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 |
|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/find-the-highest-altitude/discuss/1680106/Python-Code-or-Beginner-Friendly-or-Easy-and-Simple | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
res=0
lst=[]
lst.insert(0,0)
for i in gain:
lst.append(res+i)
res=res+i
return max(lst) | find-the-highest-altitude | Python Code | Beginner Friendly | Easy and Simple | harie22 | 0 | 57 | find the highest altitude | 1,732 | 0.787 | Easy | 25,000 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1652572/easy-to-understand | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
l = [0 for i in range(0,len(gain)+1)]
for i in range(1,len(l)):
l[i] = l[i-1] + gain[i-1]
return max(l) | find-the-highest-altitude | easy to understand | MB16biwas | 0 | 30 | find the highest altitude | 1,732 | 0.787 | Easy | 25,001 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1591172/simple-solution-or-python | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
# basic idea is that altitude == gain + previous altitude
idx, altitudes = 1, [0]
gain.insert(0,0)
while idx < len(gain):
altitudes.append(gain[idx]+altitudes[idx-1])
idx += 1
return max(altitudes) | find-the-highest-altitude | simple solution | python | anandanshul001 | 0 | 59 | find the highest altitude | 1,732 | 0.787 | Easy | 25,002 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1428137/Python-Solution%3A-95.36-efficient-and-89.05-better-in-memory | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
maximum = [0]
for height in gain:
maximum.append(maximum[-1] + height)
return max(maximum) | find-the-highest-altitude | Python Solution: 95.36% efficient and 89.05% better in memory | sayantani_s | 0 | 105 | find the highest altitude | 1,732 | 0.787 | Easy | 25,003 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1370987/Python3-Solution | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
currentAlt = 0
maxAlt = 0
for x in range(0, len(gain)):
currentAlt += gain[x]
maxAlt = max(maxAlt, currentAlt)
return maxAlt | find-the-highest-altitude | Python3 Solution | RobertObrochta | 0 | 44 | find the highest altitude | 1,732 | 0.787 | Easy | 25,004 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1276614/Easy-Python-Solution | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
n=len(gain)
ans=[0]*(n+1)
mx=0
for i in range(n):
ans[i+1]=gain[i]+ans[i]
return max(ans) | find-the-highest-altitude | Easy Python Solution | sakshigoel123 | 0 | 105 | find the highest altitude | 1,732 | 0.787 | Easy | 25,005 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1176852/Python3-simple-solution-beats-90-users | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
l = [0]
for i in range(len(gain)):
l.append(l[-1]+gain[i])
return max(l) | find-the-highest-altitude | Python3 simple solution beats 90% users | EklavyaJoshi | 0 | 53 | find the highest altitude | 1,732 | 0.787 | Easy | 25,006 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1112974/Python3-solution-88-fast-98-lite | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
high = 0
temp = 0
for i in gain:
temp = temp + i
if temp > high:
high = temp
return high | find-the-highest-altitude | Python3 solution 88% fast, 98% lite | rikeshkamra97 | 0 | 127 | find the highest altitude | 1,732 | 0.787 | Easy | 25,007 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1067126/Python3-solution | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
curr_max, curr_alt = 0, 0
for i in range(len(gain)):
curr_alt = curr_alt + gain[i]
if curr_alt > curr_max:
curr_max = curr_alt
return curr_max | find-the-highest-altitude | Python3 solution | shreyasdamle2017 | 0 | 55 | find the highest altitude | 1,732 | 0.787 | Easy | 25,008 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1041203/Python-faster-than-97.-Linear-time-O(n)-and-const-memory-O(1)-%2B-explain. | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
current_val = max_val = 0
for x in gain:
current_val += x
max_val = current_val if current_val > max_val else max_val
return max_val | find-the-highest-altitude | Python faster than 97%. Linear time O(n) and const memory O(1) + explain. | dezintegro | 0 | 90 | find the highest altitude | 1,732 | 0.787 | Easy | 25,009 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1036429/Python3-prefix-sum | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
ans = prefix = 0
for x in gain:
prefix += x
ans = max(ans, prefix)
return ans | find-the-highest-altitude | [Python3] prefix sum | ye15 | 0 | 30 | find the highest altitude | 1,732 | 0.787 | Easy | 25,010 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1033011/100-time-and-Space-without-using-library-function-or-12ms-Runtime-or-Python3 | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
last_gain = max_gain = 0
for n in gain:
last_gain += n
if last_gain > max_gain:
max_gain = last_gain
return max_gain | find-the-highest-altitude | 100% time and Space without using library function | 12ms Runtime | Python[3] | hotassun | 0 | 38 | find the highest altitude | 1,732 | 0.787 | Easy | 25,011 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1031708/Python3-clean-solution | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
anchor, res = 0, 0
for g in gain:
anchor += g
res = max(res, anchor)
return res | find-the-highest-altitude | [Python3] clean solution | hwsbjts | 0 | 45 | find the highest altitude | 1,732 | 0.787 | Easy | 25,012 |
https://leetcode.com/problems/find-the-highest-altitude/discuss/1031094/Python3-One-liner | class Solution:
def largestAltitude(self, gain: List[int]) -> int:
return max(0,max(itertools.accumulate(gain))) | find-the-highest-altitude | [Python3] One-liner | vilchinsky | 0 | 60 | find the highest altitude | 1,732 | 0.787 | Easy | 25,013 |
https://leetcode.com/problems/minimum-number-of-people-to-teach/discuss/1059885/Python3-count-properly | class Solution:
def minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:
m = len(languages)
languages = [set(x) for x in languages]
mp = {}
for u, v in friendships:
if not languages[u-1] & languages[v-1]:
for i in range(n):
if i+1 not in languages[u-1]: mp.setdefault(u-1, set()).add(i)
if i+1 not in languages[v-1]: mp.setdefault(v-1, set()).add(i)
ans = inf
for i in range(n):
val = 0
for k in range(m):
if i in mp.get(k, set()): val += 1
ans = min(ans, val)
return ans | minimum-number-of-people-to-teach | [Python3] count properly | ye15 | 1 | 107 | minimum number of people to teach | 1,733 | 0.418 | Medium | 25,014 |
https://leetcode.com/problems/minimum-number-of-people-to-teach/discuss/1059885/Python3-count-properly | class Solution:
def minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:
languages = [set(x) for x in languages]
users = set()
for u, v in friendships:
if not languages[u-1] & languages[v-1]:
users.add(u-1)
users.add(v-1)
freq = {}
for i in users:
for k in languages[i]:
freq[k] = 1 + freq.get(k, 0)
return len(users) - max(freq.values(), default=0) | minimum-number-of-people-to-teach | [Python3] count properly | ye15 | 1 | 107 | minimum number of people to teach | 1,733 | 0.418 | Medium | 25,015 |
https://leetcode.com/problems/minimum-number-of-people-to-teach/discuss/1032620/Pythonic | class Solution:
def minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:
languages = [None] + list(map(set, languages))
friendships = [[u, v] for u, v in friendships if not languages[u] & languages[v]]
# teach whoever needs it
return min([len({u
for pair in friendships
for u in pair
if lan not in languages[u]})
for lan in range(1, n+1)]) | minimum-number-of-people-to-teach | Pythonic | beautyofdeduction | -1 | 48 | minimum number of people to teach | 1,733 | 0.418 | Medium | 25,016 |
https://leetcode.com/problems/decode-xored-permutation/discuss/1031227/Python-or-Detailed-Exeplanation-by-finding-the-first-one | class Solution:
def decode(self, encoded: List[int]) -> List[int]:
n = len(encoded)+1
XOR = 0
for i in range(1,n+1):
XOR = XOR^i
s = 0
for i in range(1,n,2):
s = s^encoded[i]
res = [0]*n
res[0] = XOR^s
for j in range(1,n):
res[j] = res[j-1]^encoded[j-1]
return res | decode-xored-permutation | Python | Detailed Exeplanation by finding the first one | jmin3 | 1 | 111 | decode xored permutation | 1,734 | 0.624 | Medium | 25,017 |
https://leetcode.com/problems/decode-xored-permutation/discuss/2673976/python-easy-solution | class Solution:
def decode(self, encoded: List[int]) -> List[int]:
l=[]
x=0
n=len(encoded)+1
for i in range (1,n+1): # xor of 1 to n+1
x^=i
#print(x)
for i in range (len(encoded)): #xor of normal array whoes index is odd
if i%2==1:
x^=encoded[i]
print(x) #finding first element of an output
l.append(x)
#print(l)
for i in range (len(encoded)): #finding first element then calculating xor with encoded elements
x=l[i]^encoded[i]
l.append(x)
return l | decode-xored-permutation | python easy solution | tush18 | 0 | 5 | decode xored permutation | 1,734 | 0.624 | Medium | 25,018 |
https://leetcode.com/problems/decode-xored-permutation/discuss/1059890/Python3-xor-manipulation | class Solution:
def decode(self, encoded: List[int]) -> List[int]:
x = reduce(xor, list(range(1, len(encoded) + 2)))
for i in range(1, len(encoded), 2): x ^= encoded[i]
ans = [x]
for x in encoded: ans.append(ans[-1] ^ x)
return ans | decode-xored-permutation | [Python3] xor manipulation | ye15 | 0 | 78 | decode xored permutation | 1,734 | 0.624 | Medium | 25,019 |
https://leetcode.com/problems/count-ways-to-make-array-with-product/discuss/1355240/No-Maths-Just-Recursion-DP-we-can-come-up-with-in-interviews-greater-WA | class Solution:
def waysToFillArray(self, queries: List[List[int]]) -> List[int]:
# brute DP O(NK) where N is max(q[0]) and K is max(q[1])
@cache
def dp(n,k):
if k == 1 or n == 1: return 1
ways = 0
for factor in range(1, k+1):
if k % factor == 0:
ways += dp(n-1, k//factor) # or take the '3' part
ways %= (10**9+7)
return ways % (10**9+7)
res = [0] * len(queries)
for i,(n, k) in enumerate(queries):
res[i] = dp(n,k)
