Leetcode imports modules as wildcards, so you don't have to specify module names. There are some exceptions:
- Single
bisect()without a prefix triggersobject is not callable, usebisect.bisect()orbisect_left(). - You have to specify
re.subbecausesubwithout a prefix isoperator.sub. - Default
powis__builtins__['pow'](supports up to 3 arguments, including the modulus), notmath.pow.
For example, Leetcode header has import * from itertools, so we use comb() instead of itertools.comb():
class Solution:
def uniquePaths(self, m: int, n: int) -> int:
return comb(m+n-2, n-1)You can also use __import__('module').func for unlisted modules (i.e. "numpy", "scipy", and "sortedcontainers").
class Solution:
def checkStraightLine(self, p):
return __import__('numpy').linalg.matrix_rank([[1]+x for x in p])<3Sometimes you can save on casting of the return type, e.g. Leetcode autoconverts keys and mixed types to lists.
class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
return dict(Counter(nums).most_common(k))It also automatically evaluates generators:
class Solution:
def countBits(self, n: int) -> List[int]:
return map(int.bit_count,range(n+1))You can also return linked list of values as ListNode('a,b,...'). This one is really specific, but sometimes useful.
class Solution:
def addTwoNumbers(self, a: Optional[ListNode], b: Optional[ListNode]) -> Optional[ListNode]:
f=lambda n:n and n.val+10*f(n.next)or 0;return ListNode(','.join([*str(f(a)+f(b))][::-1]))Leetcode lists and trees also have serialize and deserialize functions:
class Solution:
def reverseList(self, h: Optional[ListNode]) -> Optional[ListNode]:
return h and h.deserialize(str(eval(h.serialize(h))[::-1]))More undocumented functions, obtainable with print(dir(ClassName)):
- ListNode:
_has_cycle,_list_node_to_array,_array_to_list_node. - TreeNode:
_has_cycle,_tree_node_to_array,_array_to_tree_node.
You can also dump the entire solution file with all the imports and the driver code:
with open(__file__, 'rt') as f:
print(f.read())The solution driver code writes all results to the user.out file, so we can use it like this:
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
from zlib import decompress
from base64 import b64decode
open('user.out', 'wb').write(decompress(b64decode('eJzdkMEVwCAIQ++dggFyEKi2zuLr/mtItZb63KAc\
kpfwuVAYFK6tCIjNPH1KncodJMuBTqWTYUGe89hNX1Kd/K2Nh1iM3mYbkMlpIaFrvvcCaVwCH+YB3FSHVu5xXDc='))),exit(0)There is no approved method to get all the test cases for problems in LeetCode. You can, however, leverage the fact that LeetCode reveals the test case that causes your code to fail. The solution above is not very reliable, because tests and environment may change, but it's pretty fast.
You can also try exploring the sandbox using shell commands, e.g.:
import subprocess
print(subprocess.run(["ls", "-la", "/"]))Fictitious (anonymous) lambdas also may be nested. E.g. you can use lambdas as parameters:
(lambda a,b,c: code)(a,b,c)becomes(lambda a,b,c: code)(lambda a: code, lamda b: code, lambda c: code)
You can't unpack lambda tuples in Python 3 since PEP 3113, however, if your lambda is flat, there is an upgrade path:
lambda (x, y): x + yin Python 2 becomeslambda xy:(lambda x,y: x+y)(*xy)in Python 3.
You can also unpack multiple tuples as lambda xy,ab:(lambda x,y,a,b: x+y+a+b)(*(xy+ab)).
class Solution:
def countVowelPermutation(self, n: int) -> int:
return sum(reduce(lambda x,_:(lambda a,e,i,o,u:(e+i+u,a+i,e+o,i,i+o))(*x),[0]*(n-1),[1]*5))
%(10**9+7)Generator expressions (x for y in z) are memory efficient since they only require memory for
the one value they yield. If you don't care about memory you can use square brackets to make it a list comprehension that automatically runs the loop.
You can also exhaust a generator using all() or any() depending on the return values.
You can also save a few chars using [*g] syntax instead of list(g) where g is a generator function.
Generator length len(list(g)) can be calculated in constant memory as sum(1 for _ in g).
Counters (collections.Counter()) can be updated, similar to dict.update(), it's much faster than a sum of counters.
E.g. c[i]+=1 is equivalent to c.update([i]), c[i]-=1 is c.update({i:-1}).
To delete a key you can use the .pop method (same as del), it's shorter than popitem().
You can easily drop zero and negative values in a counter (it's the official way, see documentation):
c = Counter({1:1,2:0,3:-1}); print(c:=+c) #{1: 1}, same as c += Counter()Since python 3.7, as a dict subclass, Counter inherited the capability to remember insertion order.
class Solution:
def reductionOperations(self, n: List[int]) -> int:
return sum(i*v for i,(_,v)in enumerate(sorted(Counter(n).items())))
class Solution:
def reductionOperations(self, n: List[int]) -> int:
return sum(i*v for i,v in enumerate(Counter(sorted(n)).values()))Since Python 3.10 you can use total() to compute sum of the counts.
class Solution:
def minSteps(self, s: str, t: str) -> int:
return sum((Counter(s)-Counter(t)).values())
class Solution:
def minSteps(self, s: str, t: str) -> int:
return(Counter(s)-Counter(t)).total()The controversial walrus operator (:=) added in Python 3.8 (PEP-572
that made Guido resign),
can be used to define or update a variable or a function (mostly used for recursive functions).
