in range(1,n): #step 1
xor.append(nums[i]^xor[i-1])
res = []
for xor_val in xor:
bin_xor_val = list(bin(xor_val)[2:])
temp = []
for i in range(maxb-len(bin_xor_val)):
temp.append('1')
for bit in bin_xor_val: #performing inverse of the xor value
if bit == '0':
temp.append('1')
else:
temp.append('0')
res.append(int(''.join(temp),2))
res.reverse()
return res | maximum-xor-for-each-query | Python Solution | SaSha59 | 0 | 54 | maximum xor for each query | 1,829 | 0.769 | Medium | 26,130 |
https://leetcode.com/problems/minimum-number-of-operations-to-make-string-sorted/discuss/1202007/Python3-math-solution | class Solution:
def makeStringSorted(self, s: str) -> int:
freq = [0]*26
for c in s: freq[ord(c) - 97] += 1
MOD = 1_000_000_007
fac = cache(lambda x: x*fac(x-1)%MOD if x else 1)
ifac = cache(lambda x: pow(fac(x), MOD-2, MOD)) # Fermat's little theorem (a**(p-1) = 1 (mod p))
ans, n = 0, len(s)
for c in s:
val = ord(c) - 97
mult = fac(n-1)
for k in range(26): mult *= ifac(freq[k])
for k in range(val): ans += freq[k] * mult
n -= 1
freq[val] -= 1
return ans % MOD | minimum-number-of-operations-to-make-string-sorted | [Python3] math solution | ye15 | 1 | 240 | minimum number of operations to make string sorted | 1,830 | 0.491 | Hard | 26,131 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2712076/Multiple-solution-in-python | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
lst=[0]*26
for i in sentence:
lst[ord(i)-ord('a')]+=1
return 0 not in lst | check-if-the-sentence-is-pangram | Multiple solution in python | shubham_1307 | 19 | 732 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,132 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2712076/Multiple-solution-in-python | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(sentence))==26 | check-if-the-sentence-is-pangram | Multiple solution in python | shubham_1307 | 19 | 732 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,133 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1166400/Simple-Python-solution_O(n) | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(sentence)) == 26 | check-if-the-sentence-is-pangram | Simple Python solution_O(n) | smaranjitghose | 11 | 732 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,134 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1277373/Python-2-different-approaches-or-beats-96-time-O(1)-solution! | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
# naive approach - 1
# freq = {}
# for i in sentence:
# freq[i] = freq.get(i, 0) + 1
# if len(freq) == 26: return True
# return False
# optimized approach - 2
occurred = 0
for i in sentence:
temp = ord(i) - ord('a')
occurred |= (1 << temp)
if occurred == (1 << 26) - 1:
return True
return False | check-if-the-sentence-is-pangram | Python 2 different approaches | beats 96% time, O(1) solution! | vanigupta20024 | 8 | 684 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,135 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2573442/SIMPLE-PYTHON3-SOLUTION-ONE-LINE-FASTEr | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return(len(set(list(sentence))) == 26) | check-if-the-sentence-is-pangram | ✅✔ SIMPLE PYTHON3 SOLUTION ✅✔ ONE-LINE FASTEr | rajukommula | 7 | 158 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,136 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2250518/Python3-O(n)-oror-O(1)-or-O(26)-Runtime%3A-40ms-71.99-oror-Memory%3A-13.9mb-54.83 | class Solution:
# O(n) || O(1) because we are dealing with lower case english alphabets O(26)
# Runtime: 40ms 71.99% || Memory: 13.9mb 54.83%
def checkIfPangram(self, sentence: str) -> bool:
allAlpha = [False] * 26
for char in sentence:
index = ord(char) - ord('a')
allAlpha[index] = True
return all(allAlpha[i] for i in range(26)) | check-if-the-sentence-is-pangram | Python3 O(n) || O(1) or O(26) # Runtime: 40ms 71.99% || Memory: 13.9mb 54.83% | arshergon | 4 | 96 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,137 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2715601/Python-solution-using-list | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
counter = [0 for _ in range(97, 123)]
for i in sentence:
counter[ord(i)-97] += 1
for i in counter:
if i == 0:
return False
return True | check-if-the-sentence-is-pangram | 📌 Python solution using list | croatoan | 2 | 55 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,138 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2714052/Easy-Python-or-One-liner | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return(len(set(sentence))==26) | check-if-the-sentence-is-pangram | Easy Python | One-liner | tusharkhanna575 | 2 | 47 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,139 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1867123/Python-solution-faster-than-95-using-sets | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
if len(set(sentence)) == 26:
return True
return False | check-if-the-sentence-is-pangram | Python solution faster than 95%, using sets | alishak1999 | 2 | 105 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,140 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1552514/TWO-LINER-SOLUTION-using-sorted()-and-set()-(faster-than-99.49) | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