return res
# better solution -> find out how many prime factors a number has.
# how many ways to group P numbers into N groups (since array has N values only)
# but you can group in lesser groups and keep 1 1 1 1 as padding in array :( | count-ways-to-make-array-with-product | No Maths Just Recursion DP we can come up with in interviews -> WA | yozaam | 1 | 326 | count ways to make array with product | 1,735 | 0.506 | Hard | 25,020 |
https://leetcode.com/problems/count-ways-to-make-array-with-product/discuss/1355240/No-Maths-Just-Recursion-DP-we-can-come-up-with-in-interviews-greater-WA | class Solution:
def waysToFillArray(self, queries, mod = 10**9+7) -> List[int]:
def getPrimeFactorFreq(n):
i = 2
while i*i <= n:
fre = 0
while n % i == 0:
fre += 1
n //= i
if fre > 0: yield fre
i += 1
if n > 1:
yield 1
@functools.cache
def combinations(n, r):
if n < r: return 0
elif r == 0 or n == r: return 1
return (combinations(n-1,r-1) + combinations(n-1,r)) % mod
def getWays(n, k):
ways = 1
for fre in getPrimeFactorFreq(k):
print('--', fre)
# how many ways to place this factor
# among N boxes, he has `fre` occurance
# https://cp-algorithms.com/combinatorics/stars_and_bars.html
ways_cur_factor = combinations(n+fre-1, fre)
ways = (ways * ways_cur_factor) % mod
return ways
return [getWays(n,k) for n,k in queries] | count-ways-to-make-array-with-product | No Maths Just Recursion DP we can come up with in interviews -> WA | yozaam | 1 | 326 | count ways to make array with product | 1,735 | 0.506 | Hard | 25,021 |
https://leetcode.com/problems/count-ways-to-make-array-with-product/discuss/1059848/Python3-via-sieve-and-comb | class Solution:
def waysToFillArray(self, queries: List[List[int]]) -> List[int]:
spf = list(range(10001)) # spf = smallest prime factor
for i in range(4, 10001, 2): spf[i] = 2
for i in range(3, int(sqrt(10001))+1):
if spf[i] == i:
for ii in range(i*i, 10001, i):
spf[ii] = min(spf[ii], i)
ans = []
for n, k in queries:
freq = {} # prime factorization via sieve
while k != 1:
freq[spf[k]] = 1 + freq.get(spf[k], 0)
k //= spf[k]
val = 1
for x in freq.values():
val *= comb(n+x-1, x)
ans.append(val % 1_000_000_007)
return ans | count-ways-to-make-array-with-product | [Python3] via sieve & comb | ye15 | 0 | 174 | count ways to make array with product | 1,735 | 0.506 | Hard | 25,022 |
https://leetcode.com/problems/count-ways-to-make-array-with-product/discuss/1059848/Python3-via-sieve-and-comb | class Solution:
def waysToFillArray(self, queries: List[List[int]]) -> List[int]:
spf = list(range(10001)) # spf = smallest prime factor
prime = []
for i in range(2, 10001):
if spf[i] == i: prime.append(i)
for x in prime:
if x <= spf[i] and i*x < 10001: spf[i * x] = x
else: break
ans = []
for n, k in queries:
freq = {} # prime factorization via sieve
while k != 1:
freq[spf[k]] = 1 + freq.get(spf[k], 0)
k //= spf[k]
val = 1
for x in freq.values():
val *= comb(n+x-1, x)
ans.append(val % 1_000_000_007)
return ans | count-ways-to-make-array-with-product | [Python3] via sieve & comb | ye15 | 0 | 174 | count ways to make array with product | 1,735 | 0.506 | Hard | 25,023 |
https://leetcode.com/problems/count-ways-to-make-array-with-product/discuss/1033612/Look-for-helpPython3-trying-to-come-up-with-a-DP-solution | class Solution:
def __init__(self):
self.dp = {}
def waysToFillArray(self, queries: List[List[int]]) -> List[int]:
mod = 10 ** 9 + 7
def dfs(n, val):
if (n, val) not in self.dp:
if n == 1: return 1
temp = 1
for k in range(val//2, 0, -1):
if val % k == 0:
temp += dfs(n-1, val // k)
self.dp[n, val] = temp % mod
return self.dp[n, val]
res = []
for n, val in queries:
res.append(dfs(n, val))
return res | count-ways-to-make-array-with-product | [Look for help][Python3] trying to come up with a DP solution | hwsbjts | 0 | 71 | count ways to make array with product | 1,735 | 0.506 | Hard | 25,024 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1032030/Python3-if-elif | class Solution:
def maximumTime(self, time: str) -> str:
time = list(time)
for i in range(len(time)):
if time[i] == "?":
if i == 0: time[i] = "2" if time[i+1] in "?0123" else "1"
elif i == 1: time[i] = "3" if time[0] == "2" else "9"
elif i == 3: time[i] = "5"
else: time[i] = "9"
return "".join(time) | latest-time-by-replacing-hidden-digits | [Python3] if-elif | ye15 | 27 | 1,500 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,025 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1032173/Python-short | class Solution:
def maximumTime(self, time: str) -> str:
maxTime = "23:59" if time[0] in "?2" and time[1] in "?0123" else "19:59"
return "".join(t if t != "?" else m for t, m in zip(time, maxTime)) | latest-time-by-replacing-hidden-digits | Python, short | blue_sky5 | 4 | 306 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,026 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1915764/Easiest-and-Simplest-Python3-Solution-with-if-conditions-or-100-Faster-or-Easy-to-Understand | class Solution:
def maximumTime(self, time: str) -> str:
s=list(time)
for i in range(len(s)):
if s[i]=='?':
if i==0:
if s[i+1] in ['0','1','2','3','?']:
s[i]='2'
else:
s[i]='1'
elif i==1:
if s[i-1]=='1' or s[i-1]=='0':
s[i]='9'
else:
s[i]='3'
elif i==3:
s[i]='5'
elif i==4:
s[i]='9'
return ''.join(s) | latest-time-by-replacing-hidden-digits | Easiest & Simplest Python3 Solution with if conditions | 100% Faster | Easy to Understand | RatnaPriya | 3 | 116 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,027 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1118360/WEEB-DOES-PYTHON-WITH-97.70-RUNTIME | class Solution:
def maximumTime(self, time: str) -> str:
memo = {"0":"9",
"1":"9",
"?":"3",
"2":"3"}
answer = ""
for idx, val in enumerate(time):
if val == "?":
if idx == 0:
if time[idx+1] == "?":
answer += "2"
else:
if int(time[idx+1]) >= 4:
answer += "1"
else: answer += "2"
if idx == 1:
answer += memo[time[idx-1]]
if idx == 3:
answer += "5"
if idx == 4:
answer += "9"
else:
answer += val
return answer | latest-time-by-replacing-hidden-digits | WEEB DOES PYTHON WITH 97.70% RUNTIME | Skywalker5423 | 3 | 308 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,028 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1701829/Python-3-Solution-%3A-Easy-and-Understandable | class Solution:
def maximumTime(self, time: str) -> str:
hr,mn = time.split(':')
hr = list(hr)
mn = list(mn)
if hr[0]=='?' and hr[1]!='?':
if 4<= int(hr[1]) <=9:
hr[0] = "1"
else:
hr[0] = "2"
if hr[1]=='?' and hr[0]!='?':
if hr[0] == "2":
hr[1] = "3"
else:
hr[1] = "9"
if hr[0]=='?' and hr[1]=='?':
hr[0] = "2"
hr[1] = "3"
if mn[0] == '?' and mn[1]!='?':
mn[0] = "5"
if mn[1] == '?' and mn[0]!='?':
mn[1] = "9"
if mn[0] == '?' and mn[1]=='?':
mn[0] = "5"
mn[1] = "9"
hr.append(':')
return "".join(hr+mn) | latest-time-by-replacing-hidden-digits | Python 3 Solution : Easy and Understandable | deleted_user | 1 | 73 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,029 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/2789220/Python-simple-greedy-solution | class Solution:
def maximumTime(self, time: str) -> str:
hr, mn = time.split(':')
if hr[0] == '?':
if hr[1] == '?':
hr = '23'
else:
if int(hr[1]) <= 3:
hr = '2'+hr[1]
else:
hr = '1'+hr[1]
elif hr[1] == '?':
if hr[0] == '2':
hr = hr[0]+'3'
else:
hr = hr[0]+'9'
if mn[0] == '?':
if mn[1] == '?':
mn = '59'
else:
mn = '5'+mn[1]
elif mn[1] == '?':
mn = mn[0]+'9'
return hr+':'+mn | latest-time-by-replacing-hidden-digits | Python simple greedy solution | ankurkumarpankaj | 0 | 6 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,030 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/2560255/Python-easy-to-understand-(runtime%3A-30-40ms) | class Solution:
def maximumTime(self, time: str) -> str:
a=['2','3',':','5','9']
b=['1','9',':','5','9']
c=['4','5','6','7','8','9']
if time[0]=='2' or (time[0]=='?' and time[1] not in c):
for i in range(5):
if time[i]!='?':
a[i]=time[i]
return(a[0]+a[1]+':'+a[3]+a[4])
else:
for i in range(5):
if time[i]!='?':
b[i]=time[i]
return(b[0]+b[1]+':'+b[3]+b[4]) | latest-time-by-replacing-hidden-digits | Python easy to understand (runtime: 30-40ms) | jakeyhe | 0 | 34 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,031 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/2352715/Python3-Brute-force-each-%22%22-one-by-one-except-%22%22 | class Solution:
def maximumTime(self, time: str) -> str:
def brute_force(substring, limit):
for i in range(10)[::-1]:
value = substring.replace('?', f'{i}')
if int(value) < limit:
return value
hh = '23' if time[:2] == '??' else brute_force(time[:2], 24)
mm = '59' if time[3:] == '??' else brute_force(time[3:], 60)
return(f'{hh}:{mm}') | latest-time-by-replacing-hidden-digits | [Python3] Brute force each "?" one by one except "??" | TomS_Ekb | 0 | 35 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,032 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1847943/Python-dollarolution | class Solution:
def maximumTime(self, time: str) -> str:
v = ['2','3','9','5','9']
s = ''
for i in range(5):
if time[i] == '?':
if i == 0 and time[i+1] not in '0,1,2,3,?':
s += '1'
elif i == 1 and s[i-1] != '2':
s += v[i+1]
else:
s += v[i]
else:
s += time[i]
return s | latest-time-by-replacing-hidden-digits | Python $olution | AakRay | 0 | 49 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,033 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1495933/Choose-max-digits-based-on-neighbor-82-speed | class Solution:
def maximumTime(self, time: str) -> str:
lst = list(time)
if lst[0] == lst[1] == "?":
lst[0], lst[1] = "2", "3"
elif lst[0] == "?":
lst[0] = "2" if lst[1] < "4" else "1"
elif lst[1] == "?":
lst[1] = "9" if lst[0] < "2" else "3"
if lst[3] == "?":
lst[3] = "5"
if lst[4] == "?":
lst[4] = "9"
return "".join(lst) | latest-time-by-replacing-hidden-digits | Choose max digits based on neighbor, 82% speed | EvgenySH | 0 | 106 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,034 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1446044/Python3-Faster-Than-95.42 | class Solution:
def maximumTime(self, time: str) -> str:
z = [time[0], time[1], time[3], time[4]]
if time[0] == '?':
if time[1] == '?':
z[0] = '2'
elif int(time[1]) >= 4:
z[0] = '1'
else:
z[0] = '2'
if time[1] == '?':
if z[0] == '2':
z[1] = '3'
else:
z[1] = '9'
if time[3] == '?':
z[2] = '5'
if time[4] == '?':
z[3] = '9'
return z[0] + z[1] + ':' + z[2] + z[3] | latest-time-by-replacing-hidden-digits | Python3 Faster Than 95.42% | Hejita | 0 | 71 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,035 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1309740/Python-fast-and-simple-solution | class Solution:
def maximumTime(self, time: str) -> str:
_time = list(time)
if _time[0] == "?":
if _time[1] == "?" or int(_time[1]) < 4:
_time[0] = "2"
else:
_time[0] = "1"
if _time[1] == "?":
if _time[0] == "2":
_time[1] = "3"
else:
_time[1] = "9"
if _time[3] == "?":
_time[3] = "5"
if _time[4] == "?":
_time[4] = "9"
return "".join(_time) | latest-time-by-replacing-hidden-digits | Python fast and simple solution | MihailP | 0 | 120 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,036 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1265779/Python3-List-out-every-possibility | class Solution:
def maximumTime(self, time: str) -> str:
res = ''
for i in range(len(time)):
if time[i] == '?':
if i == 0:
if time[1] == '?':
res += '2'
else:
if int(time[1]) > 3:
res += '1'
else:
res += '2'
elif i == 1:
if res[0] == '2':
res += '3'
else:
res += '9'
elif i == 3:
res += '5'
else:
res += '9'
else:
res += time[i]
return res | latest-time-by-replacing-hidden-digits | Python3 List out every possibility | georgeqz | 0 | 91 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,037 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1250513/Python3-simple-solution | class Solution:
def maximumTime(self, time: str) -> str:
if time[0] == '?':
if time[1] != '?'and not int(time[1]) <= 3:
time = '1' + time[1:]
else:
time = '2' + time[1:]
if time[1] == '?':
if time[0] in ['0','1']:
time = time[0] + '9' + time[2:]
else:
time = time[0] + '3' + time[2:]
if time[3] == '?':
time = time[:3] + '5' + time[4]
if time[4] == '?':
time = time[:4] + '9'
return time | latest-time-by-replacing-hidden-digits | Python3 simple solution | EklavyaJoshi | 0 | 74 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,038 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1033653/Python3-Solution-with-in-line-explanation | class Solution:
def maximumTime(self, time: str) -> str:
# first digit: if second_digit >= 4: [0, 1], else: [0, 2]
# second digit: if first_digit == 2: [0, 3], else: [0, 9]
# third digit: always [0, 5]
# fourth digit: always [0, 9]
time = list(time)
if time[0] == '?' and (time[1] == '?' or time[1] < '4'):
time[0] = '2'
elif time[0] == '?':
time[0] = '1'
if time[1] == '?' and time[0] == '2':
