Python3 reference for interview coding problems/light competitive programming. Contributions welcome!
I built this cheatsheet while teaching myself Python3 for various interviews and leetcoding for fun after not using Python for about a decade. This cheetsheet only contains code that I didn't know but needed to use to solve a specific coding problem. I did this to try to get a smaller high frequency subset of Python vs a comprehensive list of all methods. Additionally, the act of recording the syntax and algorithms helped me store it in memory and as a result I almost never actually referenced this sheet. Hopefully it helps you in your efforts or inspires you to build your own and best of luck!
I choose Python3 despite being more familiar with Javascript, Java, C++ and Golang for interviews as I felt Python had the combination of the most standard libraries available as well as syntax that resembles psuedo code, therefore being the most expressive. Python and Java both have the most examples but Python wins in this case due to being much more concise. I was able to get myself reasonably prepared with Python syntax in six weeks of practice. After picking up Python I have timed myself solving the same exercises in Golang and Python. Although I prefer Golang, I find that I can complete Python examples in half the time even accounting for +50% more bugs (approximately) that I tend to have in Python vs Go. This is optimizing for solved interview questions under pressure, when performance is considered then Go/C++ does consistently perform 1/10 the time of Python. In some rare cases, algorithms that time out in Python sometimes pass in C++/Go on Leetcode.
- Literals
- Loops
- Strings
- Slicing
- Tuples
- Sort
- Hash
- Set
- List
- Dict
- Binary Tree
- heapq
- lambda
- zip
- Random
- Constants
- Ternary Condition
- Bitwise operators
- For Else
- Modulo
- any
- all
- bisect
- math
- iter
- map
- filter
- reduce
- itertools
- regular expression
- Types
- Grids
- General Tips
- Binary Search
- Topological Sort
- Sliding Window
- Tree Tricks
- Binary Search Tree
- Anagrams
- Dynamic Programming
- Cyclic Sort
- Quick Sort
- Merge Sort
- Merge K Sorted Arrays
- Linked List
- Convert Base
- Parenthesis
- Max Profit Stock
- Shift Array Right
- Continuous Subarrays with Sum k
- Events
- Merge Meetings
- Trie
- Kadane's Algorithm - Max subarray sum
- Union Find/DSU
- Fast Power
- Fibonacci Golden
- Basic Calculator
- Reverse Polish
- Resevior Sampling
- Candy Crush
255, 0b11111111, 0o377, 0xff # Integers (decimal, binary, octal, hex)
123.0, 1.23 # Float
7 + 5j, 7j # Complex
'a', '\141', '\x61' # Character (literal, octal, hex)
'\n', '\\', '\'', '\"' # Newline, backslash, single quote, double quote
"string\n" # String of characters ending with newline
"hello"+"world" # Concatenated strings
True, False # bool constants, 1 == True, 0 == False
[1, 2, 3, 4, 5] # List
['meh', 'foo', 5] # List
(2, 4, 6, 8) # Tuple, immutable
{'name': 'a', 'age': 90} # Dict
{'a', 'e', 'i', 'o', 'u'} # Set
None # Null varGo through all elements
i = 0
while i < len(str):
i += 1equivalent
for i in range(len(message)):
print(i)Get largest number index from right
while i > 0 and nums [i-1] >= nums[i]:
i -= 1Manually reversing
l, r = i, len(nums) - 1
while l < r:
nums[l], nums[r] = nums[r], nums[l]
l += 1
r -= 1Go past the loop if we are clever with our boundry
for i in range(len(message) + 1):
if i == len(message) or message[i] == ' ':Fun with Ranges - range(start, stop, step)
for a in range(0,3): # 0,1,2
for a in reversed(range(0,3)) # 2,1,0
for i in range(3,-1,-1) # 3,2,1,0
for i in range(len(A)//2): # A = [0,1,2,3,4,5]
print(i) # 0,1,2
print(A[i]) # 0,1,2