sentence, string = sorted(set(sentence)), "abcdefghijklmnopqrstuvwxyz"
return sentence == list(string) | check-if-the-sentence-is-pangram | TWO LINER SOLUTION using sorted() and set() (faster than 99.49%) | anandanshul001 | 2 | 87 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,141 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2714393/Python-one-liner-easy-solution-93.35-faster | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(sentence))==26 | check-if-the-sentence-is-pangram | Python one-liner easy solution 93.35% faster | mg_112002 | 1 | 24 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,142 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2713272/Python-oror-O(n)-oror-Easy-Understanding | class Solution:
def checkIfPangram(self, s: str) -> bool:
l = [0]*26
for i in range(len(s)):
l[ord(s[i])-ord('a')] = 1
return sum(l)==26 | check-if-the-sentence-is-pangram | Python || O(n) || Easy Understanding | its_iterator | 1 | 24 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,143 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2712420/python3oror-one-lineoror-O(N) | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return 26==len(set(sentence)) | check-if-the-sentence-is-pangram | [python3|| one line|| O(N)] | Sneh713 | 1 | 4 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,144 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2712300/Easy-Python-1-liner.-Beats-91 | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return set('abcdefghijklmnopqrstuvwxyz') == set(sentence) | check-if-the-sentence-is-pangram | Easy Python 1 liner. Beats 91% | sukumar-satapathy | 1 | 48 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,145 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1688489/Python-One-Liner | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(sentence)) == 26 | check-if-the-sentence-is-pangram | Python One Liner | pratushah | 1 | 85 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,146 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1585063/Python-oror-28ms-oror-Simple-hashmap-oror-Method-2%3A-set | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
english = "qwertyuiopasdfghjklzxcvnmb"
di = {ch:1 for ch in english}
for ch in sentence:
di[ch]+=1
for ch in english:
if di[ch]<2:
return False
return True | check-if-the-sentence-is-pangram | Python || 28ms || Simple hashmap || Method 2: set | ana_2kacer | 1 | 114 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,147 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1585063/Python-oror-28ms-oror-Simple-hashmap-oror-Method-2%3A-set | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
check = set(sentence)
if len(check)==26:
return True
return False | check-if-the-sentence-is-pangram | Python || 28ms || Simple hashmap || Method 2: set | ana_2kacer | 1 | 114 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,148 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1520060/Another-Python-one-liner | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return set(sentence) == set('abcdefghijklmnopqrstuvwxyz') | check-if-the-sentence-is-pangram | Another Python one-liner | SmittyWerbenjagermanjensen | 1 | 63 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,149 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1387091/Python-3-Solution | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
s = "abcdefghijklmnopqrstuvwxyz"
for i in range(len(s)):
if s[i] in sentence:
flag= 1
else:
flag = 0
break
if flag==0:
return False
else:
return True | check-if-the-sentence-is-pangram | Python 3 Solution | deleted_user | 1 | 102 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,150 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1236650/ONE-LINER-PYTHON-CODE-EASY-TO-UNDERSTAND-FASTER-THAN-90 | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(list(sentence))) == 26
```` | check-if-the-sentence-is-pangram | ONE LINER PYTHON CODE EASY TO UNDERSTAND FASTER THAN 90% | shubhamdec10 | 1 | 151 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,151 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/1231889/Python-Simply-Easy-Solution-1-Line-Faster-than-97 | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(sentence)) >= 26 | check-if-the-sentence-is-pangram | Python Simply Easy Solution 1 Line Faster than 97% | Inesh | 1 | 104 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,152 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2843072/lessless-python-on-line-very-easy-solution-greatergreater | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(sentence)) == 26; | check-if-the-sentence-is-pangram | <<-- python on line very easy solution -->> | seifsoliman | 0 | 3 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,153 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2841887/Logic-way | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
d = {}
c = 0
for i in sentence:
if not d.get(i):
d[i] = True