time[1] = '3'
elif time[1] == '?':
time[1] = '9'
if time[3] == '?': time[3] = '5'
if time[4] == '?': time[4] = '9'
return ''.join(time) | latest-time-by-replacing-hidden-digits | [Python3] Solution with in-line explanation | hwsbjts | 0 | 92 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,039 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1032916/Python-2-liner | class Solution:
def maximumTime(self, time: str) -> str:
maxChar = lambda i: "23:59"[i] if time[0] in "2?" and time[1] in "0123?" else "19:59"[i]
return "".join(c if c != "?" else maxChar(i) for i, c in enumerate(time)) | latest-time-by-replacing-hidden-digits | Python 2-liner | cenkay | 0 | 188 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,040 |
https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/discuss/1032762/PYTHON-oror-EASY-oror-20MS-oror-HASHMAP | class Solution:
def maximumTime(self, time: str) -> str:
maxValues = {
'0': '2',
'1': {'0': '9', '1': '9', '2': '3'},
'3': '5',
'4': '9'
}
result = []
for i, char in enumerate(time):
if char == '?':
if i == 0 and time[i + 1] != '?' and time[i + 1] >= '4':
result.append('1')
elif i == 1:
result.append(maxValues['1'][result[i - 1]])
else:
result.append(maxValues[str(i)])
else:
result.append(char)
return ''.join(result) | latest-time-by-replacing-hidden-digits | PYTHON || EASY || 20MS || HASHMAP | akashgkrishnan | 0 | 79 | latest time by replacing hidden digits | 1,736 | 0.422 | Easy | 25,041 |
https://leetcode.com/problems/change-minimum-characters-to-satisfy-one-of-three-conditions/discuss/1032055/Python3-scan-through-a-z-w-prefix | class Solution:
def minCharacters(self, a: str, b: str) -> int:
pa, pb = [0]*26, [0]*26
for x in a: pa[ord(x)-97] += 1
for x in b: pb[ord(x)-97] += 1
ans = len(a) - max(pa) + len(b) - max(pb) # condition 3
for i in range(25):
pa[i+1] += pa[i]
pb[i+1] += pb[i]
ans = min(ans, pa[i] + len(b) - pb[i]) # condition 2
ans = min(ans, len(a) - pa[i] + pb[i]) # condition 1
return ans | change-minimum-characters-to-satisfy-one-of-three-conditions | [Python3] scan through a-z w/ prefix | ye15 | 1 | 58 | change minimum characters to satisfy one of three conditions | 1,737 | 0.352 | Medium | 25,042 |
https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/discuss/1032117/Python3-compute-xor-O(MNlog(MN))-or-O(MNlogK)-or-O(MN) | class Solution:
def kthLargestValue(self, matrix: List[List[int]], k: int) -> int:
m, n = len(matrix), len(matrix[0]) # dimensions
ans = []
for i in range(m):
for j in range(n):
if i: matrix[i][j] ^= matrix[i-1][j]
if j: matrix[i][j] ^= matrix[i][j-1]
if i and j: matrix[i][j] ^= matrix[i-1][j-1]
ans.append(matrix[i][j])
return sorted(ans)[-k] | find-kth-largest-xor-coordinate-value | [Python3] compute xor O(MNlog(MN)) | O(MNlogK) | O(MN) | ye15 | 15 | 936 | find kth largest xor coordinate value | 1,738 | 0.613 | Medium | 25,043 |
https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/discuss/1032117/Python3-compute-xor-O(MNlog(MN))-or-O(MNlogK)-or-O(MN) | class Solution:
def kthLargestValue(self, matrix: List[List[int]], k: int) -> int:
m, n = len(matrix), len(matrix[0]) # dimensions
pq = []
for i in range(m):
for j in range(n):
if i: matrix[i][j] ^= matrix[i-1][j]
if j: matrix[i][j] ^= matrix[i][j-1]
if i and j: matrix[i][j] ^= matrix[i-1][j-1]
heappush(pq, matrix[i][j])
if len(pq) > k: heappop(pq)
return pq[0] | find-kth-largest-xor-coordinate-value | [Python3] compute xor O(MNlog(MN)) | O(MNlogK) | O(MN) | ye15 | 15 | 936 | find kth largest xor coordinate value | 1,738 | 0.613 | Medium | 25,044 |
https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/discuss/1032117/Python3-compute-xor-O(MNlog(MN))-or-O(MNlogK)-or-O(MN) | class Solution:
def kthLargestValue(self, matrix: List[List[int]], k: int) -> int:
m, n = len(matrix), len(matrix[0]) # dimensions
vals = []
for i in range(m):
for j in range(n):
if i: matrix[i][j] ^= matrix[i-1][j]
if j: matrix[i][j] ^= matrix[i][j-1]
if i and j: matrix[i][j] ^= matrix[i-1][j-1]
vals.append(matrix[i][j])
def part(lo, hi):
"""Partition vals from lo (inclusive) to hi (exclusive)."""
i, j = lo+1, hi-1
while i <= j:
if vals[i] < vals[lo]: i += 1
elif vals[lo] < vals[j]: j -= 1
else:
vals[i], vals[j] = vals[j], vals[i]
i += 1
j -= 1
vals[lo], vals[j] = vals[j], vals[lo]
return j
lo, hi = 0, len(vals)
while lo < hi:
mid = part(lo, hi)
if mid + k < len(vals): lo = mid + 1
else: hi = mid
return vals[lo] | find-kth-largest-xor-coordinate-value | [Python3] compute xor O(MNlog(MN)) | O(MNlogK) | O(MN) | ye15 | 15 | 936 | find kth largest xor coordinate value | 1,738 | 0.613 | Medium | 25,045 |
https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/discuss/1563463/Python-3-Prefix-%22XOR%22-and-Quickselect-(but-why-built-in-sort()-is-faster) | class Solution:
def quickselect(self, arr, k, low, high):
"""
select k largest element
"""
def swap(arr, i, j):
arr[i], arr[j] = arr[j], arr[i]
def partition(arr, low, high):
"""
select last elem as pivot, return the pivot index after rearranged array
"""
# randomly choose index
randidx = random.randint(low,high)
pivot = arr[randidx]
swap(arr, randidx, high)
i = low
for j in range(low, high):
if arr[j] > pivot:
swap(arr, i, j)
i += 1
swap(arr, i, high)
return i
if low == high:
return arr[low]
pivot_idx = partition(arr, low, high)
if k-1 == pivot_idx:
return arr[pivot_idx]
elif k-1 > pivot_idx:
return self.quickselect(arr, k, pivot_idx+1, high)
else:
return self.quickselect(arr, k, low, pivot_idx-1)
return quickselect(nums, k, 0, len(nums)-1)
def kthLargestValue(self, matrix: List[List[int]], k: int) -> int:
"""
prefix-XOR
flatten matrix into 1d
quickselect(k)
[ ][-][+][ ]
[ ][+][x][ ]
[ ][ ][ ][ ]
[ ][ ][ ][ ]
"""
nrow, ncol = len(matrix), len(matrix[0])
prefix_XOR = [[0 for _ in range(ncol)] for _ in range(nrow)]
# init first row and first col
prefix_XOR[0][0] = matrix[0][0]
for i in range(1,ncol):
prefix_XOR[0][i] = prefix_XOR[0][i-1] ^ matrix[0][i]
for i in range(1,nrow):
prefix_XOR[i][0] = prefix_XOR[i-1][0] ^ matrix[i][0]
# prefix XOR
for i in range(1,nrow):
for j in range(1,ncol):
prefix_XOR[i][j] = matrix[i][j] ^ prefix_XOR[i-1][j] ^ prefix_XOR[i][j-1] ^ prefix_XOR[i-1][j-1]
# flatten 2d matrix into 1d
flatten = []
for i in prefix_XOR:
flatten += i
# quick select k
# res = self.quickselect(flatten, k, 0, len(flatten)-1) # why quickselect not working? (TLE)
res = sorted(flatten)[-k] # why built-in sorted() works?