print(~i) # -1,-2,-3
print(A[~i]) # 5,4,3str1.find('x') # find first location of char x and return index
str1.rfind('x') # find first int location of char x from reverseParse a log on ":"
l = "0:start:0"
tokens = l.split(":")
print(tokens) # ['0', 'start', '0']Reverse works with built in split, [::-1] and " ".join()
# s = "the sky is blue"
def reverseWords(self, s: str) -> str:
wordsWithoutWhitespace = s.split() # ['the', 'sky', 'is', 'blue']
reversedWords = wordsWithoutWhitespace[::-1] # ['blue', 'is', 'sky', 'the']
final = " ".join(reversedWords) # blue is sky theManual split based on isalpha()
def splitWords(input_string) -> list:
words = [] #
start = length = 0
for i, c in enumerate(input_string):
if c.isalpha():
if length == 0:
start = i
length += 1
else:
if length !=0:
words.append(input_string[start:start+length])
length = 0
if length > 0:
words.append(input_string[start:start+length])
return wordsTest type of char
def rotationalCipher(input, rotation_factor):
rtn = []
for c in input:
if c.isupper():
ci = ord(c) - ord('A')
ci = (ci + rotation_factor) % 26
rtn.append(chr(ord('A') + ci))
elif c.islower():
ci = ord(c) - ord('a')
ci = (ci + rotation_factor) % 26
rtn.append(chr(ord('a') + ci))
elif c.isnumeric():
ci = ord(c) - ord('0')
ci = (ci + rotation_factor) % 10
rtn.append(chr(ord('0') + ci))
else:
rtn.append(c)
return "".join(rtn)AlphaNumberic
isalnum()Get charactor index
print(ord('A')) # 65
print(ord('B')-ord('A')+1) # 2
print(chr(ord('a') + 2)) # cReplace characters or strings
def isValid(self, s: str) -> bool:
while '[]' in s or '()' in s or '{}' in s:
s = s.replace('[]','').replace('()','').replace('{}','')
return len(s) == 0Insert values in strings
txt3 = "My name is {}, I'm {}".format("John",36) # My name is John, I'm 36Multiply strings/lists with *, even booleans which map to True(1) and False(0)
'meh' * 2 # mehmeh
['meh'] * 2 # ['meh', 'meh']
['meh'] * True #['meh']
['meh'] * False #[]Find substring in string
txt = "Hello, welcome to my world."
x = txt.find("welcome") # 7startswith and endswith are very handy
str = "this is string example....wow!!!"
str.endswith("!!") # True
str.startswith("this") # True
str.endswith("is", 2, 4) # TruePython3 format strings
name = "Eric"
profession = "comedian"
affiliation = "Monty Python"
message = (
f"Hi {name}. "
f"You are a {profession}. "
f"You were in {affiliation}."
)
message
'Hi Eric. You are a comedian. You were in Monty Python.'Print string with all chars, useful for debugging
print(repr("meh\n")) # 'meh\n'Slicing intro
+---+---+---+---+---+---+
| P | y | t | h | o | n |
+---+---+---+---+---+---+
Slice position: 0 1 2 3 4 5 6
Index position: 0 1 2 3 4 5
p = ['P','y','t','h','o','n']
p[0] 'P' # indexing gives items, not lists
alpha[slice(2,4)] # equivalent to p[2:4]
p[0:1] # ['P'] Slicing gives lists
p[0:5] # ['P','y','t','h','o'] Start at beginning and count 5
p[2:4] = ['t','r'] # Slice assignment ['P','y','t','r','o','n']
p[2:4] = ['s','p','a','m'] # Slice assignment can be any size['P','y','s','p','a','m','o','n']
p[4:4] = ['x','y'] # insert slice ['P','y','t','h','x','y','o','n']
p[0:5:2] # ['P', 't', 'o'] sliceable[start:stop:step]
p[5:0:-1] # ['n', 'o', 'h', 't', 'y']Go through num and get combinations missing a member
numList = [1,2,3,4]
for i in range(len(numList)):
newList = numList[0:i] + numList[i+1:len(numList)]
print(newList) # [2, 3, 4], [1, 3, 4], [1, 2, 4], [1, 2, 3]Collection that is ordered and unchangable
thistuple = ("apple", "banana", "cherry")
print(thistuple[1]) # bananaCan be used with Dicts
def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
d = defaultdict(list)
for w in strs:
key = tuple(sorted(w))
d[key].append(w)
