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/number-of-steps-to-reduce-a-number-to-zero/discuss/2394669/Python-solution-Num-of-steps-to-reduce-a-num-to-zero | class Solution:
def numberOfSteps(self, num: int) -> int:
count = 0
while(num):
if(num%2==0):
num = num/2
else:
num = num-1
count+=1
return count | number-of-steps-to-reduce-a-number-to-zero | Python solution - Num of steps to reduce a num to zero | gurudadhich76 | 0 | 28 | number of steps to reduce a number to zero | 1,342 | 0.854 | Easy | 20,200 |
https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/discuss/2381047/Time-efficient-Python-using-bit-manipulation | class Solution:
def numberOfSteps(self, num: int) -> int:
''' Intuitive Method '''
# res = 0
# while num > 0:
# num = num // 2 if num % 2 == 0 else num - 1
# res += 1
# return res
''' Binary Method '''
binary_expression = bin(num)[2:] # to get rid of '0b'
return len(binary_expression) + binary_expression.count('1') - 1 | number-of-steps-to-reduce-a-number-to-zero | Time-efficient Python using bit manipulation | MajimaAyano | 0 | 73 | number of steps to reduce a number to zero | 1,342 | 0.854 | Easy | 20,201 |
https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/discuss/2376900/Python-recursive-solution-Fastest | class Solution:
def numberOfSteps(self, num: int) -> int:
s=0
def ops(n,s):
# print(n,s)
if n == 0:
return s
if (n & 1):
s += 1
return ops(n-1,s)
else:
s += 1
return ops(n//2,s)
return ops(num,s)
``` | number-of-steps-to-reduce-a-number-to-zero | Python recursive solution Fastest | Vigneshbabupj | 0 | 68 | number of steps to reduce a number to zero | 1,342 | 0.854 | Easy | 20,202 |
https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/discuss/2362339/Python-or-Recursive | class Solution:
def recursive_solve(self, num: int, step: int) -> int:
if num > 0:
return self.recursive_solve(num - 1 if num % 2 else num // 2, step + 1)
return step
def numberOfSteps(self, num: int) -> int:
return self.recursive_solve(num, 0) | number-of-steps-to-reduce-a-number-to-zero | Python | Recursive | c1nn4 | 0 | 25 | number of steps to reduce a number to zero | 1,342 | 0.854 | Easy | 20,203 |
https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/discuss/2354099/Faster-than-80 | class Solution:
def numberOfSteps(self, num: int) -> int:
steps = 0
while num != 0:
if num % 2 == 0:
num = num/2
else:
num -= 1
steps += 1
return steps | number-of-steps-to-reduce-a-number-to-zero | Faster than 80% | Arrstad | 0 | 29 | number of steps to reduce a number to zero | 1,342 | 0.854 | Easy | 20,204 |
https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/discuss/2294052/Highly-efficient-Python-Solution | class Solution:
def numberOfSteps(self, num: int) -> int:
steps = 0
while num:
if num % 2:
num -= 1
steps += 1
else:
num /= 2
steps += 1
return steps | number-of-steps-to-reduce-a-number-to-zero | Highly efficient Python Solution | samsub15 | 0 | 123 | number of steps to reduce a number to zero | 1,342 | 0.854 | Easy | 20,205 |
https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/discuss/2253645/Python-Simplest-Solution-With-Explanation-or-Beg-to-adv-or-Math | class Solution:
def numberOfSteps(self, num: int) -> int:
count = 0 # taking a couter to count number of steps to reduce it to zero
while num>0: # nums should not be less then 0
if num % 2 == 0: # if num is completely divide by zero
num = num//2 # to have the quotient
count+=1 # as number got reduced we will increase the counter
else:
num-=1 # and if num is not even then we subtract one out of it.
count+=1 # as number got reduced we will increase the counter
return count # returning the step occurred o reduce the number. | number-of-steps-to-reduce-a-number-to-zero | Python Simplest Solution With Explanation | Beg to adv | Math | rlakshay14 | 0 | 60 | number of steps to reduce a number to zero | 1,342 | 0.854 | Easy | 20,206 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/1816745/Python-Solution-oror-Sliding-Window | class Solution:
def numOfSubarrays(self, arr, k, threshold):
windowStart = 0
max_avg = 0
avg = 0
c=0
result = []
windowSum = 0
for windowEnd in range(len(arr)):
windowSum += arr[windowEnd]
if((windowEnd)>=k-1):
avg = windowSum//k
result.append(avg)
windowSum -= arr[windowStart]
windowStart += 1
for i in range(len(result)):
if(result[i]>=threshold):
c=c+1
return c | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Python Solution || Sliding Window | aashutoshjha21022002 | 3 | 281 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,207 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/2797739/Python-oror-91.78Faster-oror-Sliding-Window-oror-O(N)-Solution | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
c,j,n=0,0,len(arr)
s=sum(arr[:k])
if s>=k*threshold:
c=1
for i in range(k,n):
s+=arr[i]-arr[j]
if s>=k*threshold:
c+=1
j+=1
return c | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Python || 91.78%Faster || Sliding Window || O(N) Solution | DareDevil_007 | 2 | 89 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,208 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/1125577/Python3-sliding-window | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
ans = sm = 0
for i, x in enumerate(arr):
sm += x
if i >= k: sm -= arr[i-k]
if i+1 >= k and sm >= k*threshold: ans += 1
return ans | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | [Python3] sliding window | ye15 | 2 | 151 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,209 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/1714628/Python-Simple-Sliding-Window-explained | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
# start with the sum of first k elements
summ = sum(arr[:k])
