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/concatenation-of-consecutive-binary-numbers/discuss/2613324/Python-Elegant-and-Short-or-One-line-or-Reducing | class Solution:
"""
Time: O(n*log(n))
Memory: O(1)
"""
MOD = 10 ** 9 + 7
def concatenatedBinary(self, n: int) -> int:
return reduce(lambda x, y: ((x << y.bit_length()) | y) % self.MOD, range(1, n + 1)) | concatenation-of-consecutive-binary-numbers | Python Elegant & Short | One line | Reducing | Kyrylo-Ktl | 2 | 181 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,300 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/961371/Python3-1-line | class Solution:
def concatenatedBinary(self, n: int) -> int:
return int("".join(bin(i)[2:] for i in range(1, n+1)), 2) % 1_000_000_007 | concatenation-of-consecutive-binary-numbers | [Python3] 1-line | ye15 | 2 | 134 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,301 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/961371/Python3-1-line | class Solution:
def concatenatedBinary(self, n: int) -> int:
ans = k = 0
for x in range(1, n+1):
if not x & x-1: k += 1
ans = ((ans << k) + x) % 1_000_000_007
return ans | concatenation-of-consecutive-binary-numbers | [Python3] 1-line | ye15 | 2 | 134 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,302 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2613430/python-one-line-solution | class Solution:
def concatenatedBinary(self, n: int) -> int:
return int("".join([bin(i)[2:] for i in range(1, n+1)]), base=2) % 1000000007 | concatenation-of-consecutive-binary-numbers | python one-line solution | secretbit | 1 | 42 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,303 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612449/python3-or-3-lines-of-code-or-easy | class Solution:
def concatenatedBinary(self, n: int) -> int:
bin_str = ''
for i in range(1, n+1): bin_str += bin(i)[2:]
return int(bin_str, 2)%(10**9 + 7) | concatenation-of-consecutive-binary-numbers | python3 | 3 lines of code | easy | H-R-S | 1 | 154 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,304 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612294/Easy-Python-Solution-(4-lines) | class Solution:
def concatenatedBinary(self, n: int) -> int:
ans = ""
for i in range(1, n + 1):
ans += bin(i)[2:]
return int(ans, 2) % (1000000000 + 7) | concatenation-of-consecutive-binary-numbers | Easy Python Solution (4 lines) | Dayeon | 1 | 12 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,305 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612164/One-Liner-explanation-and-code-oror-Python3-Solution-and-Explanation | class Solution:
def concatenatedBinary(self, n: int) -> int:
return int("".join([bin(i)[2:] for i in range(1,n+1)]),2)%1000000007 | concatenation-of-consecutive-binary-numbers | 💚One-Liner explanation and code || Python3 Solution and Explanation | Rage-Fox | 1 | 70 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,306 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2473571/python-solution-oror-easy-to-understand-oror-beginner-friendly | class Solution:
def concatenatedBinary(self, n: int) -> int:
x=""
for i in range(1,n+1):
x+=bin(i).replace("0b","")
return int(x,2)%(10**9+7) | concatenation-of-consecutive-binary-numbers | python solution || easy to understand || beginner friendly🤞✔ | shane6123 | 1 | 54 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,307 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/1037832/PythonPython3-Concatenation-of-Consecutive-Binary-Numbers | class Solution:
def concatenatedBinary(self, n: int) -> int:
final_number = ''
for x in range(1, n+1):
final_number += bin(x)[2:]
return int(final_number, 2) % (10**9 + 7) | concatenation-of-consecutive-binary-numbers | [Python/Python3] Concatenation of Consecutive Binary Numbers | newborncoder | 1 | 123 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,308 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/961626/Python-1-liner | class Solution:
def concatenatedBinary(self, n: int) -> int:
ans=''
for i in range(n+1):
ans+=bin(i)[2:]
return int(ans,2)%(10**9+7) | concatenation-of-consecutive-binary-numbers | Python 1-liner | lokeshsenthilkumar | 1 | 106 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,309 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/961626/Python-1-liner | class Solution:
def concatenatedBinary(self, n: int) -> int:
return int(''.join([bin(i)[2:] for i in range(n+1)]),2)%(10**9+7) | concatenation-of-consecutive-binary-numbers | Python 1-liner | lokeshsenthilkumar | 1 | 106 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,310 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2837562/Brutish-Python-Answer-that-fits-in-time-limits | class Solution:
def concatenatedBinary(self, n: int) -> int:
# get modulo of the current value
modulo_value = 10**9 + 7
# build a string array
binary_array = []
# for value in range 1 to n+1 so you include n
for value in range(1, n+1) :
# take the string binary version of it from 2nd indice forward
binary_array.append(str(bin(value))[2:])
# concat them all
binary_string_of_val = ''.join(binary_array)
# return int cast of the string from binary modulo the modulo value
return int(binary_string_of_val, 2) % modulo_value | concatenation-of-consecutive-binary-numbers | Brutish Python Answer that fits in time limits | laichbr | 0 | 1 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,311 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2829182/Runtime-Beats-79.44-Memory-Beats-81.27 | class Solution:
def concatenatedBinary(self, n: int) -> int:
t = 1
c = 0
for i in range(2,n+1):
c = len(str(bin(i)))-2
tshift = t<<c
t = (tshift+i)%(10**9+7)
return t | concatenation-of-consecutive-binary-numbers | Runtime Beats 79.44% Memory Beats 81.27% | Sashaghoori | 0 | 1 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,312 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2822258/Easy-single-line-Python | class Solution:
def concatenatedBinary(self, n: int) -> int:
return int("".join([bin(x)[2:] for x in range(1, n+1)]), 2) % (10**9 + 7) | concatenation-of-consecutive-binary-numbers | Easy single-line Python | nickheyer | 0 | 1 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,313 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2724162/Simple-python-code-with-explanation | class Solution:
def concatenatedBinary(self, n):
#create a empty sting s to store the contatination of binary values from 1 to n
s = ''
#iterate forloop from 1 to n
for i in range(1,n+1):
#convert each number to binary
k= bin(i)
#for i == 2:
#k will be 0b10
#binary number starts from second index
#so add the binary number k[2:] to string (s)
s = s+ k[2:]
#after ending forloop
#s contains all the binary numbers from 1 to n
#change that string s to integer
res = int(s,2)
#value of ans will be more so do modulo (10^9 + 7)
return res%(1000000007) | concatenation-of-consecutive-binary-numbers | Simple python code with explanation | thomanani | 0 | 5 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,314 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2623313/python3-Bit-shifting-solution-for-reference. | class Solution:
def concatenatedBinary(self, n: int) -> int:
MOD = (10**9+7)
ans = 0
powerOf2 = 0
p2v = 2**powerOf2
for i in range(1, n+1):
if i >= p2v:
powerOf2 += 1
p2v = 2**powerOf2
ans = ans << powerOf2
ans += i
ans = ans % MOD
return ans % MOD | concatenation-of-consecutive-binary-numbers | [python3] Bit shifting solution for reference. | vadhri_venkat | 0 | 16 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,315 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2622373/Python-Solution | class Solution:
def concatenatedBinary(self, n: int) -> int:
l=[]
for i in range(1,n+1):
a=bin(i)[2:]
#print(a)
l.append(a)
#print(l)
n=''.join(l)
n=int(n,2)
return int(n)%((10**9)+7) | concatenation-of-consecutive-binary-numbers | Python Solution | shagun_pandey | 0 | 12 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,316 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2614561/Easy-Python3-Solution | class Solution:
def concatenatedBinary(self, n: int) -> int:
result = ""
for i in range(1,n+1):
result += bin(i)[2:]
return int(result, 2) % ((10**9)+7) | concatenation-of-consecutive-binary-numbers | Easy Python3 Solution | leetcodeninja | 0 | 25 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,317 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2614556/Python3-oror-Illegal-Solution-oror-But-Fast | class Solution:
def concatenatedBinary(self, n: int) -> int:
s = ''
for i in range(1, n+1):
s += bin(i).replace('0b', '')
return int(s, 2) % (10**9 + 7) | concatenation-of-consecutive-binary-numbers | Python3 || Illegal Solution || But Fast | Dewang_Patil | 0 | 14 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,318 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2614374/O(n)-using-bit-counting-and-simulation | class Solution:
def concatenatedBinary(self, n: int) -> int:
mod = 10**9 + 7
ans = 0
k, bits = 31, 1
for ni in range(n, 0, -1):
ans = ans + (ni * bits) % mod
while ni>>k==0:
k -= 1
bits = (bits * (1 << (k + 1))) % mod
ans = ans % mod
return ans | concatenation-of-consecutive-binary-numbers | O(n) using bit counting and simulation | dntai | 0 | 20 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,319 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2614162/Python-Solution | class Solution:
def concatenatedBinary(self, n: int) -> int:
M = 10 ** 9 + 7
s = ""
for i in range(1, n + 1):
s += "{0:b}".format(i)
return int(s, 2) % M | concatenation-of-consecutive-binary-numbers | Python Solution | Harshi0109 | 0 | 16 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,320 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2614088/Python-one-liner-easy-and-simple-solution | class Solution:
def concatenatedBinary(self, n: int) -> int:
return int("".join([format(i,"b") for i in range(1,n+1)]),2)%(10**9+7) | concatenation-of-consecutive-binary-numbers | Python one liner easy and simple solution | SouravSingh49 | 0 | 9 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,321 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2613815/Python-Readable-bit-Manipulation | class Solution:
def concatenatedBinary(self, n: int) -> int:
MOD = 10**9+7
# this is a shifted binary problem
# we need to figure out, how many binary
# digits the number has
# we can keep track of the bit length of
# the number as it will be increasing
#
# every time we encounter a number that
# is a power of two, we increase the bit
# length
result = 0
curr_but_length = 0
for num in range(1, n+1):
# check whether our bit length increased
# num & (num-1) deletes the lowest set bit
# if there is no bit left after deletion
# we have a power of two and need to increase
# our bit length
if num & (num-1) == 0:
curr_but_length += 1
# shift the result upwards for the number of binary digits
# add the current number (by just or, as there will be only
# zeros in the lower bits after shifting)
result = ((result << (curr_but_length)) | num) % MOD
return result | concatenation-of-consecutive-binary-numbers | [Python] - Readable bit Manipulation | Lucew | 0 | 6 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,322 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2613594/Python-simple-solution-2657-ms-15.8-MB | class Solution:
def concatenatedBinary(self, n: int) -> int:
binary_string = ""
for i in range(1, n+1):
binary_string += bin(i)[2:]
return int(binary_string, 2) % (10**9 + 7) | concatenation-of-consecutive-binary-numbers | Python simple solution # 2657 ms 15.8 MB | remenraj | 0 | 8 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,323 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2613566/Python-or-O(n)-97.64-or-simulation-or-bit-manipultion | class Solution:
def concatenatedBinary(self, n: int) -> int:
modulo = 10 ** 9 + 7
shift = 0 # tracking power of 2
res = 0
for i in range(1, n+1):
if i & (i - 1) == 0: # see if num reaches a greater power of 2
shift += 1
res = ((res << shift) + i) % modulo # do the simulation
return res | concatenation-of-consecutive-binary-numbers | Python | O(n), 97.64% | simulation | bit manipultion | MajimaAyano | 0 | 29 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,324 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2613468/concatenation-of-consecutive-binary-numbers | class Solution:
def concatenatedBinary(self, n: int) -> int:
mod=10**9+7
l=[bin(i)[2:] for i in range(1,n+1)]
return int(''.join(l),2)%mod | concatenation-of-consecutive-binary-numbers | concatenation-of-consecutive-binary-numbers | shivansh2001sri | 0 | 7 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,325 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2613359/Python-solution-oror-Short-and-Easy-oror-Using-Bit-manipulation | class Solution:
def concatenatedBinary(self, n: int) -> int:
places_shifted = 0
ans = 0
for num in range(1, n+1):
# places_shifted is the number of bits
if num & (num-1) == 0: places_shifted += 1
# ans will be shifted by places_shifted and num will be added
ans = ((ans << places_shifted) + num) % (10**9 + 7)
return ans | concatenation-of-consecutive-binary-numbers | Python solution || Short and Easy || Using Bit manipulation | vanshika_2507 | 0 | 12 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,326 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2613202/Python3-Solution-or-O(n)-or-Bit-Manipulation | class Solution:
def concatenatedBinary(self, n):
ans, l, mod = 0, 0, 1000_000_007
for i in range(1, n + 1):
if i & (i-1) == 0: l += 1
ans = ((ans << l) + i) % mod
return ans | concatenation-of-consecutive-binary-numbers | ✔ Python3 Solution | O(n) | Bit Manipulation | satyam2001 | 0 | 18 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,327 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2613031/Python-or-Brute-Force-One-liner-or-Do-not-do-this-at-home | class Solution:
def concatenatedBinary(self, n: int) -> int:
return int("".join(bin(i)[2:] for i in range(1, n + 1)), 2) % (10 ** 9 + 7) | concatenation-of-consecutive-binary-numbers | Python | Brute Force One-liner | Do not do this at home | sr_vrd | 0 | 3 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,328 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612989/Python-or-Simplest-Soln-or-Easy-Approch | class Solution:
def concatenatedBinary(self, n: int) -> int:
s=""
for i in range(1,n+1):
s+= str(bin(i)[2:])
return int(s,2)%(10**9+7) | concatenation-of-consecutive-binary-numbers | Python | Simplest Soln | Easy Approch 💯✅ | adarshg04 | 0 | 15 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,329 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612928/Python-Solution-Easy-to-understand-using-bin-function | class Solution:
def concatenatedBinary(self, n: int) -> int:
res = ""
# iterateing till the limit n
for i in range(n+1):
# getting the binary value for each one using bin
# storing it in the variable
# remove the 0b in front of the binary value which is used to
# contrast binary from decimal
res += bin(i).replace('0b', '')
# at last converting it into decimal using int() and 2 represents the base value
# return int(res, 2)%(10**9+7)** | concatenation-of-consecutive-binary-numbers | Python Solution Easy to understand using bin function | Sleur | 0 | 4 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,330 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612877/GolangPython-O(N)-time-or-O(1)-space | class Solution:
def concatenatedBinary(self, n: int) -> int:
s = 0
for i in range(1, n+1):
s = (s << i.bit_length() | i) % 1000000007
return s | concatenation-of-consecutive-binary-numbers | Golang/Python O(N) time | O(1) space | vtalantsev | 0 | 21 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,331 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612791/Python-3-Line-Self-Explanatory | class Solution:
def concatenatedBinary(self, n: int) -> int:
ans=''
for i in range(1,n+1):
ans+=bin(i)[2:]
return int(ans,2)%(10**9+7) | concatenation-of-consecutive-binary-numbers | 🔥🔥🔥Python ,3-Line Self Explanatory🔥🔥🔥 | lakshayne | 0 | 13 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,332 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612783/Python-or-Self-explanatory-or-easy-to-understand | class Solution:
def concatenatedBinary(self, n: int) -> int:
temp=""
for i in range(1,n+1):
temp_bin=bin(i)[2:]
temp+=temp_bin
temp=int(temp,2)
return temp % (10**9+7) | concatenation-of-consecutive-binary-numbers | Python | Self-explanatory | easy to understand | Mom94 | 0 | 11 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,333 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612726/Python-Solution-or-Simple-or-Brute-Force%3A-Accepted | class Solution:
def concatenatedBinary(self, n: int) -> int:
s=''
for i in range(1, n+1):
s+=bin(i).replace("0b", "")
# print(s)
return int(s,2)%1000000007 | concatenation-of-consecutive-binary-numbers | Python Solution | Simple | Brute Force: Accepted | Siddharth_singh | 0 | 10 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,334 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612563/python3-oror-simple-and-easy-oror-2-line | class Solution:
def concatenatedBinary(self, n: int) -> int:
x = "".join(bin(i)[2:] for i in range(1, n+1))
return (int(x, 2) % (10**9+7)) | concatenation-of-consecutive-binary-numbers | python3 || simple and easy || 2 line | Exoutia | 0 | 21 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,335 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612521/Easy-to-understand-One-line-solution | class Solution:
def concatenatedBinary(self, n: int) -> int:
return int(''.join([format(i,'b') for i in range(n+1)]),2) %(10**9+7) | concatenation-of-consecutive-binary-numbers | Easy to understand One line solution | Endale | 0 | 6 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,336 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612513/Python-Solution | class Solution:
def concatenatedBinary(self, n: int) -> int:
sr = ""
for i in range(1,n+1):
sr += bin(i)[2:]
return int(sr,2)%(10**9+7) | concatenation-of-consecutive-binary-numbers | Python Solution | a_dityamishra | 0 | 13 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,337 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612423/Python-One-Line-Solution | class Solution:
def concatenatedBinary(self, n: int) -> int:
return int("".join(bin(i+1)[2:] for i in range(n)), 2) % (10**9+7) | concatenation-of-consecutive-binary-numbers | Python One Line Solution | cheetah5 | 0 | 20 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,338 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612416/python3-easy-solution(4-line) | class Solution:
def concatenatedBinary(self, n: int) -> int:
# first get the binary value of each number and concatinate
s=""
for x in range(1,n+1):
s+=bin(x)[2:]
# now change the concatinated binary to decimal and do the moudulo operation
return(int(s,2)%(10**9+7)) | concatenation-of-consecutive-binary-numbers | python3 easy solution(4 line) | priyam_jsx | 0 | 15 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,339 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612372/python-easy-approach | class Solution:
def concatenatedBinary(self, n: int) -> int:
binary = ""
for i in range(1,n+1):
#convert the decimal to binary
bindata = format(i,"b")
#add them together
binary += bindata
#convert binary to decimal
ans = int(binary,2)
return ans % (10**9 + 7) | concatenation-of-consecutive-binary-numbers | python easy approach | Zao_Needs_Buff | 0 | 9 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,340 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/2612148/Python-Solution-or-one-liner | class Solution:
def concatenatedBinary(self, n: int) -> int:
return int(''.join('{0:b}'.format(i) for i in range(1, n+1)), 2) % (10 ** 9 + 7) | concatenation-of-consecutive-binary-numbers | Python Solution | one-liner | everydayspecial | 0 | 29 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,341 |
https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/discuss/1831270/Python-Solution | class Solution:
def concatenatedBinary(self, n: int) -> int:
s=''
for i in range(1, n+1):
s+=bin(i).replace("0b", "")
# print(s)
return int(s,2)%1000000007 | concatenation-of-consecutive-binary-numbers | Python Solution | Siddharth_singh | 0 | 38 | concatenation of consecutive binary numbers | 1,680 | 0.57 | Medium | 24,342 |
https://leetcode.com/problems/minimum-incompatibility/discuss/965262/Python3-backtracking | class Solution:
def minimumIncompatibility(self, nums: List[int], k: int) -> int:
nums.sort()
def fn(i, cand):
"""Populate stack and compute minimum incompatibility."""
nonlocal ans
if cand + len(nums) - i - sum(not x for x in stack) > ans: return
if i == len(nums): ans = cand
else:
for ii in range(k):
if len(stack[ii]) < len(nums)//k and (not stack[ii] or stack[ii][-1] != nums[i]) and (not ii or stack[ii-1] != stack[ii]):
stack[ii].append(nums[i])
if len(stack[ii]) == 1: fn(i+1, cand)
else: fn(i+1, cand + stack[ii][-1] - stack[ii][-2])
stack[ii].pop()
ans = inf
stack = [[] for _ in range(k)]
fn(0, 0)
return ans if ans < inf else -1 | minimum-incompatibility | [Python3] backtracking | ye15 | 3 | 156 | minimum incompatibility | 1,681 | 0.374 | Hard | 24,343 |
https://leetcode.com/problems/minimum-incompatibility/discuss/965262/Python3-backtracking | class Solution:
def minimumIncompatibility(self, nums: List[int], k: int) -> int:
nums.sort()
def fn(i, cand=0, ans=inf):
"""Return minimum incompatibility via backtracking."""