c += 1
return c == 26 | check-if-the-sentence-is-pangram | Logic way | Abraha111 | 0 | 1 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,154 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2826298/Beats-99-Easy-Python-Solution | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
if len(sentence) < 26:
return False
else:
for i in range(97, 123):
if chr(i) not in sentence:
return False
return True | check-if-the-sentence-is-pangram | Beats 99% - Easy Python Solution | PranavBhatt | 0 | 4 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,155 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2793295/python-one-liner | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(sentence))==26 | check-if-the-sentence-is-pangram | [python]--one liner | user9516zM | 0 | 1 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,156 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2792708/Python-or-Easy-or-Faster-than-99.33 | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
# For iterating over range of characters from a to z
for i in range(97, 123):
# Check if each character is present in sentence or not
if chr(i) not in sentence:
# If any character is not present in sentence, then return False and stop further processing
return False
# return True if all the characters are present
return True | check-if-the-sentence-is-pangram | Python | Easy | Faster than 99.33% | pawangupta | 0 | 3 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,157 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2771286/GolangRustpython-solution | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
if len(set(list(sentence))) == 26:
return True
return False | check-if-the-sentence-is-pangram | Golang,Rust,python solution | anshsharma17 | 0 | 1 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,158 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2757444/Python-solution-using-set-to-check-if-the-letters-are-unique-in-a-sentence | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
set_sentence = set(sentence)
if len(set_sentence) == 26:
return True
return False | check-if-the-sentence-is-pangram | Python solution using set to check if the letters are unique in a sentence | samanehghafouri | 0 | 2 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,159 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2754769/Python3-Solution-oror-Set-oror-One-Liner-Solution | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(sentence)) == 26 | check-if-the-sentence-is-pangram | Python3 Solution || Set || One-Liner Solution | shashank_shashi | 0 | 2 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,160 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2744562/Python-Set-Easy | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(sentence))==26 | check-if-the-sentence-is-pangram | Python Set Easy | ben_wei | 0 | 1 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,161 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2737292/Python-Two-Solutions-(Including-One-Liner) | class Solution:
#Most Time Efficient
def checkIfPangram(self, sentence: str) -> bool:
checked = []
for s in sentence:
if s not in checked:
checked.append(s)
return len(checked) == 26
#Most Space Efficient
def checkIfPangram(self, sentence: str) -> bool:
return len(set(list(sentence)))==26 | check-if-the-sentence-is-pangram | Python Two Solutions (Including One-Liner) | keioon | 0 | 3 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,162 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2731698/solution-in-python-3 | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
ascii = []
for x in range(97, 123, 1):
ascii.append(x)
for y in sentence:
if ord(y) in ascii:
ascii.remove(ord(y))
if len(ascii) != 0:
return False
else:
return True | check-if-the-sentence-is-pangram | solution in python 3 | ramana721 | 0 | 1 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,163 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2731683/Panagram-solution-with-string.ascii_lowercase-python | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
for i in string.ascii_lowercase:
if i not in sentence:
return False
return True | check-if-the-sentence-is-pangram | Panagram solution with string.ascii_lowercase, python | arshadali7860 | 0 | 2 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,164 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2724167/Simple-Python-Solution | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
alphabets = dict()
for i in range(26):
alphabets[chr(i + ord('a'))] = 0
for c in sentence:
alphabets[c] += 1
for i in range(26):
if alphabets[chr(i + ord('a'))] == 0:
return False
return True | check-if-the-sentence-is-pangram | Simple Python Solution | mansoorafzal | 0 | 3 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,165 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2721237/Python-Simple-Python-Solution-By-Checking-all-Character | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
alpha = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
check = False
for char in alpha:
if char not in sentence:
return False
return True | check-if-the-sentence-is-pangram | [ Python ] ✅✅ Simple Python Solution By Checking all Character 🥳✌👍 | ASHOK_KUMAR_MEGHVANSHI | 0 | 2 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,166 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2717355/Simple-One-Liner-or-Python | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(sentence))==26 | check-if-the-sentence-is-pangram | Simple One Liner | Python | Abhi_-_- | 0 | 3 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,167 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2715771/Python3-using-hash-table. | class Solution:
def __init__(self) -> None:
self.d = {}
def checkIfPangram(self, sentence: str) -> bool:
count = 0
for s in sentence:
if s not in self.d:
self.d[s] = True
count += 1
return count == 26 | check-if-the-sentence-is-pangram | Python3 using hash table. | parryrpy | 0 | 3 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,168 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2715688/Python3-solution-memory-beats%3A-99.80 | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
char_dict = {chr(i):0 for i in range(97,123)}
for i in sentence:
char_dict[i] += 1
return min(char_dict.values()) | check-if-the-sentence-is-pangram | Python3 solution - memory beats: 99.80% | sipi09 | 0 | 1 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,169 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2715507/Python-or-Faster-than-97-or-Easy-3-line-Solution | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
if len(sentence) < 26 : return False
res = set(list(sentence))
return not len(res) < 26 | check-if-the-sentence-is-pangram | ✔Python | Faster than 97% | Easy 3 line Solution | sarveshmantri200 | 0 | 4 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,170 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2715391/Python3!-1-Line-solution. | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(collections.Counter(sentence)) == 26 | check-if-the-sentence-is-pangram | 😎Python3! 1 Line solution. | aminjun | 0 | 4 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,171 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2715337/Beginner-Friendly-and-Fast-Solution | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
alpha = "abcdefghijklmnopqrstuvwxyz"
sentence_alpha = ""
for i in sentence:
if i not in sentence_alpha:
sentence_alpha += i
return True if alpha == ''.join(sorted(sentence_alpha)) else False | check-if-the-sentence-is-pangram | Beginner Friendly and Fast Solution | user6770yv | 0 | 2 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,172 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2714970/PYTHON-SOLUTION-USING-SET | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
if len(set(sentence))==26:
return True
else:
return False | check-if-the-sentence-is-pangram | PYTHON SOLUTION USING SET | mritunjayyy | 0 | 1 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,173 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2714907/Simple-solution-in-Python | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return (len(set(sentence))== 26 )
# occurred = 0
# for i in sentence:
# temp = ord(i) - ord('a')
# occurred |= (1 << temp)
# if occurred == (1 << 26) - 1:
# return True
# return False | check-if-the-sentence-is-pangram | Simple solution in Python | Baboolal | 0 | 2 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,174 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2714777/one-Line-code-in-python-best-Solution | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
lst = list(sentence)
return len(set(lst))==26 | check-if-the-sentence-is-pangram | one Line code in python best Solution | kartik_5051 | 0 | 1 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,175 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2714743/PYHTON-oror-ONE-LINER | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(set(list(sentence)))==26 | check-if-the-sentence-is-pangram | [PYHTON] || ONE LINER | SuganthSugi | 0 | 3 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,176 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2714642/Python-or-Beats-98-in-Time-and-Space-Complexity | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
alphabet = set()
for char in sentence:
alphabet.add(char)
return len(alphabet) >= 26 | check-if-the-sentence-is-pangram | Python | Beats 98% in Time & Space Complexity | rschevenin | 0 | 1 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,177 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2714586/python3-shortest-and-elegant-solution | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
return len(Counter(sentence))==26 | check-if-the-sentence-is-pangram | python3 shortest and elegant solution | benon | 0 | 3 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,178 |
https://leetcode.com/problems/check-if-the-sentence-is-pangram/discuss/2714542/python-easy-solution-using-set | class Solution:
def checkIfPangram(self, sentence: str) -> bool:
s=set(sentence)
if len(s)==26:
return True