You can define and call a recursive function in a single line with Y-combinator, e.g.:
return (lambda y,x:y(y,x))(lambda f,x:1 if x==0 else x*f(f,x-1),5)But the walrus operator syntax is much more concise:
return (f:=lambda x:1 if x==0 else x*f(x-1))(5)Many oneliners would be impossible to do without it (or rather, very hard, with nested lambdas).
Sometimes you don't even need extra brackets, e.g. in map(f:=x,y) or next(g,f:=x) so it may be shorter than operators separated by semicolons.
class Solution:
def numOfMinutes(self, n: int, h: int, m: List[int], t: List[int]) -> int:
return max(map(f:=cache(lambda i:~i and t[i]+f(m[i])),m))class Solution(object):
def guessNumber(self, n: int) -> int:
l,r = 1, n
while l <= r:
m = (l + r) // 2
res = guess(m)
if res == 0:
return m
elif res > 0:
l = m + 1
else:
r = m - 1
return 0
class Solution:
def guessNumber(self, n: int) -> int:
return (f:=lambda l,h:h if l+1==h else f(m,h) if guess(m:=(l+h)//2)>0 else f(l,m))(0,n)class Solution:
def reverse(self, x: int) -> int:
r, x = 0, abs(x)
while x:
r = r*10 + x%10
x //= 10
return ((x>0)-(x<0))*min(2**31, r)
class Solution:
def reverse(self, x: int) -> int:
return ((x>0)-(x<0))*min(2**31,(f:=lambda r,x:f(r*10 + x%10, x//10) if x else r)(0,abs(x)))class Solution:
def topKFrequent(self, words: List[str], k: int) -> List[str]:
return nsmallest(k,(f:=Counter(words)).keys(),lambda x:(-f[x],x))You can't use walrus operator for structures, however, you can use __setattr__ for dictionaries or __setitem__ for lists if you need an assignment
(functions return None). To set a key for the list or for the dictionary, you can also use setattr or setitem functions from the operator module,
e.g. c[x]=1 is the same as setitem(c,x,1).
class Solution:
def addOneRow(self, root: TreeNode, v: int, d: int, isLeft: bool = True) -> TreeNode:
if d == 1:
return TreeNode(v, root if isLeft else None, root if not isLeft else None)
if not root:
return None
root.left = self.addOneRow(root.left, v, d - 1, True)
root.right = self.addOneRow(root.right, v, d - 1, False)
return root
class Solution:
def addOneRow(self, root: TreeNode, v: int, d: int, isLeft: bool = True) -> TreeNode:
return TreeNode(v, root if isLeft else None, root if not isLeft else None) if d==1 else \
setattr(root,'left', self.addOneRow(root.left, v, d - 1, True)) or \
setattr(root,'right', self.addOneRow(root.right, v, d - 1, False)) or root if root else Noneclass Solution(object):
def deleteMiddle(self, head):
def f(a, b):
if not b:
return a.next
a.next = f(a.next, b.next.next) if b.next else f(a.next, b.next)
return a
return f(head, head.next)
class Solution(object):
def deleteMiddle(self, head):
return (f:=lambda a,b:setattr(a,'next',f(a.next, b.next.next) if b.next
else f(a.next, b.next)) or a if b else a.next)(head, head.next)Note that setitem also supports slices:
class Solution:
def countPrimes(self, n):
a = [0,0]+[1]*(n-2)
for i in range(2,int(n**0.5)+1):
if a[i]:
a[i*i:n:i] = [0]*len(a[i*i:n:i])
return sum(a)
class Solution:
def countPrimes(self, n):
return sum(reduce(lambda a,i:a[i] and setitem(a,slice(i*i,n,i),[0]*len(a[i*i:n:i])) or a,
range(2,int(n**0.5)+1), [0,0]+[1]*(n-2)))Note slices can extend the list implicitly, e.g.:
a = [0,1,2]
a[3:4] = [3] # [0,1,2,3]Be careful though, slicing doesn't extend list beyond the slice size:
a = [0,1]
a[3:4] = [3,4] # the result is [0,1,3,4], NOT [0,1,?,3,4] (!)Examples:
class Solution:
def longestObstacleCourseAtEachPosition(self, o: List[int]) -> List[int]:
d = []
for e in o:
i = bisect_right(d,e)
if i==len(d):
d.append(0)
d[i] = e
yield i+1
class Solution:
def longestObstacleCourseAtEachPosition(self, o: List[int]) -> List[int]:
d = []
for e in o:
i = bisect_right(d,e)
d[i:i+1] = [e]
yield i+1
class Solution:
def longestObstacleCourseAtEachPosition(self, o: List[int]) -> List[int]:
d=[];return[setitem(d,slice(i:=bisect_right(d,e),i+1),[e])or i+1for e in o]Sometimes exec is shorter than setitem.