if cand + len(nums) - i - sum(not x for x in stack) >= ans: return ans # early return
if i == len(nums): return cand
for ii in range(k):
if len(stack[ii]) < len(nums)//k and (not stack[ii] or stack[ii][-1] != nums[i]) and (not ii or stack[ii-1] != stack[ii]):
stack[ii].append(nums[i])
if len(stack[ii]) == 1: ans = fn(i+1, cand, ans)
else: ans = fn(i+1, cand + stack[ii][-1] - stack[ii][-2], ans)
stack[ii].pop()
return ans
stack = [[] for _ in range(k)]
ans = fn(0)
return ans if ans < inf else -1 | minimum-incompatibility | [Python3] backtracking | ye15 | 3 | 156 | minimum incompatibility | 1,681 | 0.374 | Hard | 24,344 |
https://leetcode.com/problems/minimum-incompatibility/discuss/962339/Python-or-2-solutions-or-1-Brute-force-or-1-Memoization-(AC) | class Solution:
def minimumIncompatibility(self, nums: List[int], k: int) -> int:
nums.sort()
x = len(nums)//k
combos = []
for i in itertools.combinations(nums, x):
if len(set(i)) == x:
combos.append(i)
result = float('inf')
for i in itertools.combinations(combos, k):
if sorted([y for x in i for y in x]) == nums:
score = sum([max(x) - min(x) for x in i])
result = min(score, result)
return result if result != float("inf") else -1 | minimum-incompatibility | Python | 2 solutions | 1 Brute force | 1 Memoization (AC) | dev-josh | 0 | 69 | minimum incompatibility | 1,681 | 0.374 | Hard | 24,345 |
https://leetcode.com/problems/minimum-incompatibility/discuss/962339/Python-or-2-solutions-or-1-Brute-force-or-1-Memoization-(AC) | class Solution:
def minimumIncompatibility(self, nums: List[int], k: int) -> int:
partition_len = len(nums) // k
@functools.lru_cache(maxsize=None)
def recurse(nums):
if not nums: return 0
result = float('inf')
for combo in itertools.combinations(nums, partition_len):
if len(set(combo)) < partition_len: continue
updated_nums = list(nums)
for i in combo:
updated_nums.remove(i)
result = min(
result,
max(combo) - min(combo) + recurse(tuple(updated_nums))
)
return result
result = recurse(tuple(nums))
return result if result != float('inf') else -1 | minimum-incompatibility | Python | 2 solutions | 1 Brute force | 1 Memoization (AC) | dev-josh | 0 | 69 | minimum incompatibility | 1,681 | 0.374 | Hard | 24,346 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1054303/Python-simple-one-liner | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
return sum(set(allowed) >= set(i) for i in words) | count-the-number-of-consistent-strings | Python - simple one liner | angelique_ | 12 | 828 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,347 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1577520/python-sol-using-sorted()-and-set() | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
count = 0
for i in range(len(words)):
if sorted(set(list(words[i] + allowed))) == list(sorted(allowed)): count += 1
return count | count-the-number-of-consistent-strings | python sol using sorted() and set() | anandanshul001 | 6 | 567 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,348 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1347474/Python-solution-%3A-EXPLAINED-%3A-with-fastest-in-time(90)-and-lowest(99.5)-in-space. | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
"""Brute Force
Time complexity : O(n*m) where m = max(len(word) in words)
Space complexity: O(1)
"""
count = 0
for word in words:
for ch in word:
if ch not in allowed:
count += 1
break
return len(words)-count
"""One liner
Set operation
"""
return sum(not set(word)-set(allowed) for word in words)
"""Set operation by subset operation
NOTE : SPACE COMPLEXITY INCREASES IN THIS APPROACH.
"""
allowed = set(allowed)
n = 0
for w in words:
if set(w).issubset(allowed):
n += 1
return n | count-the-number-of-consistent-strings | Python solution : EXPLAINED : with fastest in time(90%) and lowest(99.5%) in space. | er1shivam | 5 | 316 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,349 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1163984/Python3-Brute-Force | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
count = 0
for i in words:
f = True
for j in set(i):
if(j not in allowed):
f = False
break
if(f):
count += 1
return count | count-the-number-of-consistent-strings | [Python3] Brute Force | VoidCupboard | 4 | 212 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,350 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1056181/Python-simple-brute-force-or-95-memory-and-time | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
count = 0
allowed = set(allowed)
for i in words:
for letter in i:
if letter not in allowed:
count += 1
break
return len(words) - count | count-the-number-of-consistent-strings | Python simple brute force | 95% memory and time | vanigupta20024 | 4 | 440 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,351 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1316220/Python-3-oror-Faster-than-93oror-Easy-to-understand | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
not_allowed=0
#counts the number of not allowed words
for word in words:
for letter in word:
if letter not in allowed:
not_allowed+=1
break
#this word is not allowed move to next word
return len(words)-not_allowed | count-the-number-of-consistent-strings | Python 3 || Faster than 93%|| Easy to understand | ana_2kacer | 2 | 158 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,352 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/969684/1-liner | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
s=set(allowed)
return sum(all(c in s for c in w) for w in words) | count-the-number-of-consistent-strings | 1-liner | lokeshsenthilkumar | 2 | 252 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,353 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2478717/Python-simple-solution | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
count = 0
for word in words:
temp = set()
for i in word:
temp.add(i)
for i in temp:
if i not in allowed:
break
else:
count += 1
return count | count-the-number-of-consistent-strings | Python simple solution | aruj900 | 1 | 134 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,354 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2430112/Python-Solution | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
c,w,a = 0,words,set(allowed)
for i in range(len(w)):
if not((set(w[i]))-a):
c += 1
return c | count-the-number-of-consistent-strings | Python Solution | SouravSingh49 | 1 | 60 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,355 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2117548/easy | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
return sum(set(allowed) >= set(i) for i in words)