return False | check-if-the-sentence-is-pangram | python easy solution using set | lalli307 | 0 | 1 | check if the sentence is pangram | 1,832 | 0.839 | Easy | 26,179 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1165500/Python3-with-Explanation-100-faster-and-100-memory-efficient | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
'''
1. If the minimum of all costs is greater than amount of coins, the boy can't buy any bar, return 0
2. Else, sort the list of costs in a non-decreasing order
3. For each 'cost' in costs, if the cost is less than current coins
-increase the count of ice cream bars that can be bought by 1
-decrease the current coins amount by 'cost'
4. If the cost is greater than current coins, return the ice cream bar count value
'''
if min(costs)>coins: #minimum cost is greater than the coins available
return 0 #can't buy any ice cream bar
costs=sorted(costs) #sort the list of costs in a non-decreasing order
res = 0 #the resultant count of ice cream bars that can be bought
for cost in costs:
if cost<=coins: #in this case, the boy can buy the ice cream bar
res+=1 #increase the ice cream bar count
coins-=cost #spent an amount equal to 'cost', decrease current coins amount by cost
else:
break #not enough coins, return the bars count
return res | maximum-ice-cream-bars | Python3 with Explanation, 100% faster and 100% memory efficient | bPapan | 3 | 279 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,180 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/2490078/Easy-Python-Solution.....Faster-than-100 | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
result=0
costs.sort()
for i in costs:
if coins<i:
break
result+=1
coins-=i
return result | maximum-ice-cream-bars | Easy Python Solution.....Faster than 100% | guneet100 | 1 | 76 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,181 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1175446/Python3-simple-solution-using-two-approaches | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
return len([i for i in itertools.accumulate(sorted(costs)) if i <= coins]) | maximum-ice-cream-bars | Python3 simple solution using two approaches | EklavyaJoshi | 1 | 74 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,182 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1175446/Python3-simple-solution-using-two-approaches | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
costs.sort()
l = [costs[0]]
c = costs[0]
for i in range(1,len(costs)):
c += costs[i]
l.append(c)
return len([i for i in l if i <= coins]) | maximum-ice-cream-bars | Python3 simple solution using two approaches | EklavyaJoshi | 1 | 74 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,183 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/2834166/Simple-or-Python3-or-C%2B%2B | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
costs = sorted(costs)
total, ans = 0, 0
for cost in costs:
if total + cost <= coins:
total += cost
ans += 1
else:
break
return ans | maximum-ice-cream-bars | Simple | Python3 | C++ | joshua_mur | 0 | 1 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,184 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/2755610/Simple-and-Efficient-Python-Solution-Greedy-Approach | class Solution:
def maxIceCream(self, cost: List[int], coins: int) -> int:
cost.sort(); count = 0
for i in cost:
if (coins >= i):
count += 1; coins -= i;
else: break
return count | maximum-ice-cream-bars | Simple and Efficient Python Solution } Greedy Approach | avinashdoddi2001 | 0 | 3 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,185 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/2743204/Python3-No-direct-Sort-Unique-Costs-and-Heap | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
# we can count the elements
cn = collections.Counter(costs)
# we can do a min heap from the costs
costs = list(cn.keys())
heapq.heapify(costs)
# pop the elements
ans = 0
for _ in range(len(costs)):
# pop the costs
cost = heapq.heappop(costs)
# get the amount of costs
fit = coins // cost
# subtract the costs
if cn[cost] < fit:
ans += cn[cost]
coins -= cn[cost]*cost
else:
ans += fit
break
return ans | maximum-ice-cream-bars | [Python3] - No direct Sort - Unique Costs and Heap | Lucew | 0 | 1 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,186 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/2729452/Python-O(NlogN)-O(N) | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
heap = []
left = coins
for cost in costs:
if cost <= left:
heapq.heappush(heap, -cost)
left -= cost
continue
if heap and -heap[0] > cost:
left += -heapq.heappop(heap)
heapq.heappush(heap, -cost)
left -= cost
return len(heap) | maximum-ice-cream-bars | Python - O(NlogN), O(N) | Teecha13 | 0 | 2 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,187 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/2713955/Python3-sort%2Bgreedy | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
costs.sort()
if costs[0] > coins:
return 0
i = 0
while i < len(costs) and coins > 0:
if costs[i] > coins:
break
coins -= costs[i]
i += 1