return d.values()sorted(iterable, key=key, reverse=reverse)
Sort sorts alphabectically, from smallest to largest
print(sorted(['Ford', 'BMW', 'Volvo'])) # ['BMW', 'Ford', 'Volvo']
nums = [-4,-1,0,3,10]
print(sorted(n*n for n in nums)) # [0,1,9,16,100]cars = ['Ford', 'BMW', 'Volvo']
cars.sort() # returns None type
cars.sort(key=lambda x: len(x) ) # ['BMW', 'Ford', 'Volvo']
print(sorted(cars, key=lambda x:len(x))) # ['BMW', 'Ford', 'Volvo']Sort key by value, even when value is a list
meh = {'a':3,'b':0,'c':2,'d':-1}
print(sorted(meh, key=lambda x:meh[x])) # ['d', 'b', 'c', 'a']
meh = {'a':[0,3,'a'],'b':[-2,-3,'b'],'c':[2,3,'c'],'d':[-2,-2,'d']}
print(sorted(meh, key=lambda x:meh[x])) # ['b', 'd', 'a', 'c']def merge_sorted_lists(arr1, arr2): # built in sorted does Timsort optimized for subsection sorted lists
return sorted(arr1 + arr2)Sort an array but keep the original indexes
self.idx, self.vals = zip(*sorted([(i,v) for i,v in enumerate(nums)], key=lambda x:x[1]))Sort by tuple, 2nd element then 1st ascending
a = [(5,10), (2,20), (2,3), (0,100)]
test = sorted(a, key = lambda x: (x[1],x[0]))
print(test) # [(2, 3), (5, 10), (2, 20), (0, 100)]
test = sorted(a, key = lambda x: (-x[1],x[0]))
print(test) # [(0, 100), (2, 20), (5, 10), (2, 3)]Sort and print dict values by key
ans = {-1: [(10, 1), (3, 3)], 0: [(0, 0), (2, 2), (7, 4)], -3: [(8, 5)]}
for key, value in sorted(ans.items()): print(value)
# [(8, 5)]
# [(10, 1), (3, 3)]
# [(0, 0), (2, 2), (7, 4)]
# sorted transforms dicts to lists
sorted(ans) # [-3, -1, 0]
sorted(ans.values()) # [[(0, 0), (2, 2), (7, 4)], [(8, 5)], [(10, 1), (3, 3)]]
sorted(ans.items()) # [(-3, [(8, 5)]), (-1, [(10, 1), (3, 3)]), (0, [(0, 0), (2, 2), (7, 4)])]
# Or just sort the dict directly
[ans[i] for i in sorted(ans)]
# [[(8, 5)], [(10, 1), (3, 3)], [(0, 0), (2, 2), (7, 4)]]for c in s1: # Adds counter for c
ht[c] = ht.get(c, 0) + 1 # ht[a] = 1, ht[a]=2, etca = 3
st = set()
st.add(a) # Add to st
st.remove(a) # Remove from st
st.discard(a) # Removes from set with no error
st.add(a) # Add to st
next(iter(s)) # return 3 without removal
st.pop() # returns 3s = set('abc') # {'c', 'a', 'b'}
s |= set('cdf') # {'f', 'a', 'b', 'd', 'c'} set s with elements from new set
s &= set('bd') # {'d', 'b'} only elements from new set
s -= set('b') # {'d'} remove elements from new set
s ^= set('abd') # {'a', 'b'} elements from s or new but not bothStacks are implemented with Lists. Stacks are good for parsing and graph traversal
test = [0] * 100 # initialize list with 100 0's2D
rtn.append([])
rtn[0].append(1) # [[1]]List Comprehension
number_list = [ x for x in range(20) if x % 2 == 0]
print(number_list) # [0, 2, 4, 6, 8, 10, 12, 14, 16, 18]Reverse a list
ss = [1,2,3]
ss.reverse()
print(ss) #3,2,1Join list
list1 = ["a", "b" , "c"]
list2 = [1, 2, 3]
list3 = list1 + list2 # ['a', 'b', 'c', 1, 2, 3]Hashtables are implemented with dictionaries
d = {'key': 'value'} # Declare dict{'key': 'value'}
d['key'] = 'value' # Add Key and Value
{x:0 for x in {'a', 'b'}} # {'a': 0, 'b': 0} declare through comprehension
d['key']) # Access value
d.items() # Items as tuple list dict_items([('key', 'value')])
if 'key' in d: print("meh") # Check if value exists
par = {}
par.setdefault(1,1) # returns 1, makes par = { 1 : 1 }
par = {0:True, 1:False}
par.pop(0) # Remove key 0, Returns True, par now {1: False}
for k in d: print(k) # Iterate through keysCreate Dict of Lists that match length of list to count votes
votes = ["ABC","CBD","BCA"]
rnk = {v:[0] * len(votes[0]) for v in votes[0]}
print(rnk) # {'A': [0, 0, 0], 'B': [0, 0, 0], 'C': [0, 0, 0]}-
A tree is an undirected graph in which any two vertices are connected by exactly one path.