# set out(result) variable to 1
# if the first set of sum is valid
# i.e its avg >= threshold
out = 1 if summ//k >= threshold else 0
# start the loop from 1, since we've
# already taken into account the elements
# from i = 0.
# go till k elements less than the length
# that is the length of our window and since
# it has to be inclusive, add 1 to the range
for i in range(1, len(arr)-k+1):
# remove the last element from the sum
# and add the next element (i+k-1)
summ -= arr[i-1]
summ += arr[i+k-1]
# increment counter if avg >= threshold
if summ//k >= threshold: out += 1
return out | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | [Python] Simple Sliding Window explained | buccatini | 1 | 136 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,210 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/1617475/easy-python3-solution | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
n=len(arr)
res=0
s=sum(arr[:k])
if s/k>=threshold:
res+=1
for i in range(1,n-k+1):
s-=arr[i-1]
s+=arr[i+k-1]
if s/k>=threshold:
res+=1
return res | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | easy python3 solution | Karna61814 | 1 | 59 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,211 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/1468412/Python-Sliding-Window-approach | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
wstart = wsum = count = 0
for wend in range(len(arr)):
wsum += arr[wend]
if wend >= k:
wsum -= arr[wstart]
wstart += 1
if (wsum//k) >= threshold and (wend-wstart+1) == k:
count += 1
return count | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Python Sliding Window approach | _r_rahul_ | 1 | 142 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,212 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/2799176/Python3-or-Basic-Sliding-Window-Solution | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
i = 0;j = k-1;n = len(arr)
ans = 0
windowSum = sum(arr[:k-1])
while(j<n):
windowSum+=arr[j]
if(windowSum/k>=threshold):
ans+=1
windowSum-=arr[i]
i+=1;j+=1
return ans | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Python3 | Basic Sliding Window Solution | ty2134029 | 0 | 2 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,213 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/2710093/Clean-Sliding-Window-or-Python | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
n = len(arr)
i, j, s, res, count = 0, 0, threshold*k, 0, 0
while j<n:
res += arr[j]
if j-i+1<k:
j += 1
else:
if res>=s: count += 1
res -= arr[i]
i += 1; j += 1
return count | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Clean Sliding Window | Python | RajatGanguly | 0 | 9 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,214 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/2653616/simple-and-easy-sliding-window-solution-for-begginers | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
ws=0
we=0
sumi=0
ans=[]
c=0
for we in range(len(arr)):
sumi+=arr[we]
if we>=k-1:
ans.append(sumi//k)
sumi-=arr[ws]
ws+=1
for i in ans:
if i>=threshold:
c+=1
return c | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | simple and easy sliding window solution for begginers | insane_me12 | 0 | 3 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,215 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/2645381/Python3-Solution-oror-O(N)-Time-and-O(1)-Space-Complexity | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
count=0
sum=0
n=len(arr)
for i in range(k):
sum+=arr[i]
if sum>=threshold*k:
count+=1
for i in range(k,n):
sum=sum+arr[i]-arr[i-k]
if sum>=threshold*k:
count+=1
return count | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Python3 Solution || O(N) Time & O(1) Space Complexity | akshatkhanna37 | 0 | 3 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,216 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/2635305/Simple-and-easy-solution-using-SLIDING-WINDOW-or-Python | class Solution(object):
def numOfSubarrays(self, arr, k, threshold):
"""
:type arr: List[int]
:type k: int
:type threshold: int
:rtype: int
"""
count=0
windowSum=0
windowStart = 0
for i in range(len(arr)):
windowSum+=arr[i]
if i+1>=k:
if windowSum/k>=threshold:
count+=1
windowSum-=arr[windowStart]
windowStart+=1
return count | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Simple and easy solution using SLIDING WINDOW | Python | msherazedu | 0 | 10 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,217 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/2607918/Python-Sliding-Window-Solution-92 | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
left = 0
right = k-1
count = 0
curr_sum = sum(arr[left:right+1])
while righ < len(arr):
if curr_sum//k >= threshold:
count += 1
curr_sum -= arr[left]
left+=1
right+=1
try:
curr_sum += arr[right]
except:
pass
return count | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Python Sliding Window Solution 92% | pandish | 0 | 27 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,218 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/2607571/Straightforward-Python-3-solution | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
res = []
pos, curSum = 0, 0
while pos < k:
curSum += arr[pos]
pos += 1
res.append(curSum / k)
while pos < len(arr):
curSum += arr[pos]
curSum -= arr[pos-k]
res.append(curSum / k)
pos += 1
return len( [x for x in res if x >= threshold] ) | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Straightforward Python 3 solution | kodrevol | 0 | 15 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,219 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/2381599/Python-Solution-or-Classic-Sliding-Window-or-Two-Pointers-or-90-Faster | class Solution:
def numOfSubarrays(self, nums: List[int], k: int, threshold: int) -> int:
currSum = 0
start = 0
end = 0
count = 0
# run right pointer till end element
for end in range(len(nums)):
# update value to window
currSum += nums[end]
# check if window size achieved
if (end - start + 1) == k:
# is average > target val
if (currSum // k) >= threshold:
count += 1
# slide the window
currSum -= nums[start]
start += 1
return count | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Python Solution | Classic Sliding Window | Two Pointers | 90% Faster | Gautam_ProMax | 0 | 20 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,220 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/2110423/Python-3-Interview-Answer-or-Clean-and-Easy-Readability-or-Sliding-Window | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
result, sumWindow = 0, 0
for i, num in enumerate(arr):
sumWindow += num
if i >= k - 1:
if sumWindow / k >= threshold:
result += 1
sumWindow -= arr[i - k + 1]
return result | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | [Python 3] Interview Answer | Clean & Easy Readability | Sliding Window | Cut | 0 | 36 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,221 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/1957693/python-3-oror-O(n)-faster-than-91oror-python-oror-sliding-window | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
s=sum(arr[:k])
c=0
for i in range(k,len(arr)):
avg=s//k
if avg>=threshold:
c+=1
s=s+arr[i]-arr[i-k]
avg=(s+arr[len(arr)-1]-arr[len(arr)-1-k])//k
if avg>=threshold:
c+=1
return c | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | python 3 || O(n) faster than 91%|| python || sliding window | nileshporwal | 0 | 64 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,222 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/1710387/Easy-Solution-or-Sliding-window-or-Python | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, t: int) -> int:
add,cnt=0,0
i,j=0,0
target=(k*t)
for j in range(len(arr)):
add+=arr[j]
if j>=k-1:
if add >= target:
cnt+=1
add-=arr[i]
i+=1
return(cnt) | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Easy Solution | Sliding window | Python | shandilayasujay | 0 | 48 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,223 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/1702174/Python-simple-O(n)-time-O(1)-space-sliding-window-solution | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
n = len(arr)
target = k*threshold
if n < k:
return 0
elif n == k:
if sum(arr) >= k*threshold:
return 1
else:
return 0
res = 0
i = 0
s = sum(arr[:k])
while i < n-k: # i <= n-k-1
if s >= target:
res += 1
s -= arr[i]
s += arr[i+k]
i += 1
if s > target:
res += 1 | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Python simple O(n) time, O(1) space sliding window solution | byuns9334 | 0 | 78 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,224 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/1304161/Python3-solution-using-sliding-window | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
x = sum(arr[0:k])
i = 1
count = 0
j = k
if x/k >= threshold:
count += 1
while j < len(arr):
x = x-arr[i-1]+arr[j]
if x/k >= threshold:
count += 1
i += 1
j += 1