return res | find-kth-largest-xor-coordinate-value | [Python 3] Prefix-"XOR" and Quickselect (but why built-in sort() is faster?) | nick19981122 | 1 | 126 | find kth largest xor coordinate value | 1,738 | 0.613 | Medium | 25,046 |
https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/discuss/1197268/Python3-Solution-with-Comments-Faster-than-96 | class Solution:
def kthLargestValue(self, matrix: List[List[int]], k: int) -> int:
res = []
prefix_sum = [[0]*(len(matrix[0])+1) for _ in range(0,len(matrix)+1)] #initialize prefix sum matrix
# for each index (i,j) in matrix, calculate XOR of prefix_sum[i-1][j] and matrix[i][j]
# then XOR the result with XOR of all values from matrix[i][0] to matrix[i][j-1]
for i in range(1,len(matrix)+1):
XOR_value = 0 #initialize XOR value for row i of matrix
for j in range(1,len(matrix[0])+1):
XOR_value ^= matrix[i-1][j-1] #XOR with value at index i,j of matrix (as loops start from 1, we use indices i-1 and j-1)
prefix_sum[i][j] = XOR_value^prefix_sum[i-1][j] #update current index of prefix sum by XORing the current XOR value with the prefix sum at the upper cell
res.append(prefix_sum[i][j]) #store this resultant XOR at res
return sorted(res)[-k] #need k'th largest element, so sort it and get the value at the (k-1)'th index from the right | find-kth-largest-xor-coordinate-value | Python3 Solution with Comments, Faster than 96% | bPapan | 1 | 73 | find kth largest xor coordinate value | 1,738 | 0.613 | Medium | 25,047 |
https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/discuss/1358597/Python3-solution-beats-95-users | class Solution:
def kthLargestValue(self, matrix: List[List[int]], k: int) -> int:
dp = []
for i in range(len(matrix)):
x = []
for j in range(len(matrix[0])):
x.append(0)
dp.append(x)
for i in range(len(matrix)):
c = 0
for j in range(len(matrix[0])):
c ^= matrix[i][j]
if j == 0:
if i == 0:
dp[i][j] = matrix[i][j]
else:
dp[i][j] = dp[i-1][j] ^ matrix[i][j]
else:
if i == 0:
dp[i][j] = c
else:
dp[i][j] = c ^ dp[i-1][j]
l = []
for i in dp:
l += i
return sorted(l,reverse=True)[k-1] | find-kth-largest-xor-coordinate-value | Python3 solution beats 95% users | EklavyaJoshi | 0 | 78 | find kth largest xor coordinate value | 1,738 | 0.613 | Medium | 25,048 |
https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/discuss/1032166/Python-Solution | class Solution:
def kthLargestValue(self, matrix, k: int) -> int:
large = [matrix[0][0]]
n = len(matrix)
m = len(matrix[0])
for i in range(1, n):
matrix[i][0] ^= matrix[i - 1][0]
large.append(matrix[i][0])
for j in range(1, m):
matrix[0][j] ^= matrix[0][j - 1]
large.append(matrix[0][j])
for i in range(1, n):
for j in range(1, m):
matrix[i][j] ^= matrix[i][j - 1] ^ matrix[i - 1][j] ^ matrix[i - 1][j - 1]
large.append(matrix[i][j])
large.sort()
return large[-k] | find-kth-largest-xor-coordinate-value | Python Solution | mariandanaila01 | 0 | 69 | find kth largest xor coordinate value | 1,738 | 0.613 | Medium | 25,049 |
https://leetcode.com/problems/building-boxes/discuss/1032104/Python3-math | class Solution:
def minimumBoxes(self, n: int) -> int:
x = int((6*n)**(1/3))
if x*(x+1)*(x+2) > 6*n: x -= 1
ans = x*(x+1)//2
n -= x*(x+1)*(x+2)//6
k = 1
while n > 0:
ans += 1
n -= k
k += 1
return ans | building-boxes | [Python3] math | ye15 | 12 | 371 | building boxes | 1,739 | 0.519 | Hard | 25,050 |
https://leetcode.com/problems/building-boxes/discuss/1032104/Python3-math | class Solution:
def minimumBoxes(self, n: int) -> int:
x = int((6*n)**(1/3))
if x*(x+1)*(x+2) > 6*n: x -= 1
n -= x*(x+1)*(x+2)//6
return x*(x+1)//2 + ceil((sqrt(1+8*n)-1)/2) | building-boxes | [Python3] math | ye15 | 12 | 371 | building boxes | 1,739 | 0.519 | Hard | 25,051 |
https://leetcode.com/problems/building-boxes/discuss/1777473/python3-Fastest-and-Most-Elegant-Script | class Solution:
def minimumBoxes(self, n: int) -> int:
a = 0
b = 0
s = 0
while n > s:
a += 1
b += a
s += b
while n <= s:
s -= a
a -= 1
b -= 1
return b + 1 | building-boxes | [python3] Fastest & Most Elegant Script | wubj97 | 1 | 69 | building boxes | 1,739 | 0.519 | Hard | 25,052 |
https://leetcode.com/problems/building-boxes/discuss/2792603/Python-or-Tetrahedral-and-Triangular-Numbers-or-O(m13) | class Solution:
def minimumBoxes(self, m: int) -> int:
def cbrt(x):
return x**(1. / 3)
# Find the first tetrahedral number greater than
# or equal to m.
x = cbrt(sqrt(3)*sqrt(243*(m**2) - 1) + 27*m)
n = ceil(x/cbrt(9) + 1/(cbrt(3)*x) - 1)
# If m is the nth tetrahedral number, return the
# nth triangular number (the base).
t_n =n*(n+1)*(n+2) // 6
if m == t_n:
return n*(n+1)//2
# Otherwise, we must adjust the answer.
ans = n*(n+1)//2
j = t_n + 1
while m < j:
j -= n
ans -= 1
n -= 1
return ans + 1 | building-boxes | Python | Tetrahedral and Triangular Numbers | O(m^1/3) | on_danse_encore_on_rit_encore | 0 | 1 | building boxes | 1,739 | 0.519 | Hard | 25,053 |
https://leetcode.com/problems/building-boxes/discuss/2111487/Python-or-Detailed-Explanation | class Solution:
def minimumBoxes(self, n: int) -> int:
r = 0
while (n_upper := r*(r+1)*(r+2)//6) < n:
r += 1
m = r*(r+1)//2
for i in range(r, 0, -1):
if (n_upper - i) < n:
break
n_upper -= i
m -= 1
return m | building-boxes | Python | Detailed Explanation | Lindelt | 0 | 137 | building boxes | 1,739 | 0.519 | Hard | 25,054 |
https://leetcode.com/problems/building-boxes/discuss/1032028/Python3.-Build-pyramid.-Easy-with-explanation. | class Solution:
def minimumBoxes(self, n: int) -> int:
boxesPlaced = 0
maxFloor = 1
boxesOnFloor = 0
while boxesPlaced < n:
for i in range(1, maxFloor + 1):
boxesPlaced += i
boxesOnFloor += 1
if boxesPlaced >= n:
break
maxFloor += 1
return boxesOnFloor | building-boxes | Python3. Build pyramid. Easy with explanation. | yaroslav-repeta | 0 | 74 | building boxes | 1,739 | 0.519 | Hard | 25,055 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1042922/Python3-freq-table | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
freq = defaultdict(int)
for x in range(lowLimit, highLimit+1):
freq[sum(int(xx) for xx in str(x))] += 1
return max(freq.values()) | maximum-number-of-balls-in-a-box | [Python3] freq table | ye15 | 11 | 1,300 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,056 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1043417/Python-3-Easy-to-understand-COMMENTED-solution. | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
boxes = [0] * 100
for i in range(lowLimit, highLimit + 1):