return i | maximum-ice-cream-bars | Python3 sort+greedy | skrrtttt | 0 | 2 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,188 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/2673594/Python-solution | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
costs.sort()
count = 0
for i in costs:
if i <= coins:
count+=1
coins = coins-i
return count | maximum-ice-cream-bars | Python solution | divyanshikathuria | 0 | 3 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,189 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/2610626/Python-easy-solution | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
costs.sort()
res = 0
for i in range(len(costs)):
if coins >= costs[i]:
coins -= costs[i]
res +=1
return res | maximum-ice-cream-bars | Python easy solution | anshsharma17 | 0 | 6 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,190 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/2538119/easy-python-solution | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
if sum(costs) <= coins :
return len(costs)
else :
costs.sort()
counter = 0
for ice in costs :
if ice > coins :
return counter
counter += 1
coins -= ice
return counter | maximum-ice-cream-bars | easy python solution | sghorai | 0 | 14 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,191 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1500808/Python-O(nlogn)-time-O(1)-space-solution | class Solution(object):
def maxIceCream(self, arr, target):
"""
:type costs: List[int]
:type coins: int
:rtype: int
"""
n = len(arr)
arr.sort(reverse=True)
res = 0
while len(arr) > 0 and target > 0:
p = arr.pop()
if p <= target:
target -= p
res += 1
else:
break
return res | maximum-ice-cream-bars | Python O(nlogn) time, O(1) space solution | byuns9334 | 0 | 84 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,192 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1420038/Python3-or-explained-or-easy-solution | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
# you have to get as much ice cream you can buy
# so the best idea is to sort the list in order to have the smallest values from the start
# count them while iterating and subtract the price after buying the ice cream
costs.sort()
n = len(costs)
count = 0
for i in range(n):
if costs[i] <= coins:
count += 1
coins -= costs[i]
else:
break
return count | maximum-ice-cream-bars | Python3 | explained | easy solution | FlorinnC1 | 0 | 86 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,193 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1353114/Sort-and-loop-84-speed | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
spend_coins = 0
costs.sort()
for i, cost in enumerate(costs):
if spend_coins + cost <= coins:
spend_coins += cost
else:
return i
return len(costs) | maximum-ice-cream-bars | Sort and loop, 84% speed | EvgenySH | 0 | 43 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,194 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1206168/python3-easy-solution | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
costs.sort()
count=0
for i in range(len(costs)):
coins-=costs[i]
if coins>=0:
count+=1
else:
break
return count | maximum-ice-cream-bars | python3 easy solution | akashmaurya001 | 0 | 68 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,195 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1199204/Simple-Python-Sorting-and-Greedy-approach-O(n)-time-and-O(1)-space | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
costs.sort() # sort the array in increasing order
# starting from 0th elem greedily pick ice-cream bars untill we run out of coins
n, idx = len(costs), 0
while(idx < n and coins-costs[idx] >= 0):
coins -= costs[idx]
idx+=1
return idx | maximum-ice-cream-bars | Simple Python Sorting and Greedy approach O(n) time and O(1) space | thecoder_elite | 0 | 59 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,196 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1166920/Python3-greedy | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
for i, x in enumerate(sorted(costs)):
if x <= coins: coins -= x
else: return i
return len(costs) | maximum-ice-cream-bars | [Python3] greedy | ye15 | 0 | 42 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,197 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1166920/Python3-greedy | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
ans = 0
for x in sorted(costs):
if x <= coins:
ans += 1
coins -= x
else: break
return ans | maximum-ice-cream-bars | [Python3] greedy | ye15 | 0 | 42 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,198 |
https://leetcode.com/problems/maximum-ice-cream-bars/discuss/1164270/Python3-Heap-Solution | class Solution:
def maxIceCream(self, costs: List[int], coins: int) -> int:
result = 0
min_heap = []
heapq.heapify(min_heap)
for cost in costs:
heapq.heappush(min_heap,cost)
while min_heap and coins:
temp = heapq.heappop(min_heap)
if temp <= coins:
coins -= temp
result += 1
return result | maximum-ice-cream-bars | [Python3] Heap Solution | pratushah | 0 | 42 | maximum ice cream bars | 1,833 | 0.656 | Medium | 26,199 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.