ParkingSystem=type('',(),{'__init__':lambda s,a,b,c:setattr(s,'p',[0,a,b,c]),'addCar':lambda s,t:\
setitem(s.p,t,s.p[t]-1)or s.p[t]>=0})
ParkingSystem=type('',(),{'__init__':lambda s,a,b,c:setattr(s,'p',[0,a,b,c]),'addCar':lambda s,t:\
exec('s.p[t]-=1')or s.p[t]>=0})Use the usual bracket notation [] or dict.get(key,default) (where needed).
Getitem used to construct a bisect comparator object, now we have the key parameter (since Python 3.10).
class Solution:
def guessNumber(self, n: int) -> int:
return bisect_left(type('',(),{'__getitem__':lambda _,i: -guess(i)})(), 0, 1, n)
class Solution:
def guessNumber(self, n: int) -> int:
return bisect_left(range(n), 0, key=lambda num: -guess(num))You can also write a subclass in one line, if needed.
MyHashSet=type('',(set,),{'remove':set.discard,'contains':set.__contains__})MyStack=type('',(list,),{'push':list.append,'top':lambda s:s[-1],'empty':lambda s:not s})с=Counter;с.insert=lambda s,x:s.update({x})or s[x]<2;с.remove=lambda s,x:s.pop(x,0);
с.getRandom=lambda s:choice([*s]);RandomizedSet=сSometimes (not always) you can skip __init__ and use static attributes.
UndergroundSystem=type('',(),{'h':{},'m':{},
'checkIn':lambda s,i,v,t:setitem(s.m,i,(v,t)),
'checkOut':lambda s,i,d,w:(v:=s.m[i][0])and setitem(s.h,(v,d),
[*map(sum,zip(s.h.pop((v,d),(0,0)),(w-s.m[i][1],1)))]),
'getAverageTime':lambda s,v,d:truediv(*s.h[v,d])})While loops are not very oneliner-friendly. You can use count() generator with next().
Note that next default parameter gets initialized first so you can use it for the startup code (but can't use to calculate result).
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
seen = {}
for i,x in enumerate(nums):
if target-x in seen:
return seen[target-x], i
seen[x] = i
return False
class Solution:
def twoSum(self, n: List[int], t: int) -> List[int]:
return next(((m[t-x],i)for i,x in enumerate(n)if t-x in m or setitem(m,x,i)),m:={})class Solution:
def breakPalindrome(self, s: str) -> str:
for i in range(len(s) // 2):
if s[i] != 'a':
return s[:i] + 'a' + s[i + 1:]
return s[:-1] + 'b' if s[:-1] else ''
class Solution:
def breakPalindrome(self, s: str) -> str:
return next((s[:i]+'a'+s[i+1:]for i in range(len(s)//2)if s[i]!='a'),s[:-1]and s[:-1]+'b')class Solution:
def isPossible(self, target: List[int]) -> bool:
s = sum(target)
q = [-a for a in target]
heapify(q)
while True:
x = -heappop(q)
if x==1:
return True
if s==x:
return False
d = 1 + (x-1) % (s-x)
if x==d:
return False
s = s - x + d
heappush(q, -d)
class Solution:
def isPossible(self, target: List[int]) -> bool:
return (s:=sum(target),q:=[-a for a in target],heapify(q)) and next((x==1 for _ in count()
if (x:=-heappop(q))==1 or s==x or (d:=1+(x-1)%(s-x))==x or not (s:=s-x+d,heappush(q,-d))),1)You can also use takewhile(), it's also a generator, so you need to expand it (e.g. with repeat(0)).
class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
r, d = [], deque()
for i, n in enumerate(nums):
while d and n>=nums[d[-1]]:
d.pop()
d.append(i)
if d[0] == i-k:
d.popleft()
r.append(nums[d[0]])
return r[k-1:]
class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
return (d:=deque()) or reduce(lambda r,p:(
any(takewhile(lambda _:d and p[1]>=nums[d[-1]] and d.pop(), repeat(0))),
d.append(p[0]), d[0]==p[0]-k and d.popleft(), r.append(nums[d[0]])) and r,
enumerate(nums), [])[k-1:]You could also try any() or all() as a while loop instead of next(), it may be shorter.
You can assure that expression never returns None, using [] ([None] evaluates to True).
class Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
stones.sort()
while len(stones) > 1:
insort(stones,stones.pop() - stones.pop())
return stones[0]
class Solution:
def lastStoneWeight(self, s: List[int]) -> int:
return next((s[0] for _ in count() if not s[1:] or insort(s,s.pop()-s.pop())),s.sort())
class Solution:
def lastStoneWeight(self, s: List[int]) -> int:
return (s.sort(),all(s[1:] and [insort(s,s.pop()-s.pop())] for _ in count()),s[0])[2]You can also evalulate multiline code with exec. Unlike eval, is not limited to a single string.
class Solution:
def minimumOneBitOperations(self, n: int) -> int:
return next((r for _ in count()if not(n and(r:=r^n,n:=n//2))),r:=0)
class Solution:
def minimumOneBitOperations(self, n: int) -> int:
r=[0];exec('while n:\n r[0]^=n\n n//=2');return r[0]
class Solution:
def minimumOneBitOperations(self, n: int) -> int:
return(f:=lambda n:n and n^f(n//2))(n)To swap values you can use either exec (inline version of a,b=b,a) or a temporary variable (t:=a,a:=b,b:=t).