return count | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | Python3 solution using sliding window | EklavyaJoshi | 0 | 34 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,225 |
https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/discuss/1128952/Python-3-Straight-forward-sliding-window | class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
window_start, total, res = 0, 0, 0
for i in range(k - 1):
total += arr[i]
for window_end in range(k - 1, len(arr)):
total += arr[window_end]
if window_end - window_start + 1 > k:
total -= arr[window_start]
window_start += 1
if total / k >= threshold:
res += 1
return res | number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold | [Python 3] Straight forward sliding window | elliottbarbeau | 0 | 100 | number of sub arrays of size k and average greater than or equal to threshold | 1,343 | 0.676 | Medium | 20,226 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/1911342/Python-one-line-solution-based-on-aptitude-formula | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
return min(abs(30*hour-5.5*minutes),360-abs(30*hour-5.5*minutes)) | angle-between-hands-of-a-clock | Python one line solution based on aptitude formula | amannarayansingh10 | 2 | 49 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,227 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/1889556/Python-Easiest-to-understand!-Comments-Clear-and-Concise | class Solution:
def angleClock(self, h, m):
# Convert the hour hand to another minute hand
m2 = (h%12 + m/60)*5
# Calculate the difference between the two minute hands
diff = abs(m-m2)
# Convert the difference to an angle
ang = diff*(360/60)
# Return the smallest angle
return min(360-ang, ang) | angle-between-hands-of-a-clock | Python - Easiest to understand! Comments - Clear and Concise | domthedeveloper | 2 | 81 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,228 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/2102837/PYTHON-oror-NAIVE-APPROACH-oror-EXPLAINED-WITH-COMMENTS | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
# Normalizing minute position in range (1-12)
min_clock = minutes/5
# If time is 12.00 returning 0
if minutes == 0 and hour*30 == 360:
return 0
# If minutes is 0 then multiply hour by 30 degree as each hour consists of 30 degree
elif minutes == 0:
only_hour = hour*30
# Check whether it is shorter in opposite direction
if only_hour > 180:
return 360-(only_hour)
return only_hour
else:
# Finding the degree between minute hand and closest hour of the hour hand
time = abs(hour-min_clock)*30
# Finding the difference that needs to added/subtracted
diff = 30/(60/minutes)
# Subtracting when minute hand is at greater value that hour hand
if min_clock > hour:
fin_time = time-diff
# Adding when minute hand is at lesser value that hour hand
else:
fin_time = time+diff
# Check the shorter direction
if fin_time > 180:
diff = fin_time-180
return abs(180-diff)
else:
return abs(fin_time) | angle-between-hands-of-a-clock | PYTHON || NAIVE APPROACH || EXPLAINED WITH COMMENTS | klmsathish | 1 | 41 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,229 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/2667553/Python-solution | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
angle = abs((hour + (minutes / 60)) - (minutes/5)) * 30
if angle > 180:
angle = 360 - angle
return angle | angle-between-hands-of-a-clock | Python solution | hyroas | 0 | 5 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,230 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/2155318/Simple-Math-Solution | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
minutes_angle = minutes / 60 * 360
hours_angle = hour / 12 * 360 + minutes / 60 * 30
if minutes_angle > hours_angle:
angle = minutes_angle - hours_angle
return min(angle, 360 - angle)
else:
angle = hours_angle - minutes_angle
return min(angle, 360 - angle) | angle-between-hands-of-a-clock | Simple Math Solution | Vayne1994 | 0 | 24 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,231 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/2097253/Python-One-liner-with-detailed-explanation | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
return min(360 - abs(30 * hour - 11/2 * minutes), abs(30 * hour - 11/2 * minutes)) | angle-between-hands-of-a-clock | Python One liner with detailed explanation | Coconut0727 | 0 | 48 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,232 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/1979348/Python-easy-to-read-and-understand-or-math | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
hr_hand = 30*hour + 0.5*minutes
min_hand = 0*hour + 6*minutes
res = 0
if min_hand > hr_hand:
res = min_hand-hr_hand
else:
res = hr_hand-min_hand
return res if res < 180 else 360-res | angle-between-hands-of-a-clock | Python easy to read and understand | math | sanial2001 | 0 | 48 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,233 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/1593235/Python-Straightforward-Solution-or-Easy-to-Understand | class Solution:
def angleClock(self, hr: int, mint: int) -> float:
# Formula: angle = (30*Hours) - (11/2 * Minutes)
res = abs(30*hr-5.5*mint)
if res > 180:
return 360-res
return res | angle-between-hands-of-a-clock | Python Straightforward Solution | Easy to Understand | leet_satyam | 0 | 54 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,234 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/1570664/Python3-Solution-with-using-math | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
degree_per_minute = 6
degree_per_hour = 30
minute_hand = degree_per_minute * minutes
hour_hand = degree_per_hour * (hour % 12)
ratio = 0
if minute_hand != 0:
ratio = degree_per_hour / (360 / minute_hand)
angle = abs(minute_hand - hour_hand - ratio)
return angle if angle <= 180.0 else 360.0 - angle # min(angle, 360 - angle) | angle-between-hands-of-a-clock | [Python3] Solution with using math | maosipov11 | 0 | 73 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,235 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/1380248/Python3-solution | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
minute_angle = minutes*6
if hour == 12:
hour_angle = minutes*0.5
else:
hour_angle = hour*30+minutes*0.5
return min(abs(minute_angle - hour_angle),360-abs(minute_angle - hour_angle)) | angle-between-hands-of-a-clock | Python3 solution | EklavyaJoshi | 0 | 40 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,236 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/1297137/Python-solution | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
min_degree = minutes / 60 * 360
hour_degree = (hour / 12) * 360 + (minutes / 60) * 30
diff = abs(min_degree-hour_degree)
return min(diff, 360-diff) | angle-between-hands-of-a-clock | Python solution | 5tigerjelly | 0 | 50 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,237 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/1052724/Ultra-Simple-CPPJava-Solution-or-0ms-or-Suggestions-or-optimization-are-welcomed-or | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
if hour==12:
hour=0
mins= 6*minutes
hrs=hour*30+int(minutes%60)*0.5
ans=0
if mins>=hrs:
ans=mins-hrs
if hrs>mins:
ans=hrs-mins;
if 360-mins+hrs<mins-hrs:
ans=360-mins+hrs
if 360-hrs+mins<hrs-mins:
ans=360-hrs+mins
return ans; | angle-between-hands-of-a-clock | Ultra Simple CPP/Java Solution | 0ms | Suggestions or optimization are welcomed | | angiras_rohit | 0 | 56 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,238 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/738439/Intuitive-approach | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
hour = 0 if hour == 12 else hour
h_degree = 30 * hour + 30 * (minutes/60)
m_degree = 6 * minutes if minutes < 60 else 0
diff_degree = abs(h_degree-m_degree)
return min(diff_degree, 360 - diff_degree) | angle-between-hands-of-a-clock | Intuitive approach | puremonkey2001 | 0 | 25 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,239 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/736043/Python3%3A-Easy-To-Understand | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
hour_angle = 0.5*(60*hour + minutes)
minute_angle = 6*minutes
return min(abs(hour_angle-minute_angle),
360-abs(hour_angle-minute_angle)) | angle-between-hands-of-a-clock | Python3: Easy To Understand | pranavd2895 | 0 | 57 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,240 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/735352/Python-or-Easy-Solution-for-Beginners-with-Comments-beats-around-80 | class Solution(object):
def angleClock(self, hour, minutes):
# Converting Minutes To Angle Degree
#[There are 60min and 360 degree each min counts for six]
t_min = 6*minutes
# Converting Hours To Angle Degree
#[There are 12 hours and 360 degree each hour acccounts for 30 degree]
# Partial Angle of Hour is also added depanding upon the minutes
t_hour = 30*hour + 30*(minutes/float(60))