-
Any connected graph who has n nodes with n-1 edges is a tree.
-
The degree of a vertex is the number of edges connected to the vertex.
-
A leaf is a vertex of degree 1. An internal vertex is a vertex of degree at least 2.
-
A path graph is a tree with two or more vertices with no branches, degree of 2 except for leaves which have degree of 1
-
Any two vertices in G can be connected by a unique simple path.
-
G is acyclic, and a simple cycle is formed if any edge is added to G.
-
G is connected and has no cycles.
-
G is connected but would become disconnected if any single edge is removed from G.
DFS Pre, In Order, and Post order Traversal
- Preorder
- encounters roots before leaves
- Create copy
- Inorder
- flatten tree back to original sequence
- Get values in non-decreasing order in BST
- Post order
- encounter leaves before roots
- Helpful for deleting
Recursive
"""
1
/ \
2 3
/ \
4 5
"""
# PostOrder 4 5 2 3 1 (Left-Right-Root)
def postOrder(node):
if node is None:
return
postorder(node.left)
postorder(node.right)
print(node.value, end=' ')Iterative PreOrder
# PreOrder 1 2 4 5 3 (Root-Left-Right)
def preOrder(tree_root):
stack = [(tree_root, False)]
while stack:
node, visited = stack.pop()
if node:
if visited:
print(node.value, end=' ')
else:
stack.append((node.right, False))
stack.append((node.left, False))
stack.append((node, True))Iterative InOrder
# InOrder 4 2 5 1 3 (Left-Root-Right)
def inOrder(tree_root):
stack = [(tree_root, False)]
while stack:
node, visited = stack.pop()
if node:
if visited:
print(node.value, end=' ')
else:
stack.append((node.right, False))
stack.append((node, True))
stack.append((node.left, False))Iterative PostOrder
# PostOrder 4 5 2 3 1 (Left-Right-Root)
def postOrder(tree_root):
stack = [(tree_root, False)]
while stack:
node, visited = stack.pop()
if node:
if visited:
print(node.value, end=' ')
else:
stack.append((node, True))
stack.append((node.right, False))
stack.append((node.left, False))Iterative BFS(LevelOrder)
from collections import deque
#BFS levelOrder 1 2 3 4 5
def levelOrder(tree_root):
queue = deque([tree_root])
while queue:
node = queue.popleft()
if node:
print(node.value, end=' ')
queue.append(node.left)
queue.append(node.right)
def levelOrderStack(tree_root):
stk = [(tree_root, 0)]
rtn = []
while stk:
node, depth = stk.pop()
if node:
if len(rtn) < depth + 1:
rtn.append([])
rtn[depth].append(node.value)
stk.append((node.right, depth+1))
stk.append((node.left, depth+1))
print(rtn)
return True
def levelOrderStackRec(tree_root):
rtn = []
def helper(node, depth):
if len(rtn) == depth:
rtn.append([])
rtn[depth].append(node.value)
if node.left:
helper(node.left, depth + 1)
if node.right:
helper(node.right, depth + 1)
helper(tree_root, 0)
print(rtn)
return rtnTraversing data types as a graph, for example BFS
def removeInvalidParentheses(self, s: str) -> List[str]:
rtn = []
v = set()
v.add(s)
if len(s) == 0: return [""]
while True:
for n in v:
if self.isValid(n):
rtn.append(n)
if len(rtn) > 0: break
level = set()
for n in v:
for i, c in enumerate(n):
if c == '(' or c == ')':
sub = n[0:i] + n[i + 1:len(n)]
level.add(sub)
v = level
return rtnReconstructing binary trees
- Binary tree could be constructed from preorder and inorder traversal
- Inorder traversal of BST is an array sorted in the ascending order
Convert tree to array and then to balanced tree
def balanceBST(self, root: TreeNode) -> TreeNode:
self.inorder = []
def getOrder(node):
if node is None:
return
getOrder(node.left)
self.inorder.append(node.val)
getOrder(node.right)
# Get inorder treenode ["1,2,3,4"]
getOrder(root)