Note that eval accepts only a single expression, and returns the value of the given expression,
whereas exec ignores the return value from its code, and always returns None, its use has no effect
on the compiled bytecode of the function where it is used. It does however affect existing variables.
Example:
class Solution:
def sortColors(self, nums: List[int]) -> None:
def fn(t,b):
red, white, blue = t
return (swap:=lambda a,x,y:exec('a[x],a[y]=a[y],a[x]'),(swap(nums,red,white),
(red+1,white+1,blue))[1] if nums[white]==0 else ((red,white+1,blue) if nums[white]==1
else (swap(nums,white,blue),(red,white,blue-1))[1]))[1]
reduce(fn, nums, [0,0,len(nums)-1])
class Solution:
def sortColors(self, nums: List[int]) -> None:
(s:=lambda a,x,y:(t:=a[x],setitem(a,x,a[y]),setitem(a,y,t),a)[3],
f:=lambda a,i,j,k:(f(s(a,i,j),i+1,j+1,k) if a[j]==0 else f(a,i,j+1,k) if a[j]==1
else f(s(a,j,k),i,j,k-1)) if i<=j<=k else None)[1](nums,0,0,len(nums)-1)Also you can try a swap function here (but it's pretty long, I don't use it):
swap = lambda a,x,y:(lambda f=a.__setitem__:(f(x,(a[x],a[y])),f(y,a[x][0]),f(x,a[x][1])))()You can use map for a lot of things, for example to traverse through adjacent cells.
class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
def dfs(i,j):
if 0<=i<len(grid) and 0<=j<len(grid[0]) and grid[i][j]:
grid[i][j] = 0
return 1 + sum(map(dfs,(i+1,i,i-1,i),(j,j+1,j,j-1)))
return 0
return max(dfs(i,j) for i in range(len(grid)) for j in range(len(grid[0])))
class Solution:
def maxAreaOfIsland(self, g: List[List[int]]) -> int:
return max((f:=lambda i,j:setitem(g[i],j,0) or 1 + sum(map(f,(i+1,i,i-1,i),(j,j+1,j,j-1)))
if 0<=i<len(g) and 0<=j<len(g[0]) and g[i][j] else 0)(i,j)
for i in range(len(g)) for j in range(len(g[0])))Though it's shorter to use complex numbers for 2d maps.
class Solution:
def maxAreaOfIsland(self, grid):
grid = {i + j*1j: val for i, row in enumerate(grid) for j, val in enumerate(row)}
def area(z):
return grid.pop(z, 0) and 1 + sum(area(z + 1j**k) for k in range(4))
return max(map(area, set(grid)))
class Solution:
def maxAreaOfIsland(self, grid):
return max(map(a:=lambda z: g.pop(z, 0) and 1 + sum(a(z + 1j**k) for k in range(4)),
set(g:= {i + j*1j: val for i, row in enumerate(grid) for j, val in enumerate(row)})))Complex numbers in general are very useful as 2d coordinates:
class Solution:
def isPathCrossing(self, p: str) -> bool:
z=0;return len(p)>=len({0,*{z:=z+1j**'NESW'.find(c)for c in p}})You can convert lists or tuples to True with !=0 instead of bool() (3 chars shorter).
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
grid = {i + j*1j:int(val) for i,row in enumerate(grid) for j,val in enumerate(row)}
def f(z):
return grid.pop(z,0) and bool([f(z + 1j**k) for k in range(4)])
return sum(map(f, set(grid)))
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
return sum(map(f:=lambda z:g.pop(z,0) and [f(z + 1j**k) for k in range(4)]!=0,
set(g:={i + j*1j:int(x) for i,row in enumerate(grid) for j,x in enumerate(row)})))class Solution:
def closedIsland(self, grid: List[List[str]]) -> int:
g = {i+j*1j:1-x for i,r in enumerate(grid) for j,x in enumerate(r)}
f = lambda z:g.pop(z,0) and [f(z+1j**k) for k in range(4)]!=0
sum(f(z) for z in set(g) if not(0<z.real<len(grid)-1 and 0<z.imag<len(grid[0])-1))
return sum(map(f,set(g)))
class Solution:
def closedIsland(self, grid: List[List[str]]) -> int:
return (g:={i+j*1j:1-x for i,r in enumerate(grid) for j,x in enumerate(r)},
f:=lambda z:g.pop(z,0) and [f(z+1j**k) for k in range(4)]!=0,[f(z) for z in set(g)
if not(0<z.real<len(grid)-1 and 0<z.imag<len(grid[0])-1)]) and sum(map(f,set(g)))class Solution:
def uniquePathsIII(self, grid: List[List[int]]) -> int:
def f(z,r):
if x:=g.pop(z,0):
if x==3 and not g:
r = r + 1
for k in range(4):
r = f(z + 1j**k, r)
g.update({z:x})
return r
g = {i + j*1j:x+1 for i, row in enumerate(grid) for j,x in enumerate(row) if x!=-1}
return f(next(z for z,x in g.items() if x==2),0)
class Solution:
def uniquePathsIII(self, grid: List[List[int]]) -> int:
return (g:={i + j*1j:x+1 for i, row in enumerate(grid) for j,x in enumerate(row)
if x!=-1}) and (f:=lambda z,r:[(x:=g.pop(z,0)) and (x==3 and not g and (r:=r+1),
[r:=f(z + 1j**k,r) for k in range(4)],g.update({z:x}))] and r)
(next(z for z,x in g.items() if x==2), 0)Unicode find (NOT Union Find) is the greatest trick of all time to solve graph problems. The idea is to use string replace in a Unicode space. Introduced by Stephan Pochmann.