# This if Condition is when time is between 12 and 1.
#Because degree will exceed more than 360 in that case
if t_hour>=360:
t_hour = t_hour-360
x = abs(t_hour-t_min)
# This if loop is to provide the smallest value from the both the sides.
if x>180:
return 360-x
else:
return x | angle-between-hands-of-a-clock | [Python] | Easy Solution for Beginners with Comments beats around 80% | rachitsxn292 | 0 | 33 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,241 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/735352/Python-or-Easy-Solution-for-Beginners-with-Comments-beats-around-80 | class Solution(object):
def angleClock(self, hour, minutes):
t_min = 6*minutes
t_hour = 30*hour + 30*(minutes/float(60))
if t_hour>=360:
t_hour = t_hour-360
x = abs(t_hour-t_min)
if x>180:
return 360-x
else:
return x | angle-between-hands-of-a-clock | [Python] | Easy Solution for Beginners with Comments beats around 80% | rachitsxn292 | 0 | 33 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,242 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/607258/python3-32-ms-easy-newbie-solution | class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
if hour==12:
degh=(12-hour)*30+(minutes*0.5)
degm=minutes*6
else:
degh=(hour*30)+(minutes*0.5)
degm=minutes*6
diff=max(degh,degm)-min(degh,degm)
return min(diff,360-diff)
``` | angle-between-hands-of-a-clock | python3 32 ms easy newbie solution | Hordlaw | 0 | 34 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,243 |
https://leetcode.com/problems/angle-between-hands-of-a-clock/discuss/505123/Python-3-(one-line)-(beats-~97) | class Solution:
def angleClock(self, H: int, M: int) -> float:
return min(abs(30*H-11*M/2),360-abs(30*H-11*M/2))
- Junaid Mansuri
- Chicago, IL | angle-between-hands-of-a-clock | Python 3 (one line) (beats ~97%) | junaidmansuri | 0 | 150 | angle between hands of a clock | 1,344 | 0.634 | Medium | 20,244 |
https://leetcode.com/problems/jump-game-iv/discuss/1691093/Python3-RECURSIVE-BFS-(_)-Explained | class Solution:
def minJumps(self, arr: List[int]) -> int:
N, groups = len(arr), defaultdict(list)
for i, el in enumerate(arr):
groups[el].append(i)
vis, vis_groups = set(), set()
def bfs(lvl, dist):
nextLvl = set()
for i in lvl:
if i in vis: continue
if i == N-1: return dist
vis.add(i)
if i: nextLvl.add(i-1)
if i+1 < N: nextLvl.add(i+1)
if not arr[i] in vis_groups:
vis_groups.add(arr[i])
nextLvl.update(groups[arr[i]])
return bfs(nextLvl, dist + 1)
return bfs(set([0]), 0) | jump-game-iv | ✔️ [Python3] RECURSIVE BFS (҂◡_◡) ᕤ, Explained 💥 | artod | 9 | 321 | jump game iv | 1,345 | 0.44 | Hard | 20,245 |
https://leetcode.com/problems/jump-game-iv/discuss/2658643/Python3-Solution-ororFaster-than-56-ororBFSoror-Bahut-TEZ | class Solution:
def minJumps(self, arr: List[int]) -> int:
h={}
for i,e in enumerate(arr):
if e not in h:
h[e] = []
h[e].append(i)
q = [(0,0)]
while q:
n,d = q.pop(0)
if n == len(arr)-1:
return d
if n+1 == len(arr)-1:
return d+1
if n+1 < len(arr) and h.get(arr[n+1]):
q.append((n+1,d+1))
if n-1 >= 0 and h.get(arr[n-1]):
q.append((n-1,d+1))
for i in h[arr[n]]:
if i != n:
q.append((i,d+1))
if i == len(arr)-1:
return d+1
h[arr[n]] = [] | jump-game-iv | Python3 Solution ||Faster than 56% ||BFS|| Bahut TEZ | hoo__mann | 4 | 125 | jump game iv | 1,345 | 0.44 | Hard | 20,246 |
https://leetcode.com/problems/jump-game-iv/discuss/1775035/Python-easy-to-read-and-understand-or-BFS | class Solution:
def minJumps(self, arr: List[int]) -> int:
n = len(arr)
d = defaultdict(list)
for i, val in enumerate(arr):
d[val].append(i)
visited = [False for _ in range(n)]
q = [0]
visited[0] = True
ans = 0
while q:
for i in range(len(q)):
ind = q.pop(0)
#print(ind)
if ind == n-1:
return ans
if ind + 1 < n and visited[ind+1] == False:
visited[ind+1] = True
q.append(ind+1)
if ind - 1 > 0 and visited[ind-1] == False:
visited[ind-1] = True
q.append(ind-1)
for nei in d[arr[ind]]:
if visited[nei] == False:
visited[nei] = True
q.append(nei)
del d[arr[ind]]
ans += 1 | jump-game-iv | Python easy to read and understand | BFS | sanial2001 | 1 | 98 | jump game iv | 1,345 | 0.44 | Hard | 20,247 |
https://leetcode.com/problems/jump-game-iv/discuss/1692042/Python-Solution-using-Dictionary-and-BFS-! | class Solution:
def minJumps(self, arr: List[int]) -> int:
n=len(arr)
if n<2:
return 0
graph={}
for i in range(n):
if arr[i] not in graph:
graph[arr[i]]=[i]
else:
graph[arr[i]].append(i)
current = [0]
visited = {0}
step = 0
while current:
next=[]
for node in current:
if node == n-1:
return step
for child in graph[arr[node]]:
if child not in visited:
visited.add(child)
next.append(child)
graph[arr[node]].clear()
for child in [node-1 , node+1]:
if 0<= child <len(arr) and child not in visited:
visited.add(child)
next.append(child)
current = next
step = step + 1
return -1
# If It is Usefull to Understand Please UpVote 🙏🙏 | jump-game-iv | Python Solution using Dictionary and BFS ! | ASHOK_KUMAR_MEGHVANSHI | 1 | 76 | jump game iv | 1,345 | 0.44 | Hard | 20,248 |
https://leetcode.com/problems/jump-game-iv/discuss/1690774/Python3-or-BFS-or-Memory-limit-solution | class Solution:
def minJumps(self, arr: List[int]) -> int:
n = len(arr)
visited = set()
if n <= 1:
return 0
same_value = dict()
for i, v in enumerate(arr):
if v not in same_value.keys():
same_value[v]= [i]
else:
same_value[v].append(i)
stack = list()
level = 0
stack.append((0,0))
total = -1
while stack:
level,node = stack.pop(0)
visited.add(node)
if node == n-1:
return level
else:
for a in same_value[arr[node]] :
if a != node and a not in visited:
stack.append((level+1,a))
same_value[arr[node]].clear()
if node+1 < n and node+1 not in visited:
stack.append((level+1,node+1))
if node-1 >=0 and node-1 not in visited:
stack.append((level+1,node-1))
return level | jump-game-iv | Python3 | BFS | Memory limit solution | letyrodri | 1 | 55 | jump game iv | 1,345 | 0.44 | Hard | 20,249 |
https://leetcode.com/problems/jump-game-iv/discuss/989599/Python3-bfs-by-level | class Solution:
def minJumps(self, arr: List[int]) -> int:
loc = defaultdict(list)
for i, x in enumerate(arr): loc[x].append(i)
ans = 0
seen = {0}
queue = deque([0])
while queue:
for _ in range(len(queue)):
i = queue.popleft()
if i+1 == len(arr): return ans
for ii in [i-1, i+1] + loc[arr[i]]:
if 0 <= ii < len(arr) and ii not in seen:
seen.add(ii)
queue.append(ii)
loc.pop(arr[i])
ans += 1 | jump-game-iv | [Python3] bfs by level | ye15 | 1 | 108 | jump game iv | 1,345 | 0.44 | Hard | 20,250 |
https://leetcode.com/problems/jump-game-iv/discuss/2838297/Python-(Simple-BFS) | class Solution:
def minJumps(self, arr):
n, dict1 = len(arr), defaultdict(list)
for i,j in enumerate(arr):
dict1[j].append(i)
stack, visited, visited_group = [(0,0)], set(), set()
visited.add(0)
while stack:
idx, jump = stack.pop(0)
if idx == n-1:
return jump
if idx+1 < n and idx+1 not in visited:
visited.add(idx+1)
stack.append((idx+1,jump+1))
if idx-1 >= 0 and idx-1 not in visited:
visited.add(idx-1)
stack.append((idx-1,jump+1))
if arr[idx] not in visited_group:
for j in dict1[arr[idx]]:
if j not in visited:
visited.add(j)
stack.append((j,jump+1))
visited_group.add(arr[idx]) | jump-game-iv | Python (Simple BFS) | rnotappl | 0 | 2 | jump game iv | 1,345 | 0.44 | Hard | 20,251 |
https://leetcode.com/problems/jump-game-iv/discuss/2057794/BFS-Python-Solution | class Solution:
def minJumps(self, arr: List[int]) -> int:
n=len(arr) ; graph=defaultdict(list) ; layer=[0] ; visited={0} ; step=0
if n<=1: return 0
for i,v in enumerate(arr): graph[v].append(i)
while layer:
next_layer=[]
for node in layer:
if node==n-1: return step
for child in graph[arr[node]]:
if child not in visited: visited.add(child) ; next_layer.append(child)
del graph[arr[node]]
for child in {node-1,node+1}:
if 0<=child<len(arr) and child not in visited: visited.add(child) ; next_layer.append(child)
layer=next_layer
step+=1 | jump-game-iv | BFS Python Solution | Taha-C | 0 | 49 | jump game iv | 1,345 | 0.44 | Hard | 20,252 |
https://leetcode.com/problems/jump-game-iv/discuss/1692196/Simple-BFS-Solution-or-Challenge-January-Day-15 | class Solution:
def minJumps(self, arr: List[int]) -> int:
N = len(arr)
num_dict = defaultdict(list)
for i, n in enumerate(arr):
num_dict[n].append(i)
visited = defaultdict(bool)
dist = [math.inf] * N
queue = deque([0])
dist[0] = 0
while dist[N - 1] == math.inf:
u = queue.popleft()
num = arr[u]
if u - 1 >= 0 and dist[u - 1] > dist[u] + 1:
dist[u - 1] = dist[u] + 1
queue.append(u - 1)
if u + 1 < N and dist[u + 1] > dist[u] + 1:
dist[u + 1] = dist[u] + 1
queue.append(u + 1)
if visited[num]:
continue
visited[num] = True
for v in num_dict[num]:
if dist[v] > dist[u] + 1:
dist[v] = dist[u] + 1
queue.append(v)
return dist[N-1] | jump-game-iv | Simple BFS Solution | Challenge January, Day 15 | atiq1589 | 0 | 43 | jump game iv | 1,345 | 0.44 | Hard | 20,253 |
https://leetcode.com/problems/jump-game-iv/discuss/1692061/Python3-Solution | class Solution:
def minJumps(self, arr: List[int]) -> int:
N, grps = len(arr), defaultdict(list)
for i, el in enumerate(arr):
grps[el].append(i)
vis, vis_grps = set(), set()
def bfs(lvl, dist):
nextLvl = set()
for i in lvl:
if i in vis: continue
if i == N-1: return dist
vis.add(i)
if i: nextLvl.add(i-1)
if i+1 < N: nextLvl.add(i+1)
if not arr[i] in vis_grps:
vis_grps.add(arr[i])
nextLvl.update(grps[arr[i]])
return bfs(nextLvl, dist + 1)
return bfs(set([0]), 0) | jump-game-iv | Python3 Solution | nomanaasif9 | 0 | 46 | jump game iv | 1,345 | 0.44 | Hard | 20,254 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/503507/Python-3-(five-lines)-(beats-100) | class Solution:
def checkIfExist(self, A: List[int]) -> bool:
if A.count(0) > 1: return True
S = set(A) - {0}
for a in A:
if 2*a in S: return True
return False
- Junaid Mansuri
- Chicago, IL | check-if-n-and-its-double-exist | Python 3 (five lines) (beats 100%) | junaidmansuri | 20 | 4,100 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,255 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2197827/Python3-solution-using-hashMap | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
hashMap = {}
for i in arr:
if(hashMap.get(i+i)):
return True
if(i%2 == 0 and hashMap.get(i//2)):
return True
hashMap[i] = True
return False | check-if-n-and-its-double-exist | 📌 Python3 solution using hashMap | Dark_wolf_jss | 6 | 152 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,256 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2074775/Explanation-oror-(-N-*-LOG-N-)-oror-BINARY-SEARCH | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
arr.sort()
n=len(arr)
for i in range(n):
k=arr[i]
# binary search for negatives
if k<0:
lo=0
hi=i
while lo<hi:
mid=(lo+hi)//2
if arr[mid]==(2*k):
return True
elif arr[mid]<(2*k):
lo=mid+1
else:
hi=mid
# binary seach for non negatives
else:
lo=i+1
hi=n
while lo<hi:
mid=(lo+hi)//2
if arr[mid]==(k*2):
return True
elif arr[mid]<(k*2):
lo=mid+1
else:
hi=mid
return False | check-if-n-and-its-double-exist | Explanation || ( N * LOG N ) || BINARY SEARCH | karan_8082 | 5 | 312 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,257 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2074764/PYTHON-oror-O(-n-logn-)-oror-Binary-Search | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
arr.sort()
n=len(arr)
for i in range(n):
k=arr[i]
# binary search for negatives
if k<0:
lo=0
hi=i
while lo<hi:
mid=(lo+hi)//2
if arr[mid]==(2*k):
return True
elif arr[mid]<(2*k):
lo=mid+1
else:
hi=mid
# binary seach for non negatives
else:
lo=i+1
hi=n
while lo<hi:
mid=(lo+hi)//2
if arr[mid]==(k*2):
return True
elif arr[mid]<(k*2):
lo=mid+1
else:
hi=mid
return False | check-if-n-and-its-double-exist | PYTHON || O( n logn ) || Binary Search | karan_8082 | 5 | 255 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,258 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/503455/JavaPython3-using-hash-set | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
seen = set()
for x in arr:
if 2*x in seen or x/2 in seen: return True
seen.add(x)
return False | check-if-n-and-its-double-exist | [Java/Python3] using hash set | ye15 | 4 | 262 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,259 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2340623/Python-Simple-Python-Solution-Using-Binary-Search | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
def BinarySearch(array, target):
low = 0
high = len(array) - 1
while low <= high:
mid = (low + high ) // 2
if array[mid] == target:
return mid
elif array[mid] < target:
low = mid + 1
elif array[mid] > target:
high = mid - 1
return -1
array = sorted(arr)
for i in range(len(array)):