# Convert to Tree
# 2
# 1 3
# 4
def bst(listTree):
if not listTree:
return None
mid = len(listTree) // 2
root = TreeNode(listTree[mid])
root.left = bst(listTree[:mid])
root.right = bst(listTree[mid+1:])
return root
return bst(self.inorder)Build an adjecency graph from edges list
# N = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
graph = [[] for _ in range(N)]
for u,v in edges:
graph[u].append(v)
graph[v].append(u)
# [[1, 2], [0], [0, 3, 4, 5], [2], [2], [2]]Build adjecency graph from traditional tree
adj = collections.defaultdict(list)
def dfs(node):
if node.left:
adj[node].append(node.left)
adj[node.left].append(node)
dfs(node.left)
if node.right:
adj[node].append(node.right)
adj[node.right].append(node)
dfs(node.right)
dfs(root)Traverse Tree in graph notation
# [[1, 2], [0], [0, 3, 4, 5], [2], [2], [2]]
def dfs(node, par=-1):
for nei in graph[node]:
if nei != par:
res = dfs(nei, node)
dfs(0) # 1->2->3->4->5 1
/ \
2 3
/ \ / \
5 6 8 7
- Implemented as complete binary tree, which has all levels as full excepted deepest
- In a heap tree the node is smaller than its children
def maximumProduct(self, nums: List[int]) -> int:
l = heapq.nlargest(3, nums)
s = heapq.nsmallest(3, nums)
return max(l[0]*l[1]*l[2],s[0]*s[1]*l[0])Heap elements can be tuples, heappop() frees the smallest element (flip sign to pop largest)
def kClosest(self, points: List[List[int]], K: int) -> List[List[int]]:
heap = []
for p in points:
distance = sqrt(p[0]* p[0] + p[1]*p[1])
heapq.heappush(heap,(-distance, p))
if len(heap) > K:
heapq.heappop(heap)
return ([h[1] for h in heap])nsmallest can take a lambda argument
def kClosest(self, points: List[List[int]], K: int) -> List[List[int]]:
return heapq.nsmallest(K, points, lambda x: x[0]*x[0] + x[1]*x[1])The key can be a function as well in nsmallest/nlargest
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
count = Counter(nums)
return heapq.nlargest(k, count, count.get)Tuple sort, 1st/2nd element. increasing frequency then decreasing order
def topKFrequent(self, words: List[str], k: int) -> List[str]:
freq = Counter(words)
return heapq.nsmallest(k, freq.keys(), lambda x:(-freq[x], x))Can be used with (list).sort(), sorted(), min(), max(), (heapq).nlargest,nsmallest(), map()
# a=3,b=8,target=10
min((b,a), key=lambda x: abs(target - x)) # 8>>> ids = ['id1', 'id2', 'id30', 'id3', 'id22', 'id100']
>>> print(sorted(ids)) # Lexicographic sort
['id1', 'id100', 'id2', 'id22', 'id3', 'id30']
>>> sorted_ids = sorted(ids, key=lambda x: int(x[2:])) # Integer sort
>>> print(sorted_ids)
['id1', 'id2', 'id3', 'id22', 'id30', 'id100']trans = lambda x: list(al[i] for i in x) # apple, a->0..
print(trans(words[0])) # [0, 15, 15, 11, 4]Lambda can sort by 1st, 2nd element in tuple
sorted([('abc', 121),('bbb',23),('abc', 148),('bbb', 24)], key=lambda x: (x[0],x[1]))
# [('abc', 121), ('abc', 148), ('bbb', 23), ('bbb', 24)]Combine two dicts or lists
s1 = {2, 3, 1}
s2 = {'b', 'a', 'c'}
list(zip(s1, s2)) # [(1, 'a'), (2, 'c'), (3, 'b')]Traverse in Parallel
letters = ['a', 'b', 'c']
numbers = [0, 1, 2]
for l, n in zip(letters, numbers):
print(f'Letter: {l}') # a,b,c
print(f'Number: {n}') # 0,1,2Empty in one list is ignored
letters = ['a', 'b', 'c']
numbers = []
for l, n in zip(letters, numbers):
print(f'Letter: {l}') #
print(f'Number: {n}') #Compare characters of alternating words
for a, b in zip(words, words[1:]):
for c1, c2 in zip(a,b):
print("c1", c1, end=" ")
print("c2", c2, end="\n ")Passing in * unpacks a list or other iterable, making each of its elements a separate argument.