# For the current number "i", convert it into a list of its digits.
# Compute its sum and increment the count in the frequency table.
boxes[sum([int(j) for j in str(i)])] += 1
return max(boxes) | maximum-number-of-balls-in-a-box | [Python 3] - Easy to understand COMMENTED solution. | mb557x | 7 | 675 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,057 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1692190/Python-Solution-(Dictionary-based) | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
box = {}
for balls in range(lowLimit,highLimit+1):
boxNo = self.sumOfDigits(balls)
if boxNo in box:
box[boxNo] += 1
else:
box[boxNo] = 1
max = 0
for key in box:
if box[key] > max:
max = box[key]
return max
def sumOfDigits(self,n):
ans = 0
while n > 0:
ans += n % 10
n = n // 10
return ans | maximum-number-of-balls-in-a-box | Python Solution (Dictionary based) | s_m_d_29 | 1 | 123 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,058 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1043081/Python-Dictionary-or-easy-to-understand | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
res = 0
boxes = collections.defaultdict(int)
for num in range(lowLimit, highLimit+1):
box = 0
while num:
digit = num%10
num = num//10
box += digit
boxes[box] +=1
res = max(res, boxes[box])
return res | maximum-number-of-balls-in-a-box | [Python] Dictionary | easy to understand | SonicM | 1 | 229 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,059 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/2781705/ONLY-ARRAY-oror-NO-HASHING-oror-2ms-SOULTION | class Solution:
def countBalls(self, lowlimit: int, highlimit: int) -> int:
a=[]
a=[0 for i in range(highlimit+1)]
for i in range(lowlimit,highlimit+1):
sum=0
while(i>0):
b=i%10
i=i//10
sum+=b
a[sum]=a[sum]+1
c=max(a)
return(c) | maximum-number-of-balls-in-a-box | ONLY ARRAY || NO HASHING || 2ms SOULTION | Ujjwal_2458 | 0 | 2 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,060 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/2654509/Python3_Easy_Understand_Method | class Solution:
def countBalls(self, lowLimit, highLimit):
dic_A = defaultdict(int)
for i in range(lowLimit, highLimit+1):
count = 0
for j in str(i):
count += int(j)
dic_A[count] = dic_A[count]+1
return max(dic_A.values()) | maximum-number-of-balls-in-a-box | Python3_Easy_Understand_Method | Hsien_Chiu | 0 | 1 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,061 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/2507451/Python3-Beats-99.7-with-Dictionary-O(n)-Time-O(n)-Space | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
h = defaultdict(int)
# find the first box number
s = 0
temp = lowLimit
while temp > 0:
s += temp % 10
temp = temp // 10
h[s] += 1
x = lowLimit + 1
while x <= highLimit:
# check rightmost digit
if 1 <= x % 10 <= 9:
s = s+1
else:
s = s - 8
temp = x // 10
while temp > 0 and temp % 10 == 0:
s -= 9
temp = temp // 10
h[s] += 1
x += 1
return max(h.values()) | maximum-number-of-balls-in-a-box | [Python3] Beats 99.7% with Dictionary - O(n) Time, O(n) Space | rt500 | 0 | 46 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,062 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/2348005/Python-dictionary-solution | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
hashmap = {}
for i in range(lowLimit, highLimit +1):
digit_sum = 0
while i:
digit_sum += i%10
i = i//10
if digit_sum not in hashmap:
hashmap[digit_sum] = 1
else:
hashmap[digit_sum] += 1
return max(hashmap.values()) | maximum-number-of-balls-in-a-box | Python dictionary solution | hochunlin | 0 | 49 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,063 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/2290269/Dictionary-self-explaining-code | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
boxes = {}
for ball_number in range(lowLimit, highLimit + 1):
ball_sum = sum([int(str_number) for str_number in str(ball_number)])
if ball_sum in boxes:
boxes[ball_sum] += 1
else:
boxes[ball_sum] = 1
return max([count for box, count in boxes.items()]) | maximum-number-of-balls-in-a-box | Dictionary, self-explaining code | Simzalabim | 0 | 24 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,064 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/2270443/Maximum-Number-of-Balls-in-a-Box | class Solution:
def countSum(self,x):
y=0
while(x!=0 ):
y+=x%10
x=x//10
return y
def countBalls(self, lowLimit: int, highLimit: int) -> int:
x={}
count=0
for i in range(lowLimit,highLimit+1):
y = self.countSum(i)
if y in x:
x[y] +=1
else:
x[y]=1
return max([i for i in x.values()]) | maximum-number-of-balls-in-a-box | Maximum Number of Balls in a Box | dhananjayaduttmishra | 0 | 22 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,065 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/2179957/Python-simple-solution | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
ans = [0 for x in range(47)]
for i in range(lowLimit, highLimit+1):
ans[sum(map(int,list(str(i))))] += 1
return max(ans) | maximum-number-of-balls-in-a-box | Python simple solution | StikS32 | 0 | 53 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,066 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/2095436/PYTHON-or-Simple-python-solution | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
boxMap = {}
def getSum(n):
sum = 0
for digit in str(n):
sum += int(digit)
return sum
for i in range(lowLimit, highLimit + 1):
boxMap[getSum(i)] = 1 + boxMap.get(getSum(i), 0)
res = 0
for i in boxMap:
res = max(res, boxMap[i])
return res | maximum-number-of-balls-in-a-box | PYTHON | Simple python solution | shreeruparel | 0 | 103 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,067 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1913734/Python-solution-memory-less-than-74 | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
freq = [0] * 45
for i in range(lowLimit, highLimit+1):
dig_sum = sum([int(x) for x in str(i)])
freq[dig_sum - 1] += 1
return max(freq) | maximum-number-of-balls-in-a-box | Python solution memory less than 74% | alishak1999 | 0 | 93 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,068 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1848002/Python-dollarolution | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
d, maximum = {}, 1
for i in range(lowLimit, highLimit+1):
count = 0
while i != 0:
count += i%10
i //= 10
if count in d:
d[count] += 1
maximum = max(maximum,d[count])
else:
d[count] = 1
return maximum | maximum-number-of-balls-in-a-box | Python $olution | AakRay | 0 | 63 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,069 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1845703/3-Lines-Python-Solution-oror-50-Faster-oror-Memory-less-than-85 | class Solution:
def countBalls(self, l: int, h: int) -> int:
ans=[0]*46
for i in range(l,h+1): ans[sum([int(j) for j in str(i)])]+=1
return max(ans) | maximum-number-of-balls-in-a-box | 3-Lines Python Solution || 50% Faster || Memory less than 85% | Taha-C | 0 | 73 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,070 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1845703/3-Lines-Python-Solution-oror-50-Faster-oror-Memory-less-than-85 | class Solution:
def countBalls(self, l: int, h: int) -> int:
ans=defaultdict(int)
for i in range(l,h+1): s=sum([int(i) for i in str(i)]) ; ans[s]+=1
return max(ans.values()) | maximum-number-of-balls-in-a-box | 3-Lines Python Solution || 50% Faster || Memory less than 85% | Taha-C | 0 | 73 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,071 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1324917/Python3-solution | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
d = {}
res = 0
for i in range(lowLimit, highLimit+1, 1):
tmp = 0
while i >0:
tmp += i%10
i = i//10
if tmp not in d:
d[tmp] =1
else:
d[tmp]+=1
if d[tmp] > res:
res = d[tmp]
return res | maximum-number-of-balls-in-a-box | Python3 solution | Wyhever | 0 | 98 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,072 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1114976/Simple-solution-in-normal-programming-way | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
a=[0]*46
for i in range(lowLimit,highLimit+1):
a[self.noofballs(i)] += 1
return (max(a))
def noofballs(self,n:int):
sum=0
while n>0:
sum=sum+n%(10)
n=n//10
return sum | maximum-number-of-balls-in-a-box | Simple solution in normal programming way | ashish87 | 0 | 103 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,073 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1112626/simple-python-solution | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
box=[0]*(max(lowLimit,highLimit))
for i in range(lowLimit,highLimit+1):
s=sum(list(map(int,str(i))))
box[s-1]+=1
return max(box)
``` | maximum-number-of-balls-in-a-box | simple python solution | kalluri_sumanth | 0 | 184 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,074 |
https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1043185/Python3-simple-solution-using-dictionary | class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
d = {}
for i in range(lowLimit, highLimit+1):
d[sum(list(map(int,list(str(i)))))] = d.get(sum(list(map(int,list(str(i))))),0) + 1
return max(d.values()) | maximum-number-of-balls-in-a-box | Python3 simple solution using dictionary | EklavyaJoshi | 0 | 60 | maximum number of balls in a box | 1,742 | 0.739 | Easy | 25,075 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/1042939/Python3-graph | class Solution:
def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
graph = {}
for u, v in adjacentPairs:
graph.setdefault(u, []).append(v)
graph.setdefault(v, []).append(u)
ans = []
seen = set()
stack = [next(x for x in graph if len(graph[x]) == 1)]
while stack:
n = stack.pop()
ans.append(n)
seen.add(n)
for nn in graph[n]:
if nn not in seen: stack.append(nn)
return ans | restore-the-array-from-adjacent-pairs | [Python3] graph | ye15 | 14 | 1,300 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,076 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/1043117/Python3-Graph-%2B-DFS-or-easy-to-understand | class Solution:
def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
# create the map
adj = collections.defaultdict(list)
for a, b in adjacentPairs:
adj[a].append(b)
adj[b].append(a)
# find the start num
start = adjacentPairs[0][0]
for k, v in adj.items():
if len(v) ==1:
start = k
break
# dfs to connect the graph
nums=[]
seen = set()
def dfs(num):
seen.add(num)
for next_num in adj[num]:
if next_num in seen: continue
dfs(next_num)
nums.append(num)
dfs(start)
return nums | restore-the-array-from-adjacent-pairs | [Python3] Graph + DFS | easy to understand | SonicM | 8 | 1,000 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,077 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/1915756/Python3-Easy-DFS-Solution | class Solution:
def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
adjList = defaultdict(list)
visited = set()
res = []
for a, b in adjacentPairs:
adjList[a].append(b)
adjList[b].append(a)
def dfs(element):
visited.add(element)
res.append(element)
for nei in adjList[element]:
if nei not in visited:
dfs(nei)
for start in adjList.keys():
if len(adjList[start]) == 1:
dfs(start)
break
return res | restore-the-array-from-adjacent-pairs | [Python3] Easy DFS Solution | ochen24 | 2 | 152 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,078 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/2601489/Python-Readable-Iterative-Solution-using-HashSet | class Solution:
def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
# make the connections
conns = collections.defaultdict(set)
# connect the connections
for a, b in adjacentPairs:
conns[a].add(b)
conns[b].add(a)
# find the start value
for node, conn in conns.items():
if len(conn) == 1:
break
# remove the first node from the first connection
result = [node]
while conns[node]:
# get the next element
# (there can only be one, as we always remove one)
new = conns[node].pop()
# append the new element to the list
result.append(new)
# delete node from the next connections
conns[new].remove(node)
# set the new element
node = new
return result | restore-the-array-from-adjacent-pairs | [Python] - Readable Iterative Solution using HashSet | Lucew | 1 | 141 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,079 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/2177279/python3-simple-dfs-solution | class Solution:
def restoreArray(self, adj: List[List[int]]) -> List[int]:
## RC ##
## APPROACH : GRAPH / DFS ##
graph = collections.defaultdict(list)
for u,v in adj:
graph[u].append(v)
graph[v].append(u)
first = None
for u in graph:
if len(graph[u]) == 1:
first = u
break
res = []
visited = set()
def dfs(node):
if node in visited:
return
visited.add(node)
res.append(node)
for nei in graph[node]:
dfs(nei)
dfs(first)
return res | restore-the-array-from-adjacent-pairs | [python3] simple dfs solution | 101leetcode | 1 | 57 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,080 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/1316177/Python3-solution-using-dictionary-and-dfs | class Solution:
def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
d = {}
for i,j in adjacentPairs:
d[i] = d.get(i,[])+[j]
d[j] = d.get(j,[])+[i]
for i in d:
if len(d[i]) == 1:
start = i
break
def dfs(s,r,v,d):
r.append(s)
v.add(s)
for i in d[s]:
if i not in v:
v.add(i)
dfs(i,r,v,d)
visited = set()
res = []
dfs(start,res,visited,d)
return res | restore-the-array-from-adjacent-pairs | Python3 solution using dictionary and dfs | EklavyaJoshi | 1 | 107 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,081 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/1043896/Python3-simple-solution | class Solution:
def restoreArray(self, ap: List[List[int]]) -> List[int]:
nei = collections.defaultdict(list)
for x, y in ap:
nei[x].append(y)
nei[y].append(x)
res = []
for k, v in nei.items():
if len(v) == 1:
res = [k, v[0]]
break
while len(res) < len(ap)+1:
for x in nei[res[-1]]:
if x != res[-2]:
res.append(x)
break
return res | restore-the-array-from-adjacent-pairs | [Python3] simple solution | joysword | 1 | 111 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,082 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/1043889/Python-Solution-with-Detailed-Comments | class Solution:
def restoreArray(self, A: List[List[int]]) -> List[int]:
"""
Build an edge-list/graph.