index = BinarySearch(array, array[i] * 2)
if index != -1 and index != i:
return True
return False | check-if-n-and-its-double-exist | [ Python ] ✅✅ Simple Python Solution Using Binary Search 🔥✌👍 | ASHOK_KUMAR_MEGHVANSHI | 3 | 152 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,260 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2597037/PYTHON-Simple-Clean-approach-beat-95-Easy-T%3A-O(N) | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
d = {}
for val in arr:
if d.get(val*2,0) or d.get(val/2,0): return True
d[val] = 1
return False | check-if-n-and-its-double-exist | ✅ [PYTHON] Simple, Clean approach beat 95% Easy T: O(N) | girraj_14581 | 2 | 275 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,261 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2838923/4-LINES-oror-PYTHON-SOLUTION-oror-USING-SETSoror-40-MSororEASY | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
if arr.count(0) > 1: return 1
S = set(arr) - {0}
for i in arr:
if 2*i in S: return 1
return 0 | check-if-n-and-its-double-exist | 4 LINES || PYTHON SOLUTION || USING SETS|| 40 MS||EASY | thezealott | 1 | 34 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,262 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2818128/python3-beat-85-5-lines-code | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
l = len(arr)
for i in range(l-1):
if arr[i] * 2 in arr[0:i] + arr[i+1:] or arr[i] / 2 in arr[0:i]+ arr[i+1:]:
return True
return False | check-if-n-and-its-double-exist | python3, beat 85% ; 5 lines code | dongdong2022 | 1 | 55 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,263 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2384444/Python-Faster-Solution-using-Set-and-Bitwise-Ops-oror-Documented | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
s = set() # lookup table
for n in arr:
if n << 1 in s or (n & 1 == 0 and (n >> 1) in s): # Means: 2*n in s or ( n%2 == 0 and n//2 in s)
return True
s.add(n)
return False | check-if-n-and-its-double-exist | [Python] Faster Solution using Set & Bitwise Ops || Documented | Buntynara | 1 | 30 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,264 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1947511/Python3-or-simple | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
for i, n in enumerate(arr):
if 2*n in arr and arr.index(2*n) != i:
return True
return False | check-if-n-and-its-double-exist | Python3 | simple | user0270as | 1 | 59 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,265 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1810969/3-Lines-Python-Solution-oror-93-Faster-oror-Memory-less-than-76 | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
for i in range(len(arr)):
if (arr[i]*2 in arr[i+1:]) or (arr[i]/2 in arr[i+1:]): return True
return False | check-if-n-and-its-double-exist | 3-Lines Python Solution || 93% Faster || Memory less than 76% | Taha-C | 1 | 130 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,266 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1547762/PYTHON%3A-Very-easy-and-fast-solution-(99)-using-hashing(dictionary) | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
arrDict ={}
for num in arr:
if num * 2 in arrDict or num/2 in arrDict:
return True
if num not in arrDict:
arrDict[num] = None
return False | check-if-n-and-its-double-exist | PYTHON: Very easy and fast solution (99%) using hashing(dictionary) | shubhamrana | 1 | 236 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,267 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1533282/Python-97%2B-Faster-Solution | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
for i in range(len(arr)):
if arr[i] == False:
try:
if arr[i+1] == False:
return True
except:
pass
if arr[i] + arr[i] in arr and arr[i] != False:
return True
else:
return False | check-if-n-and-its-double-exist | Python 97%+ Faster Solution | aaffriya | 1 | 171 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,268 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1280883/Python-Easy-Solution-90faster | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
for num in arr:
if num==0 and arr.count(num)>1:
return True
elif num!=0 and (num/2 in arr or num*2 in arr):
return True
return False | check-if-n-and-its-double-exist | [Python] Easy Solution 90%faster | arkumari2000 | 1 | 188 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,269 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1269876/Python3-simple-solution-using-single-for-loop | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
for i,j in enumerate(arr):
if i < len(arr)-1 and j*2 in arr[i+1:] or (j%2==0 and j//2 in arr[i+1:]):
return True
return False | check-if-n-and-its-double-exist | Python3 simple solution using single for loop | EklavyaJoshi | 1 | 127 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,270 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/930522/Python-Solution-HashMap | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
hashset = {v:k for k,v in enumerate(arr)}
for a in range(len(arr)):
if 2*arr[a] in hashset and hashset[2*arr[a]]!=a:
return True
return False | check-if-n-and-its-double-exist | Python Solution HashMap | bharatgg | 1 | 127 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,271 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2813986/Easy-Python | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
a = []
b = []
for i in arr:
if i!=0:
a.append(2*i)
else:
b.append(i)
for i in a:
if i in arr:
return True
for j in b:
if j in arr and len(b)>1:
return True
break
return False | check-if-n-and-its-double-exist | Easy Python | khanismail_1 | 0 | 5 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,272 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2809865/simple-python-beats-90 | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
arr.sort()
for i in range(len(arr)):
if arr[i]==0:
if arr[i+1]==0:
return True
continue
low=0
high=len(arr)
while(low<high):
mid=(low+high)//2
if arr[i]*2 == arr[mid]:
return True
elif arr[mid] > arr[i]*2:
high = mid
else:
low = mid+1
return False | check-if-n-and-its-double-exist | simple python beats 90% | sudharsan1000m | 0 | 3 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,273 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2789591/Easy-Python-Solution-Faster-than-99.53 | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