class Solution:
def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
t = ''.join(map(chr, range(1001)))
for u,v in edges:
if t[u]==t[v]:
return [u,v]
t = t.replace(t[u],t[v])
class Solution:
def findRedundantConnection(self, e: List[List[int]]) -> List[int]:
t=''.join(map(chr,range(1001)));
return next((u,v)for u,v in e if t[u]==t[v]or not(t:=t.replace(t[u],t[v])))Another example:
class Solution:
def swimInWater(self, g: List[List[int]]) -> int:
n = len(g)
t,r = ''.join(map(chr,range(n*n))),range(n)
for w,i,j in sorted((g[i][j],i,j)for i,j in product(r,r)):
for x,y in ((i+1,j),(i-1,j),(i,j+1),(i,j-1)):
if n>y>=0<=x<n and g[x][y]<=w:
t = t.replace(t[i*n+j],t[x*n+y])
if t[0]==t[-1]:
return w
return 0
class Solution:
def swimInWater(self, g: List[List[int]]) -> int:
n=len(g);t,r=''.join(map(chr,range(n*n))),range(n);return next((w for w,i,j in
sorted((g[i][j],i,j)for i,j in product(r,r))if[t:=t.replace(t[i*n+j],t[x*n+y]) for x,y
in((i+1,j),(i-1,j),(i,j+1),(i,j-1)) if n>y>=0<=x<n and g[x][y]<=w]and t[0]==t[-1]),0)Another example (Q4 at https://leetcode.com/contest/weekly-contest-392):
class Solution:
def minimumCost(self, n: int, edges: List[List[int]], query: List[List[int]]) -> List[int]:
t,c = ''.join(map(chr,range(n))),{}
for u,v,w in edges:
t = t.replace(t[u],t[v])
for u,v,w in edges:
c[t[u]] = c.get(t[u],w)&w
return [0 if u==v else c[t[u]] if t[u]==t[v] else -1 for u,v in query]
class Solution:
def minimumCost(self, n: int, e: List[List[int]], q: List[List[int]]) -> List[int]:
t,c=''.join(map(chr,range(n))),{};all(t:=t.replace(t[u],t[v])for u,v,_ in e);
[setitem(c,t[u],c.get(t[u],w)&w)for u,v,w in e];
return[u!=v and t[u]!=t[v]and-1or c[t[u]]for u,v in q]Cache decorator, @lru_cache or @cache (since Python 3.9) may be used as an inline function cache(lambda ...).
class Solution:
def maxProfit(self, k: int, prices: List[int]) -> int:
@cache
def dfs(i, k, sell):
return 0 if k==0 or i==len(prices) \
else max(dfs(i+1, k-1, 0) + prices[i], dfs(i+1, k, 1)) if sell \
else max(dfs(i+1, k, 1)-prices[i], dfs(i+1, k, sell))
return dfs(0, k, 0)
class Solution:
def maxProfit(self, k: int, prices: List[int]) -> int:
return (f:=cache(lambda i,k,s:0 if k==0 or i==len(prices)
else max(f(i+1,k-s,1-s)+prices[i]*(2*s-1),f(i+1,k,s))))(0,k,0)class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
@cache
def f(n):
return min([1 + f(n-c) for c in coins]) if n>0 else 0 if n==0 else inf
x = f(amount)
return x if x!=inf else -1
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
return (lambda x:x if x!=inf else -1)((f:=cache(lambda n:
min([1+f(n-c) for c in coins]) if n>0 else 0 if n==0 else inf))(amount))It is sometimes necessary to reset cache with cache_clear between tests to avoid Memory Limit Exceeded error.
class Solution:
def lengthOfLongestSubsequence(self, a: List[int], t: int) -> int:
return(a.sort(),r:=(f:=cache(lambda i,b:b and -inf if b<0 or i<0 else
max(1+f(i-1,b-a[i]),f(i-1,b))))(len(a)-1,t),f.cache_clear())and(-1,r)[r>0]Use it to flatten a loop.