Just do print(graph) if you're not sure what's
happening here.
"""
graph = collections.defaultdict(list)
for i, j in A:
graph[i].append(j)
graph[j].append(i)
"""
If you print(graph), you can see that the start
and end nodes always point to ONLY one node,
while middle nodes point to two nodes (as
a doubly linked list).
"""
start = None
for i in graph.keys():
if len(graph[i]) == 1:
start = i
"""
DFS code.
This is a "greedy" DFS as we only call it once.
We know the path we choose will be successful
since we are guarunteed to start at the end of
the array. That's why we can keep the result &
seen 'outside' of the dfs function.
"""
result, seen = [], set()
def dfs(i):
seen.add(i)
for next_node in graph[i]:
if next_node not in seen:
dfs(next_node)
result.append(i)
"""
Now, we just call dfs & return!
"""
dfs(start)
return result | restore-the-array-from-adjacent-pairs | Python Solution with Detailed Comments | dev-josh | 1 | 156 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,083 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/2203282/python-3-or-space-optimized-dfs | class Solution:
def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
graph = collections.defaultdict(list)
for u, v in adjacentPairs:
graph[u].append(v)
graph[v].append(u)
nums = []
for num, adj in graph.items():
if len(adj) == 1:
nums.append(num)
nums.append(adj[0])
break
for _ in range(len(graph) - 2):
if graph[nums[-1]][0] != nums[-2]:
nums.append(graph[nums[-1]][0])
else:
nums.append(graph[nums[-1]][1])
return nums | restore-the-array-from-adjacent-pairs | python 3 | space optimized dfs | dereky4 | 0 | 48 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,084 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/1526225/Python3-solution-with-comments | class Solution:
def restoreArray(self, arr: List[List[int]]) -> List[int]:
s = set()
d = {}
for i in range(len(arr)):
for j in range(len(arr[0])):
if arr[i][j] not in d: # getting a dict to store every element and a list with positions where he could go
if j == 0:
d[arr[i][j]] = [arr[i][1]]
else:
d[arr[i][j]] = [arr[i][0]]
else:
if j == 0:
d[arr[i][j]].append(arr[i][1])
else:
d[arr[i][j]].append(arr[i][0])
if arr[i][j] not in s: # get the 2 edges of the array (they appear only once)
s.add(arr[i][j])
else:
s.remove(arr[i][j])
s = list(s)
seen = set() # creating a set to mark where we've been
count = 1
res = [s[0]]
element = s[0]
seen.add(element)
while count != len(arr)+1: # the number of elements in our list will be len(arr) + 1
if len(d[element]) == 1 and count > 1: # get out because we found the other edge (the edge will have 1 element in his list)
break
if d[element][0] not in seen:
res.append(d[element][0])
element = d[element][0]
seen.add(element) # mark every element we added to our list
elif d[element][1] not in seen:
res.append(d[element][1])
element = d[element][1]
seen.add(element) # mark every element we added to our list
count += 1
return res | restore-the-array-from-adjacent-pairs | Python3 solution with comments | FlorinnC1 | 0 | 141 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,085 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/1526138/Python3-Solution-with-using-dfs | class Solution:
def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
graph = {}
# build graph
for u, v in adjacentPairs:
if u not in graph:
graph[u] = []
graph[u].append(v)
if v not in graph:
graph[v] = []
graph[v].append(u)
visited = set()
ans = []
# find begin of array
cur = 0
for v in graph:
if len(graph[v]) == 1:
cur = v
visited.add(cur)
ans.append(cur)
break
# dfs
while len(ans) < len(graph):
neighbors = graph[cur]
for neighbor in neighbors:
if neighbor not in visited:
cur = neighbor
visited.add(cur)
ans.append(cur)
return ans | restore-the-array-from-adjacent-pairs | [Python3] Solution with using dfs | maosipov11 | 0 | 60 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,086 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/1512178/Python-O(n)-time-O(n)-space-solution | class Solution:
def restoreArray(self, pairs: List[List[int]]) -> List[int]:
h = defaultdict(set)
n = 0
for x, y in pairs:
h[x].add(y)
h[y].add(x)
n += 1
n += 1 # n distinct numbers
p = -1
for v in h:
if len(h[v]) ==1:
p = v
res = []
cnt = 0
while len(res) < n:
res.append(p)
if len(h[p]) == 1:
val = sum(h[p]) ## takes O(1), cuz h[p] has only 1 element
else:
break
h[p].remove(val)
h[val].remove(p)
p = val
cnt += 1
return res | restore-the-array-from-adjacent-pairs | Python O(n) time, O(n) space solution | byuns9334 | 0 | 156 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,087 |
https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/discuss/1055499/Python3-or-Just-map-the-Relations | class Solution:
def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
# map the relations between two elements using dictionary
# key=element, value=[elements it is paired with]
dict1={}
for ii in adjacentPairs:
if ii[0] in dict1.keys():
dict1[ii[0]].append(ii[1])
if ii[0] not in dict1.keys():
dict1[ii[0]]=[ii[1]]
if ii[1] in dict1.keys():
dict1[ii[1]].append(ii[0])
if ii[1] not in dict1.keys():
dict1[ii[1]]=[ii[0]]
# 1st and last element of series will have only pairing element in the list
# finding those 2 elements
#temp array to store those 2 elements
gajab=[]
for ii in dict1.keys():
if len(dict1[ii])==1: # if len==1 it can be start or ending elemnt of list
gajab.append(ii)
# Initializing Solution array with all elemnts as Zero
# len of solution array = len(pairs)+1
ans=[0]*((len(adjacentPairs))+1)
#Placing 1st and last element in array.You can make any of two element as STARTING or ENDING
ans[0]=gajab[0]
ans[-1]=gajab[1]
#using 1st element in ANS array to find its relation element and then so on...