hash = set()
for i in arr:
if i*2 in hash or (i/2 in hash):
return True
else:
hash.add(i)
return False | check-if-n-and-its-double-exist | Easy Python Solution Faster than 99.53% | Aayush3014 | 0 | 13 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,274 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2711064/Easy-solution-oror-5-liner-oror-Beats-98.36 | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
if arr.count(0) > 1: return True
for i in arr:
if i*2 in arr and i != 0:
return True
return False | check-if-n-and-its-double-exist | Easy solution || 5 liner || Beats 98.36% | MockingJay37 | 0 | 5 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,275 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2675983/Easy-Python-Solution-or-Loops-or-O(n) | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
arr.sort()
for i in range(len(arr)):
if arr[i] == 0 and arr.count(0) < 2:
continue
if arr[i] % 2 == 0:
if arr[i] // 2 in arr:
return True
return False | check-if-n-and-its-double-exist | Easy Python Solution | Loops | O(n) | atharva77 | 0 | 3 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,276 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2675386/98-Accepted-or-Easy-to-Understand-or-Python | class Solution(object):
def checkIfExist(self, arr):
hashT = {}
for n in arr:
if n not in hashT: hashT[n] = 1
else: hashT[n] += 1
for n in arr:
if n == 0:
if hashT[n] >= 2: return True
else: continue
if (n * 2) in hashT: return True
return False | check-if-n-and-its-double-exist | 98% Accepted | Easy to Understand | Python | its_krish_here | 0 | 23 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,277 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2577971/One-pass-over-array | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
s = set()
for x in arr:
if (2*x in s) or (x % 2 == 0 and (x // 2 in s)):
return True
s.add(x)
return False | check-if-n-and-its-double-exist | One pass over array | semochka | 0 | 32 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,278 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2550848/EASY-PYTHON3-SOLUTION | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
ans = False
for i in range(len(arr)):
for j in range(i,len(arr)):
if i==j:
continue
if (arr[i] == arr[j] * 2) or (arr[j] == arr[i] * 2):
ans = True
break
return ans | check-if-n-and-its-double-exist | ✅✔🔥 EASY PYTHON3 SOLUTION 🔥✅✔ | rajukommula | 0 | 59 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,279 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2502907/PYTHON-SOLUTION-FASTER-THAN-92 | class Solution:
def checkIfExist(self, arr: list[int]) -> bool:
for i in arr:
if i == 0 and arr.count(0) == 1:
continue
elif i == 0 and arr.count(0) >= 2:
return True
elif i * 2 in arr:
return True
return False | check-if-n-and-its-double-exist | PYTHON SOLUTION FASTER THAN 92% | gkarthik923 | 0 | 27 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,280 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2473821/easy | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
for i in range(len(arr)):
pr = arr[0:i]+arr[i+1:len(arr)]
if arr[i]/2 in pr:
return True
return False | check-if-n-and-its-double-exist | easy | rohannayar8 | 0 | 19 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,281 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2450017/C%2B%2BPython-O(Nlog(N))-best-Optimized-Approach | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
def BinarySearch(array, target):
low = 0
high = len(array) - 1
while low <= high:
mid = (low + high ) // 2
if array[mid] == target:
return mid
elif array[mid] < target:
low = mid + 1
elif array[mid] > target:
high = mid - 1
return -1
array = sorted(arr)
for i in range(len(array)):
index = BinarySearch(array, array[i] * 2)
if index != -1 and index != i:
return True
return False | check-if-n-and-its-double-exist | C++/Python O(Nlog(N)) best Optimized Approach | arpit3043 | 0 | 56 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,282 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2418617/simple-python3-solution-with-93-faster | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
idx = 0
while idx < len(arr):
if arr[idx] * 2 in (arr[:idx]+arr[idx+1:]):
return True
else:
idx += 1
return False
# TC = O(n)
# SC = O(1) | check-if-n-and-its-double-exist | simple python3 solution with 93% faster | hwf87 | 0 | 12 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,283 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2417996/75-ms-faster-than-70.02-of-Python3 | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
res = []
for num in arr:
if num*2 in res or num/2 in res:
return True
else:
res.append(num)
return False | check-if-n-and-its-double-exist | 75 ms, faster than 70.02% of Python3 | Harshi_Tyagi | 0 | 18 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,284 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2122038/Python-or-Binary-Search-or-SHORT-AND-SWEET | class Solution:
def checkIfExist(self, nums: list[int]) -> bool:
nums.sort()
for i in range(len(nums)):
target = nums[i] * 2 if nums[i] > 0 else nums[i] / 2
if (j := bisect_right(a=nums, x=target, lo=i + 1) - 1) != i and nums[j] == target:
return True
return False | check-if-n-and-its-double-exist | 🐍 Python | Binary Search | SHORT AND SWEET | miguel_v | 0 | 87 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,285 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/2104267/To-finish-binary-search-study-plan-O(nlogn)-time-andand-O(1)-space | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
arr.sort()
length=len(arr)
for i in range(length):
target=arr[i]*2
m,n=0,length-1
while m<=n:
mid=(m+n)>>1
if target==arr[mid] and mid!=i:
return True
elif target>arr[mid]:
m=mid+1
else:
n=mid-1
return False | check-if-n-and-its-double-exist | To finish binary search study plan, O(nlogn) time && O(1) space | xsank | 0 | 67 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,286 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1995294/Python-3-Solution-using-sorting-and-binary-search | class Solution:
def binary_search(self, nums: List[int], target: int) -> bool:
low, high = 0, len(nums) - 1
while low <= high:
mid = (low + high) // 2