# set(allowed) >= set(i) all characters in i are also in allowed | count-the-number-of-consistent-strings | easy | writemeom | 1 | 43 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,356 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2015631/Pythin-easy-to-read-and-understand | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
ans = 0
for word in words:
ans += 1
for ch in word:
if ch not in allowed:
ans -= 1
break
return ans | count-the-number-of-consistent-strings | Pythin easy to read and understand | sanial2001 | 1 | 118 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,357 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1731928/8-lines-of-Python | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
allowed = list(allowed)
count = len(words)
for s in words:
for c in s:
if(c not in allowed):
count -= 1
break
return count | count-the-number-of-consistent-strings | 8 lines of Python | Depender | 1 | 179 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,358 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1306687/Easy-Python-Solution(98.46) | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
c=0
for i in words:
for j in i:
if(j not in allowed):
c+=1
break
return len(words)-c | count-the-number-of-consistent-strings | Easy Python Solution(98.46%) | Sneh17029 | 1 | 335 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,359 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1243574/Python3-easy-solution-with-90-best-time | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
count = 0
for word in words:
for char in word:
if char not in allowed:
count += 1
break
return len(words) - count | count-the-number-of-consistent-strings | Python3 easy solution with 90% best time | albezx0 | 1 | 75 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,360 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1188680/Python-list-comp-and-sets | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
return len([1 for word in words if set(word).issubset(allowed)]) | count-the-number-of-consistent-strings | Python list comp and sets | fast_typer | 1 | 196 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,361 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1135783/Python3-easy-to-understand. | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
count = 0
checked = 0 # Will be used to see if all the letters of word[i] == allowed
for i in words:
for j in i:
if j in allowed:
checked +=1
if checked == len(i):
count+=1
checked=0
return count | count-the-number-of-consistent-strings | Python3 easy to understand. | lukaigrutinovic | 1 | 200 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,362 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1026929/Python3-1-line | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
return sum(all(c in allowed for c in word) for word in words) | count-the-number-of-consistent-strings | [Python3] 1-line | ye15 | 1 | 74 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,363 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1026929/Python3-1-line | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
return sum(not set(word) - set(allowed) for word in words) | count-the-number-of-consistent-strings | [Python3] 1-line | ye15 | 1 | 74 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,364 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2846656/Python-or-2-Liner-or-Using-Set | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
set_allowed = set(allowed)
return sum(set(word).issubset(set_allowed) for word in words) | count-the-number-of-consistent-strings | Python | 2 Liner | Using Set | pawangupta | 0 | 1 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,365 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2804491/Python-Clean-Code | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
ans = 0
lookup = set(allowed)
for string in words:
is_allowed = True
for character in string:
if character not in lookup:
is_allowed = False
break
if is_allowed: ans += 1
return ans
# TC: O(N); SC: O(1) | count-the-number-of-consistent-strings | Python Clean Code | okhadeanimesh | 0 | 3 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,366 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2804233/Memory-97oror-Python3 | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
k=0
for i in words:
x=False
for j in i:
if j in allowed:
x=True
else:
x=False
break
if x:
k+=1
return k | count-the-number-of-consistent-strings | Memory 97%|| Python3 | Samandar_Abduvaliyev | 0 | 7 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,367 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2803818/Python-Easy-Understand-Solution | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
consistent_string_count = 0
for word in words:
for char in word:
if char not in allowed:
break
else:
consistent_string_count += 1
return consistent_string_count | count-the-number-of-consistent-strings | Python Easy Understand Solution | namashin | 0 | 2 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,368 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2783466/Simple-Python-Solution-oror-Easy-and-Explained | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
allowed = set(allowed)
count = 0
for word in words:
for letter in word:
if letter not in allowed:
count += 1
break
return len(words) - count | count-the-number-of-consistent-strings | Simple Python Solution || Easy & Explained | dnvavinash | 0 | 2 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,369 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2767671/Python-3-Using-Bit-Manipulation | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
pat = 0
for i in allowed:
pat |= 1<<(ord(i)-ord("a"))
ans = 0
for word in words:
temp = 0
for i in word:
temp |= 1<<(ord(i)-ord("a"))
if temp|pat==pat:
ans += 1
return ans
``` | count-the-number-of-consistent-strings | [Python 3] Using Bit Manipulation | user2667O | 0 | 3 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,370 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2752114/Fastest-1-line-Python-3-solution-100-faster-than-any-other-codes | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
allowed = set(allowed)
count = 0
for word in words:
for letter in word:
if letter not in allowed:
count += 1
break
return len(words) - count | count-the-number-of-consistent-strings | Fastest 1 line Python 3 solution 100 faster than any other codes | avs-abhishek123 | 0 | 4 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,371 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2744865/O(n)-or-Python-or-2-pointer-approach | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
not_allowed_word_count = 0
w, c = 0, 0
while w < len(words):
if words[w][c] not in allowed:
not_allowed_word_count += 1
w += 1
c = 0
continue
if c == len(words[w]) - 1:
w += 1
c = 0
continue
c += 1
return len(words) - not_allowed_word_count | count-the-number-of-consistent-strings | O(n) | Python | 2 pointer approach | kawamataryo | 0 | 4 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,372 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2730477/Python3-Solution-beats-98.1 | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
count_consistent = len(words)
for word in words:
for w in word:
if w not in allowed:
count_consistent -= 1
break
return count_consistent | count-the-number-of-consistent-strings | Python3 Solution - beats 98.1% | sipi09 | 0 | 2 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,373 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2547168/EASY-PYTHON3-SOLUTION | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
res=0
for i in words:
if (w := set(i)) & set(allowed) == w: #If the set of "i" is equal to set of "allowed", + 1 to the output
res+=1
return (res) | count-the-number-of-consistent-strings | 🔥 EASY PYTHON3 SOLUTION 🔥 | rajukommula | 0 | 42 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,374 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2509787/Python-or-One-liner-or-Intersection | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
return sum([1 for word in words if (w := set(word)) & set(allowed) == w]) | count-the-number-of-consistent-strings | Python | One-liner | Intersection | Wartem | 0 | 65 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,375 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2415520/Easy-Python3-or-A.issubset(B) | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
allowed = set(allowed)
count = 0
for el in words:
if set(el).issubset(allowed) == True:
count += 1
return count | count-the-number-of-consistent-strings | Easy Python3 | A.issubset(B) | Sergei_Gusev | 0 | 22 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,376 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2408685/1684.-Count-the-Number-of-Consistent-Strings%3A-One-Liner | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
return sum([1 for element in words if set(element).issubset(allowed)]) | count-the-number-of-consistent-strings | 1684. Count the Number of Consistent Strings: One Liner | rogerfvieira | 0 | 27 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,377 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2381183/Memory-Usage-less-than-97.93-and-faster-than-74.20 | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
count = 0
for word in words:
leftover = set(word) - set(allowed)
if len(leftover) == 0:
count += 1
return count | count-the-number-of-consistent-strings | Memory Usage less than 97.93% and faster than 74.20% | samanehghafouri | 0 | 43 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,378 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2342797/python3-solution | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
res = 0
for word in words:
for char in set(word):
if char not in allowed:
res += 1
break
return len(words) - res | count-the-number-of-consistent-strings | python3 solution | jd_mahmud | 0 | 58 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,379 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2326330/Python3-Solution-with-using-hashset | class Solution:
def _helper(self, word, allowed_set):
for char in word:
if char not in allowed_set:
return False
return True
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
allowed_set = set(allowed)
res = 0
for word in words:
if self._helper(word, allowed_set):
res += 1
return res | count-the-number-of-consistent-strings | [Python3] Solution with using hashset | maosipov11 | 0 | 22 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,380 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2180558/Python-Solution-Easy-to-Understand | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
c=0
for i in range(len(words)):
for j in range(len(allowed)):
if allowed[j] in words[i]:
words[i]=words[i].replace(allowed[j],"")
if len(words[i])==0:
c+=1
break
return c | count-the-number-of-consistent-strings | Python Solution - Easy to Understand | T1n1_B0x1 | 0 | 39 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,381 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2123145/Easy-python-solution | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
c = 0
at = Counter(allowed)
for word in words:
ct = Counter(word)
if set(list(ct.keys())).issubset(set(list(at.keys()))):
c += 1
return c | count-the-number-of-consistent-strings | Easy python solution | nikhitamore | 0 | 101 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,382 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2121826/Python-simple-solution | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
ans = 0
for word in words:
for letter in word:
if letter not in allowed:
break
else:
ans += 1
return ans | count-the-number-of-consistent-strings | Python simple solution | StikS32 | 0 | 74 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,383 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2068683/Python3-SImple-Beats-99 | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
allowed_set = set(allowed)
count = 0
for word in words:
flag = True
for letter in word:
if letter not in allowed_set:
flag = False
break
if flag:
count+=1
return count
Runtime: 212 ms, faster than 99.45% of Python3 online submissions for Count the Number of Consistent Strings.
Memory Usage: 16 MB, less than 89.80% of Python3 online submissions for Count the Number of Consistent Strings. | count-the-number-of-consistent-strings | Python3 SImple Beats 99% | emerald19 | 0 | 99 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,384 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2027204/Python-Clean-and-Concise! | class Solution:
def countConsistentStrings(self, allowed, words):
allowed = set(allowed)
return sum(all(c in allowed for c in set(word)) for word in words) | count-the-number-of-consistent-strings | Python - Clean and Concise! | domthedeveloper | 0 | 85 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,385 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/2014246/Python-Solution-or-Using-Set-Intersection | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
allowed = set(allowed)
count = 0
for word in words:
intsct = set(word).intersection(allowed)
if set(word) == intsct:
count += 1
return count | count-the-number-of-consistent-strings | Python Solution | Using Set Intersection | ssshekhu53 | 0 | 49 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,386 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1891284/Python3-One-Loop-Using-Set-and-Sorted-Solution | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