for ii in range(len(ans)-1):
pp=dict1[ans[ii]][0]
dict1[pp].remove(ans[ii]) # removing the element relation that is already in ANS array
ans[ii+1]=pp # adding the element in ANS array at correct position
return(ans) | restore-the-array-from-adjacent-pairs | Python3 | Just map the Relations | vishal9994 | 0 | 110 | restore the array from adjacent pairs | 1,743 | 0.687 | Medium | 25,088 |
https://leetcode.com/problems/can-you-eat-your-favorite-candy-on-your-favorite-day/discuss/1042952/Python3-greedy | class Solution:
def canEat(self, candiesCount: List[int], queries: List[List[int]]) -> List[bool]:
prefix = [0]
for x in candiesCount: prefix.append(prefix[-1] + x) # prefix sum
return [prefix[t] < (day+1)*cap and day < prefix[t+1] for t, day, cap in queries] | can-you-eat-your-favorite-candy-on-your-favorite-day | [Python3] greedy | ye15 | 6 | 279 | can you eat your favorite candy on your favorite day | 1,744 | 0.329 | Medium | 25,089 |
https://leetcode.com/problems/palindrome-partitioning-iv/discuss/1042964/Python3-dp | class Solution:
def checkPartitioning(self, s: str) -> bool:
mp = {}
for i in range(2*len(s)-1):
lo, hi = i//2, (i+1)//2
while 0 <= lo <= hi < len(s) and s[lo] == s[hi]:
mp.setdefault(lo, set()).add(hi)
lo -= 1
hi += 1
@lru_cache(None)
def fn(i, k):
"""Return True if s[i:] can be split into k palindromic substrings."""
if k < 0: return False
if i == len(s): return k == 0
return any(fn(ii+1, k-1) for ii in mp[i])
return fn(0, 3) | palindrome-partitioning-iv | [Python3] dp | ye15 | 10 | 725 | palindrome partitioning iv | 1,745 | 0.459 | Hard | 25,090 |
https://leetcode.com/problems/palindrome-partitioning-iv/discuss/1042964/Python3-dp | class Solution:
def checkPartitioning(self, s: str) -> bool:
mp = defaultdict(set)
for i in range(2*len(s)-1):
lo, hi = i//2, (i+1)//2
while 0 <= lo <= hi < len(s) and s[lo] == s[hi]:
mp[lo].add(hi)
lo, hi = lo-1, hi+1
for i in range(len(s)):
for j in range(i+1, len(s)):
if i-1 in mp[0] and j-1 in mp[i] and len(s)-1 in mp[j]: return True
return False | palindrome-partitioning-iv | [Python3] dp | ye15 | 10 | 725 | palindrome partitioning iv | 1,745 | 0.459 | Hard | 25,091 |
https://leetcode.com/problems/palindrome-partitioning-iv/discuss/1043049/Python-High-Speed-Permutation-Finding | class Solution:
def checkPartitioning(self, s: str) -> bool:
# Add bogus characters to s, so we only have to worry about odd palindromes.
s = "|" + "|".join(s) + "|"
# p[i] is the length of the longest palindrome centered at i, minus 2.
p = [0] * len(s)
# c is the center of the longest palindrome found so far that ends at r.
# r is the index of the rightmost element in that palindrome; c + p[c] == r.
c = r = 0
for i in range(len(s)):
# The mirror of i about c.
mirror = c - (i - c)
# If there is a palindrome centered on the mirror that is also within the
# c-palindrome, then we dont have to check a few characters as we
# know they will match already.
p[i] = max(min(r - i, p[mirror]), 0)
# We expand the palindrome centered at i.
a = i + (p[i] + 1)
b = i - (p[i] + 1)
while a < len(s) and b >= 0 and s[a] == s[b]:
p[i] += 1
a += 1
b -= 1
# If the palindrome centered at i extends beyond r, we update c to i.
if i + p[i] > r:
c = i
r = i + p[i]
@lru_cache(None)
def dfs(i, k):
# can s[i:] be decomposed into k palindromes
mid = i + ((len(p) - i) // 2)
if k == 1:
# s[i:] itself must be a palindrome
return p[mid] == mid - i
for j in range(mid, i - 1, -1):
# p[j] == j - i means s[i:j + p[j]] is a palindrome
if p[j] == j - i and dfs(j + p[j], k - 1):
return True
return False
return dfs(0, 3) | palindrome-partitioning-iv | Python - High Speed Permutation Finding | mildog8 | 2 | 199 | palindrome partitioning iv | 1,745 | 0.459 | Hard | 25,092 |
https://leetcode.com/problems/palindrome-partitioning-iv/discuss/2812702/Python3-Solution-or-DP-or-Bit-Manipulation-or-O(n2) | class Solution:
def checkPartitioning(self, S):
N = len(S)
dp = [1] + [0] * N
for i in range(2 * N - 1):
l = i // 2
r = l + (i & 1)
while 0 <= l and r < N and S[l] == S[r]:
dp[r + 1] |= (dp[l] << 1)
l -= 1
r += 1
return bool(dp[-1] & (1 << 3)) | palindrome-partitioning-iv | ✔ Python3 Solution | DP | Bit Manipulation | O(n^2) | satyam2001 | 1 | 14 | palindrome partitioning iv | 1,745 | 0.459 | Hard | 25,093 |
https://leetcode.com/problems/palindrome-partitioning-iv/discuss/1833531/Simple-python-solution-O(N2)-or-Easy-to-understand | class Solution:
def checkPartitioning(self, s: str) -> bool:
n=len(s)
pal=[[False]*n for i in range(n)]
for i in range(n):
pal[i][i]=True
for i in range(n-1,-1,-1):
for j in range(i+1,n):
if i+1==j and s[i]==s[j]:
pal[i][j]=True
if pal[i+1][j-1] and s[i]==s[j]:
pal[i][j]=True
dp=[[False for i in range(4)] for j in range(n+1)]
dp[n][3]=True
for ind in range(n-1,-1,-1):
for i in range(ind,n):
if pal[ind][i]:
dp[ind][2] |= dp[i+1][3]
dp[ind][1] |= dp[i+1][2]
dp[ind][0] |= dp[i+1][1]
return dp[0][0] | palindrome-partitioning-iv | Simple python solution O(N^2) | Easy to understand | _YASH_ | 0 | 71 | palindrome partitioning iv | 1,745 | 0.459 | Hard | 25,094 |
https://leetcode.com/problems/sum-of-unique-elements/discuss/1103188/Runtime-97-or-Python-easy-hashmap-solution | class Solution:
def sumOfUnique(self, nums: List[int]) -> int:
hashmap = {}
for i in nums:
if i in hashmap.keys():
hashmap[i] += 1
else:
hashmap[i] = 1
sum = 0
for k, v in hashmap.items():
if v == 1: sum += k
return sum | sum-of-unique-elements | Runtime 97% | Python easy hashmap solution | vanigupta20024 | 19 | 1,900 | sum of unique elements | 1,748 | 0.757 | Easy | 25,095 |
https://leetcode.com/problems/sum-of-unique-elements/discuss/1135187/Python-faster-than-99-(so-they-say) | class Solution:
def sumOfUnique(self, nums: List[int]) -> int:
uniq = []
[uniq.append(num) for num in nums if nums.count(num) == 1]
return sum(uniq) | sum-of-unique-elements | Python faster than 99% (so they say) | 111110100 | 7 | 853 | sum of unique elements | 1,748 | 0.757 | Easy | 25,096 |
https://leetcode.com/problems/sum-of-unique-elements/discuss/1504051/Python-Runtime-98-or-Set | class Solution:
def sumOfUnique(self, nums: List[int]) -> int:
s = set(nums)
for i in nums:
if nums.count(i) > 1 and i in s:
s.remove(i)
return sum(s) | sum-of-unique-elements | [Python] Runtime 98% | Set | deleted_user | 3 | 336 | sum of unique elements | 1,748 | 0.757 | Easy | 25,097 |
https://leetcode.com/problems/sum-of-unique-elements/discuss/1200613/Easy-simple-Python-3-solution-using-counter | class Solution:
def sumOfUnique(self, nums: List[int]) -> int:
count = Counter(nums)
ans = 0
for index,value in enumerate(nums):
if count[value]==1:
ans+=value
return ans | sum-of-unique-elements | Easy simple Python 3 solution using counter | Sanyamx1x | 3 | 261 | sum of unique elements | 1,748 | 0.757 | Easy | 25,098 |
https://leetcode.com/problems/sum-of-unique-elements/discuss/1056649/Python3-freq-table | class Solution:
def sumOfUnique(self, nums: List[int]) -> int:
freq = {}
for x in nums: freq[x] = 1 + freq.get(x, 0)
return sum(x for x in nums if freq[x] == 1) | sum-of-unique-elements | [Python3] freq table | ye15 | 2 | 84 | sum of unique elements | 1,748 | 0.757 | Easy | 25,099 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.