if nums[mid] == target:
return True
elif nums[mid] > target:
high = mid - 1
else:
low = mid + 1
return False
def checkIfExist(self, arr: List[int]) -> bool:
arr.sort()
for i in range(len(arr)):
if arr[i] >= 0 and self.binary_search(arr[i+1:], arr[i] * 2):
return True
if arr[i] < 0 and arr[i] % 2 == 0 and self.binary_search(arr[i+1:], arr[i] // 2):
return True
return False | check-if-n-and-its-double-exist | Python 3 Solution using sorting and binary search | AprDev2011 | 0 | 109 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,287 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1928865/Python3-98-faster-with-explanation | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
lmap = {}
for x in arr:
if 2 * x in lmap or x / 2 in lmap:
return True
if x not in lmap:
lmap[x] = 1
return False | check-if-n-and-its-double-exist | Python3, 98% faster with explanation | cvelazquez322 | 0 | 94 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,288 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1927359/Python-Clean-and-Simple!-Multiple-Solutions! | class Solution:
def checkIfExist(self, arr):
return any(i*2 == j for i,j in permutations(arr, 2)) | check-if-n-and-its-double-exist | Python - Clean and Simple! Multiple Solutions! | domthedeveloper | 0 | 71 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,289 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1927359/Python-Clean-and-Simple!-Multiple-Solutions! | class Solution:
def checkIfExist(self, arr):
find = set()
for n in arr:
if n in find: return True
else:
if n % 2 == 0: find.add(n//2)
find.add(n*2)
return False | check-if-n-and-its-double-exist | Python - Clean and Simple! Multiple Solutions! | domthedeveloper | 0 | 71 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,290 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1927359/Python-Clean-and-Simple!-Multiple-Solutions! | class Solution:
def checkIfExist(self, arr):
seen = set()
for n in arr:
if n*2 in seen or n % 2 == 0 and n//2 in seen: return True
else: seen.add(n)
return False | check-if-n-and-its-double-exist | Python - Clean and Simple! Multiple Solutions! | domthedeveloper | 0 | 71 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,291 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1927359/Python-Clean-and-Simple!-Multiple-Solutions! | class Solution:
def checkIfExist(self, arr):
s1 = set(arr)
s2 = {i*2 for i in s1}
# handle the 0 edge case
if 0 in s1:
z = arr.count(0)
if z >= 2: return True
s1, s2 = s1 - {0}, s2 - {0}
return s1 & s2 | check-if-n-and-its-double-exist | Python - Clean and Simple! Multiple Solutions! | domthedeveloper | 0 | 71 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,292 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1906259/Python-Solution-or-HashMap-Based-or-Over-95-Faster-or-Clean-Code | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
store = defaultdict(int)
for idx,ele in enumerate(arr):
if store:
if ele % 2 == 0 and ele / 2 in store:
return True
if ele * 2 in store:
return True
store[ele] = idx
return False | check-if-n-and-its-double-exist | Python Solution | HashMap Based | Over 95% Faster | Clean Code | Gautam_ProMax | 0 | 51 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,293 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1824657/Python-using-dict-O(n) | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
cnt_dict = dict()
for x in arr:
cnt_dict[x] = cnt_dict[x] + 1 if x in cnt_dict else 1
for x in arr:
if x * 2 in cnt_dict:
if x != 0 or cnt_dict[0] >= 2: # edge case for zero
return True
return False | check-if-n-and-its-double-exist | Python using dict O(n) | nashvenn | 0 | 83 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,294 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1737763/Python-dollarolution | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
if arr.count(0) == 2:
return True
for i in arr:
if i == 0:
continue
if i/2 in arr or i*2 in arr:
return True
return False | check-if-n-and-its-double-exist | Python $olution | AakRay | 0 | 125 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,295 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1631015/pyhton3-time-O(n)-space-O(n) | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
m=defaultdict(int)
for num in arr:
#if (num%2==0 and (num//2) in m) or num*2 in m or num in m:
if num in m:
return True
if num%2==0:
m[num//2]=1
m[num*2]=1
#print(m)
return False | check-if-n-and-its-double-exist | pyhton3 time-O(n) space-O(n) | Rohit_Patil | 0 | 138 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,296 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1578796/Python-fast-with-explanation | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
if len(arr) < 2:
return False
ht = {}
for i in range(len(arr)):
ht[arr[i]] = i
for i in range(len(arr)):
num = arr[i]
if num * 2 in ht and ht[num*2] != i:
return True
return False | check-if-n-and-its-double-exist | Python fast with explanation | SleeplessChallenger | 0 | 93 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,297 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1376936/Python3oror-optimised-O(n)oror-using-Hashmap-Dictionaryoror-faster-than-90 | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
#the rqd double can only be an even number.
#so for an even number check at di[number//2]
di={}
for x in arr:
di[x]=di.get(x,0)+1
for k,v in di.items():
if k!=0 and k%2==0 and di.get(k//2,0)>0:
return True
elif k==0 and di[k]>1:
return True
return False | check-if-n-and-its-double-exist | Python3|| optimised O(n)|| using Hashmap Dictionary|| faster than 90% | ana_2kacer | 0 | 165 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,298 |
https://leetcode.com/problems/check-if-n-and-its-double-exist/discuss/1340234/python3-simple-solution-for-beginners-99-fast | class Solution:
def checkIfExist(self, arr: List[int]) -> bool:
for i in range(0,len(arr)):
if arr[i]==0:
for i in range (i+1,len(arr)):
if arr[i]==0:
return True
else:
if (2*arr[i]) in arr:
return True
if arr[i]/2 in arr:
return True
return False | check-if-n-and-its-double-exist | python3 simple solution for beginners 99% fast | minato_namikaze | 0 | 98 | check if n and its double exist | 1,346 | 0.362 | Easy | 20,299 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.