res = 0
allowed = "".join(sorted(allowed))
for k in words:
if allowed == "".join(sorted(set(k+allowed))):
res += 1
else:
return res | count-the-number-of-consistent-strings | ✔Python3 One Loop Using Set and Sorted Solution | khRay13 | 0 | 96 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,387 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1865886/Python-solution-using-sets | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
allowed_set = set(allowed)
count = 0
for i in words:
if not(set(i).difference(allowed_set)):
count += 1
return count | count-the-number-of-consistent-strings | Python solution using sets | alishak1999 | 0 | 119 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,388 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1839407/Easy-Python-Solution-or-Faster-than-77.84 | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
a=set(list(allowed))
ans=0
for i in words:
t=set(list(i))
if t.issubset(a):
ans+=1
return ans
``` | count-the-number-of-consistent-strings | Easy Python Solution | Faster than 77.84 % | manav023 | 0 | 106 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,389 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1801210/Easy-Python-Loop | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
c=0
for i in words:
for j in set(i):
if j not in allowed:
c+=1
# print(c)
break
return len(words)-c | count-the-number-of-consistent-strings | Easy Python Loop | adityabaner | 0 | 113 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,390 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1654662/Runtime%3A-512-ms-faster-than-5.06-of-Python3-online-submissions | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
length = 0
for i in words:
str1 = ""
for j in range(len(i)):
if i[j] in allowed:
str1 += i[j]
if len(i) == len(str1):
length += 1
return length | count-the-number-of-consistent-strings | Runtime: 512 ms, faster than 5.06% of Python3 online submissions | piyushkumarpiyushkumar6 | 0 | 29 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,391 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1561404/Easy-Python3-solution | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
allowed_set: set = set(list(allowed))
consistent: int = 0
for word in words:
for char in word:
if char not in allowed_set:
break
else:
consistent += 1
return consistent | count-the-number-of-consistent-strings | Easy Python3 solution | sirenescx | 0 | 144 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,392 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1556335/Python-Easy-Solution-or-Faster-than-98 | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
count = len(words)
for word in words:
for w in word:
if w not in allowed:
count -= 1
break
return count | count-the-number-of-consistent-strings | Python Easy Solution | Faster than 98% | leet_satyam | 0 | 142 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,393 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1355273/python-3-solution | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
count=0
for i in range (0,len(words)):
for j in range(0,len(words[i])):
if words[i][j] not in allowed:
count=count+1
break
return len(words)-count | count-the-number-of-consistent-strings | python 3 solution | minato_namikaze | 0 | 99 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,394 |
https://leetcode.com/problems/count-the-number-of-consistent-strings/discuss/1348465/Python-3-oror-Using-Single-For-Loop | class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
consistentStringCount = 0
setAllowed = set(allowed)
for word in words:
setWord = set(word)
if len(setWord - setAllowed) == 0:
consistentStringCount +=1
return consistentStringCount | count-the-number-of-consistent-strings | Python 3 || Using Single For Loop | SP_Roger | 0 | 87 | count the number of consistent strings | 1,684 | 0.819 | Easy | 24,395 |
https://leetcode.com/problems/sum-of-absolute-differences-in-a-sorted-array/discuss/1439047/Python-3-or-O(N)-Prefix-Sum-Clean-or-Explanation | class Solution:
def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:
pre_sum = [0]
for num in nums: # calculate prefix sum
pre_sum.append(pre_sum[-1] + num)
n = len(nums) # render the output
return [(num*(i+1) - pre_sum[i+1]) + (pre_sum[-1]-pre_sum[i] - (n-i)*num) for i, num in enumerate(nums)] | sum-of-absolute-differences-in-a-sorted-array | Python 3 | O(N), Prefix Sum, Clean | Explanation | idontknoooo | 3 | 327 | sum of absolute differences in a sorted array | 1,685 | 0.649 | Medium | 24,396 |
https://leetcode.com/problems/sum-of-absolute-differences-in-a-sorted-array/discuss/2821092/O(n)-time-complexity-beats-97.5-python | class Solution:
def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:
total_sum = sum(nums)
N = len(nums)
ans = []
l_sum = 0
l_covered = 0
for i in range(N):
r_size = N-i-1
r_sum = total_sum - l_sum - nums[i]
s = l_covered*nums[i] - l_sum + r_sum - r_size*nums[i]
# print(i, l_covered*nums[i] - l_sum, r_sum - r_size*nums[i])
ans.append(s)
l_sum += nums[i]
l_covered += 1
return ans | sum-of-absolute-differences-in-a-sorted-array | O(n) time complexity, beats 97.5%, python | Rohit_Annamaneni | 0 | 2 | sum of absolute differences in a sorted array | 1,685 | 0.649 | Medium | 24,397 |
https://leetcode.com/problems/sum-of-absolute-differences-in-a-sorted-array/discuss/2758888/Python-3%3A-Linear-time-complexity | class Solution:
def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:
length = len(nums)
output = length*[0]
pos= sum([abs(item - nums[0]) for item in nums])
neg = 0
output[0] = pos
for i in range(1,length):
diff = nums[i] - nums[i-1]
neg += diff*i
pos -= diff*(length-i)
output[i] = pos + neg
return output
return summand_lst | sum-of-absolute-differences-in-a-sorted-array | Python 3: Linear time complexity | cleon082 | 0 | 12 | sum of absolute differences in a sorted array | 1,685 | 0.649 | Medium | 24,398 |
https://leetcode.com/problems/sum-of-absolute-differences-in-a-sorted-array/discuss/2738667/Simple-and-Easy-to-Understand-or-O(n)-Time-Complexity-or-Python | class Solution(object):
def getSumAbsoluteDifferences(self, nums):
lSum, rSum, l, r = 0, sum(nums), 0, len(nums) - 1
ans = []
for n in nums:
rSum -= n
ans.append((n * l) - lSum + rSum - (n * r))
lSum += n
l += 1
r -= 1
return ans | sum-of-absolute-differences-in-a-sorted-array | Simple and Easy to Understand | O(n) Time Complexity | Python | its_krish_here | 0 | 22 | sum of absolute differences in a sorted array | 1,685 | 0.649 | Medium | 24,399 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.