class Solution:
def lengthOfLongestSubstring(self, s):
start, res, h = 0, 0, {}
for i, c in enumerate(s):
start = max(start, h.get(c,0))
res = max(res, i - start + 1)
h[c] = i + 1
return res
class Solution:
def lengthOfLongestSubstring(self, s):
def fn(a,b):
start, res, h = a
i, c = b
start = max(start, h.get(c,0))
res = max(res, i - start + 1)
h[c] = i + 1
return start,res,h
return reduce(fn,enumerate(s),[0,0,{}])[1]
class Solution:
def lengthOfLongestSubstring(self, s):
return reduce(lambda a,b:(s:=max(a[0],a[2].get(b[1],0)),max(a[1],b[0]-s+1),
{**a[2],b[1]:b[0]+1}),enumerate(s),(0,0,{}))[1]
class Solution:
def lengthOfLongestSubstring(self, s):
return reduce(lambda a,b:(lambda t,r,h,i,c:(s:=max(t,h.get(c,0)),max(r,i-s+1),
{**h,c:i+1}))(*a,*b),enumerate(s),(0,0,{}))[1]Another example:
class Solution:
def longestValidParentheses(self, s: str) -> int:
def fn(a,b):
r, s = a
i, p = b
return (max(r,i-s[-2][0]), s[:-1]) if p==')' and s[-1][1]=='(' else (r, s+[(i,p)])
return reduce(fn, enumerate(s), (0,[(-1, ')')]))[0]
class Solution:
def longestValidParentheses(self, s: str) -> int:
return reduce(lambda a,b:(max(a[0],b[0]-a[1][-2][0]),a[1][:-1]) if b[1]==')'
and a[1][-1][1]=='(' else (a[0],a[1]+[b]),enumerate(s),(0,[(-1,')')]))[0]Nobody will stop you from using semicolons, but you'd still have to convert while and for loops.
Example:
class Solution:
def swapNodes(self, h: Optional[ListNode], k: int) -> Optional[ListNode]:
q = h
i = 1
d = {}
while q:
d[i] = q
q = q.next
i += 1
d[k].val, d[i-k].val = d[i-k].val, d[k].val
return h
class Solution:
def swapNodes(self, h: Optional[ListNode], k: int) -> Optional[ListNode]:
q=h;i=1;d={};all(q and(setitem(d,i,q),q:=q.next,i:=i+1) for _
in count());d[k].val,d[i-k].val=d[i-k].val,d[k].val;return h
class Solution:
def swapNodes(self, h: Optional[ListNode], k: int) -> Optional[ListNode]:
l=[h]+[h:=h.next for _ in[1]*10**5if h];a,b=l[k-1],l[~k];a.val,b.val=b.val,a.val;return l[0]Many leetcode problems use Fibonacci sequence that can be calculated using a variety of different methods:
- https://r-knott.surrey.ac.uk/Fibonacci/fibFormula.html
- https://en.wikipedia.org/wiki/Generating_function
- https://mathworld.wolfram.com/TribonacciNumber.html
Examples:
class Solution:
def fib(self, n: int) -> int:
a,b = 0,1
for _ in range(n):
a,b = b,a+b
return a
class Solution:
def fib(self, n: int) -> int:
return pow(x:=2<<n,n+1,x*x+~x)%xclass Solution:
def climbStairs(self, n: int) -> int:
a=b=1
for _ in range(n):
a,b = b,a+b
return a
class Solution:
def climbStairs(self, n):
return pow(x:=2<<n,n+2,x*x+~x)%xclass Solution:
def tribonacci(self, n):
a,b,c = 1,0,0
for _ in range(n):
a,b,c = b,c,a+b+c
return c
class Solution:
def tribonacci(self, n: int) -> int:
return pow(x:=2<<n,n+2,~-x*x*x+~x)%xMany problems can be solved with a single regex:
class Solution:
def makeGood(self, s: str) -> str:
[s:=re.sub(r'(.)(?!\1)(?i:\1)','',s)for _ in s];return sclass Solution:
def sortVowels(self, s: str) -> str:
return re.sub(t:='(?i)[aeiou]',lambda m,v=sorted(findall(t,s)):heappop(v),s)class Solution:
def isValid(self, w: str) -> bool:
return match('^(?=.*[aeiou])(?=.*[^0-9aeiou])[a-z0-9]{3,}$',w,I)Note that key=itemgetter(n) is the same length as key=lambda x:x[n] but a little bit clearer to read.
Sometimes you can skip key=itemgetter(0) by converting an argument to a tuple (15 characters shorter).
class Solution:
def jobScheduling(self, s: List[int], e: List[int], p: List[int]) -> int:
a=sorted(zip(s,e,p));return(f:=cache(lambda i:i-len(a)and max(f(
bisect_left(a,a[i][1],key=itemgetter(0)))+a[i][2],f(i+1))))(0)
class Solution:
def jobScheduling(self, s: List[int], e: List[int], p: List[int]) -> int:
a=sorted(zip(s,e,p));return(f:=cache(lambda i:i-len(a)and max(f(
bisect_left(a,(a[i][1],)))+a[i][2],f(i+1))))(0)You could also use map(list.pop, v) instead of [x[-1] for x in v] to collect the last elements of the list.
class Solution:
def findDiagonalOrder(self, n: List[List[int]]) -> List[int]:
return map(list.pop,sorted([i+j,j,t]for i,r in enumerate(n)for j,t in enumerate(r)))Using zip to get elements from the list of tuples is usually shorter, but not always:
class Solution:
def findSmallestSetOfVertices(self, n: int, edges: List[List[int]]) -> List[int]:
return {*range(n)}-{*[*zip(*edges)][1]}
class Solution:
def findSmallestSetOfVertices(self, n: int, edges: List[List[int]]) -> List[int]:
return {*range(n)}-{j for _,j in edges}You can can use a!=b!=c in a single boolean condition, similar to 0<=i<n and m>j>=0<=i<n.
class Solution:
def expressiveWords(self, s: str, words: List[str]) -> int:
def f(v,w,j=0):
for i in range(len(v)):
if j<len(w) and v[i]==w[j]:
j += 1
elif v[i-1:i+2] != v[i]*3 != v[i-2:i+1]:
return False
return j==len(w)
return sum(f(s,w) for w in words)
class Solution:
def expressiveWords(self, s: str, words: List[str]) -> int:
return sum((f:=lambda v,w,j=0:next((0 for i in range(len(v)) if not(j<len(w) and v[i]==w[j]
and (j:=j+1))and v[i-1:i+2]!=v[i]*3!=v[i-2:i+1]),1) and j==len(w))(s,w) for w in words)Python handles comparisons in the same order as and operator (e.g. log throws exception if n==0):
class Solution:
def isPowerOfFour(self, n: int) -> bool:
return n>0 and log(n,4)%1==0
class Solution:
def isPowerOfFour(self, n: int) -> bool:
return n>0==log(n,4)%1~ reverts every bit. Therefore, ~x means -x-1. You can use it as reversed index, i.e. for i=0, a[~i] means a[-1], etc. or just replace -x-1 with ~x.
For integer n, you can write n+1 as -~n, n-1 as ~-n. This uses the same number of characters, but can indirectly cut spaces or parens for operator precedence.
class Solution:
def generateMatrix(self, n: int) -> List[List[int]]:
r=range(n);return[[4*(n-(a:=min(min(i,n-i-1),min(j,n-j-1))))
*a+(i+j-2*a+1,4*(n-2*a-1)-(i+j-2*a)+1)[i>j] for j in r] for i in r]
class Solution:
def generateMatrix(self, n: int) -> List[List[int]]:
r=range(n);return[[4*(n-(a:=min(i,j,~i+n,~j+n)))
*a+(i+j-2*a+1,4*n-6*a-i-j-3)[i>j]for j in r]for i in r]You can replace 0 if x==y else z with x-y and z, it's a little bit counterintuitive, but shorter.
Condition x if c else y can be written as c and x or y, it's shorter but depends on x (x should not be 0).
class Solution:
def snakesAndLadders(self, board: List[List[int]]) -> int:
n,v,q = len(board),{1:0},[1]
def f(i):
x = (i - 1)%n
y = (i - 1)//n
c = board[~y][~x if y%2 else x]
return c if c>0 else i
for i in q:
for j in range(i+1, i+7):
k = f(j)
if k==n*n:
return v[i]+1
if k not in v:
v[k] = v[i]+1
q.append(k)
return -1
class Solution:
def snakesAndLadders(self, board: List[List[int]]) -> int:
return (n:=len(board),v:={1:0},q:=[1]) and next((v[i]+1 for i in q for j in range(i+1,i+7)
if (k:=(x:=(j-1)%n,y:=(j-1)//n) and ((c:=board[~y][y%2 and ~x or x])>0 and c or j))==n*n
or (k not in v and (v.update({k:v[i]+1}) or q.append(k)))),-1)You can check if any of the numbers is negative as x|y<0 or if both numbers are non-zero as x|y.
class Solution:
def minPathSum(self, grid: List[List[int]]) -> int:
return (f:=cache(lambda i,j:i|j<0 and inf or grid[i][j]+(i|j and min(f(i,j-1),f(i-1,j)))))
(len(grid)-1,len(grid[0])-1)You can use bitwise &,| instead of and,or where possible. If x is 0..2, using x&1 is shorter than x==1.
class Solution:
def isScramble(self, s1: str, s2: str) -> bool:
return (f:=cache(lambda a,b:a==b or any((f(a[:i],b[:i]) and f(a[i:],b[i:]))
or (f(a[i:],b[:-i]) and f(a[:i],b[-i:])) for i in range(1,len(a)))))(s1,s2)
class Solution:
def isScramble(self, s1: str, s2: str) -> bool:
return (f:=cache(lambda a,b:a==b or any((f(a[:i],b[:i])&f(a[i:],b[i:]))
|(f(a[i:],b[:-i])&f(a[:i],b[-i:])) for i in range(1,len(a)))))(s1,s2)You can use booleans as indexes in lists, even nested: (a,(b,c)[u==w])[x==y], or you can multiply by a boolean.
class Solution:
def removeStars(self, s: str) -> str:
return reduce(lambda r,c:(r[:-1],r+c)[c>'*'],s)class Solution:
def simplifyPath(self, path: str) -> str:
return'/'+'/'.join(reduce(lambda r,p:(r+[p]*('.'!=p!=''),r[:-1])[p=='..'],path.split('/'),[]))Python 3 lacks cmp (3-way compare) and sign function (copysign(bool(x),x) is too long), but you can use (x>0)-(x<0) for sign(x)
and (a>b)-(a<b) for cmp(a,b). Note you can use -1,0,1 indexes for Python lists natively.
class Solution:
def stoneGameIII(self, v: List[int]) -> str:
f=cache(lambda i:i<len(v)and max(sum(v[i:i+k])-f(i+k)for k in(1,2,3)));x=f(0);
return('Tie','Alice','Bob')[(x>0)-(x<0)]
# return(('Tie','Bob')[x<0],'Alice')[x>0] # or like this (1 char shorter)You can replace cmp written as lambda x:(x>0)-(x<0) with 0..__le__ or .0.__le__ (11 characters shorter).
class Solution:
def rearrangeArray(self, n: List[int]) -> List[int]:
n.sort(key=lambda x:(x>0)-(x<0));return chain(*zip(n[len(n)//2:],n))
class Solution:
def rearrangeArray(self, n: List[int]) -> List[int]:
n.sort(key=0..__le__);return chain(*zip(n[len(n)//2:],n)) You can save a few characters using asterisk operator *.
One * means "expand this as a list", two ** means "expand this as a dictionary".
class Solution:
def checkStraightLine(self, p):
(a,b),(c,d)=p[:2];return all((x-a)*(d-b)==(c-a)*(y-b)for x,y in p)
class Solution:
def checkStraightLine(self, p):
(a,b),(c,d),*_=p;return all((x-a)*(d-b)==(c-a)*(y-b)for x,y in p)class Solution:
def findMaxAverage(self, n: List[int], k: int) -> float:
s=[0]+[*accumulate(n)];return max(map(sub,s[k:],s))/k
class Solution:
def findMaxAverage(self, n: List[int], k: int) -> float:
s=[0,*accumulate(n)];return max(map(sub,s[k:],s))/kQuite a few things become shorter with statistics.mode (most common value of discrete or nominal data).
class Solution:
def findErrorNums(self, nums: List[int]) -> List[int]:
t=sum({*nums});return sum(nums)-t,comb(len(nums)+1,2)-t
class Solution:
def findErrorNums(self, nums: List[int]) -> List[int]:
return mode(nums),comb(len(nums)+1,2)-sum({*nums})class Solution:
def majorityElement(self, nums: List[int]) -> int:
return sorted(nums)[len(nums)//2]
class Solution:
def majorityElement(self, nums: List[int]) -> int:
return mode(nums)# https://youtu.be/pKO9UjSeLew (Joma Tech: If Programming Was An Anime)
class Solution:
def findDuplicate(self, nums: List[int]) -> int:
tortoise = hare = nums[0]
while True:
tortoise = nums[tortoise]
hare = nums[nums[hare]]
if tortoise == hare:
break
tortoise = nums[0]
while tortoise != hare:
tortoise = nums[tortoise]
hare = nums[hare]
return hare
class Solution:
def findDuplicate(self, nums: List[int]) -> int:
return mode(nums)Notes:
- Unless the following token starts with e or E. You can remove the space following a number. E.g.
i==4 and j==4becomesi==4and j==4. - There's a nice way to convert an iterable to list using star operator, e.g.
x=[*g]equals*x,=g(1 char shorter). You can also use this syntax to unpack iterables, e.g.a,*b,c=range(5)meansa=1;b=[2,3,4];c=5. - Instead of range(x), you can use the * operator on a list of anything, e.g.
[1]*8can replacerange(8)(unless you really need the counter value). - Conditions like
if i<len(r)may be replaced withif r[i:], it's 3 characters shorter. - You can replace
set(n)with{*n}(2 characters shorter). - You can convert bool with
~~()instead ofint()(as in js) or prepend with a single+(5 characters shorter). - You can subtract 1 or replace
notoperator with bitwise negation~-to save on space (1-5 characters shorter). - You can check for set membership with
{x}&sinstead ofx in s(1 character shorter). - An expression like
x&(x-1)==0determines if unsignedxis power of 2 or 0 (Kernighan, tests rightmost bit). - Very often
x==0can be replaced withx<1(1 character shorter). - Generator expansion
[*g]can use a traling comma*g,in the initialization section (1 character shorter). - A condition like
h>i>=0<=j<wcan be written ash>i>-1<j<w(1 character shorter). - You can replace
q and q[-1]==cwithq[-1:]==[c](3 characters shorter).
- They're not charaters at all, they're just brilliant repositories of fantastic, killer one-liners (Stephen Fry interview)
- Don't Piss down my back and tell me it's raining: Notebook for Sarcastic, Witty, and Sharp Tongued One Liners
- Python One-Liners: Write Concise, Eloquent Python Like a Professional
- Python One-Liners - Concise Python Code
- Tips for Golfing in Python (Stack Overflow)
- One-line coder makes me depressed (Reddit)