a = [[1,2],[3,4]]
test = zip(*a)
print(list(test)) # [(1, 3) (2, 4)]
matrix = [[1,2,3],[4,5,6],[7,8,9]]
test = zip(*matrix)
print(*test) # (1, 4, 7) (2, 5, 8) (3, 6, 9)Useful when rotating a matrix
# matrix = [[1,2,3],[4,5,6],[7,8,9]]
matrix[:] = [list(t) for t in zip(*matrix[::-1])] # [[7,4,1],[8,5,2],[9,6,3]]Iterate through chars in a list of strs
strs = ["cir","car","caa"]
for i, l in enumerate(zip(*strs)):
print(l)
# ('c', 'c', 'c')
# ('i', 'a', 'a')
# ('r', 'r', 'a')Diagonals can be traversed with the help of a list
"""
[[1,2,3],
[4,5,6],
[7,8,9],
[10,11,12]]
"""
def printDiagonalMatrix(self, matrix: List[List[int]]) -> bool:
R = len(matrix)
C = len(matrix[0])
tmp = [[] for _ in range(R+C-1)]
for r in range(R):
for c in range(C):
tmp[r+c].append(matrix[r][c])
for t in tmp:
for n in t:
print(n, end=' ')
print("")
"""
1,
2,4
3,5,7
6,8,10
9,11
12
"""for i, l in enumerate(shuffle):
r = random.randrange(0+i, len(shuffle))
shuffle[i], shuffle[r] = shuffle[r], shuffle[i]
return shuffleOther random generators
import random
ints = [0,1,2]
random.choice(ints) # 0,1,2
random.choices([1,2,3],[1,1,10]) # 3, heavily weighted
random.randint(0,2) # 0,1, 2
random.randint(0,0) # 0
random.randrange(0,0) # error
random.randrange(0,2) # 0,1max = float('-inf')
min = float('inf')a if condition else b
test = stk.pop() if stk else '#''0b{:04b}'.format(0b1100 & 0b1010) # '0b1000' and
'0b{:04b}'.format(0b1100 | 0b1010) # '0b1110' or
'0b{:04b}'.format(0b1100 ^ 0b1010) # '0b0110' exclusive or
'0b{:04b}'.format(0b1100 >> 2) # '0b0011' shift right
'0b{:04b}'.format(0b0011 << 2) # '0b1100' shift leftElse condition on for loops if break is not called
for w1, w2 in zip(words, words[1:]): #abc, ab
for c1, c2 in zip(w1, w2):
if c1 != c2:
adj[c1].append(c2)
degrees[c2] += 1
break
else: # nobreak
if len(w1) > len(w2):
return "" # Triggers since ab should be before abc, not afterfor n in range(-8,8):
print n, n//4, n%4
-8 -2 0
-7 -2 1
-6 -2 2
-5 -2 3
-4 -1 0
-3 -1 1
-2 -1 2
-1 -1 3
0 0 0
1 0 1
2 0 2
3 0 3
4 1 0
5 1 1
6 1 2
7 1 3if any element of the iterable is True
def any(iterable):
for element in iterable:
if element:
return True
return Falsedef all(iterable):
for element in iterable:
if not element:
return False
return True- bisect.bisect_left returns the leftmost place in the sorted list to insert the given element
- bisect.bisect_right returns the rightmost place in the sorted list to insert the given element
import bisect
bisect.bisect_left([1,2,3,4,5], 2) # 1
bisect.bisect_right([1,2,3,4,5], 2) # 2
bisect.bisect_left([1,2,3,4,5], 7) # 5
bisect.bisect_right([1,2,3,4,5], 7) # 5Insert x in a in sorted order. This is equivalent to a.insert(bisect.bisect_left(a, x, lo, hi), x) assuming that a is already sorted. Search is binary search O(logn) and insert is O(n)
import bisect
l = [1, 3, 7, 5, 6, 4, 9, 8, 2]
result = []
for e in l:
bisect.insort(result, e)
print(result) # [1, 2, 3, 4, 5, 6, 7, 8, 9]li1 = [1, 3, 4, 4, 4, 6, 7] # [1, 3, 4, 4, 4, 5, 6, 7]
bisect.insort(li1, 5) #Bisect can give two ends of a range, if the array is sorted of course
s = bisect.bisect_left(nums, target)
e = bisect.bisect(nums, target) -1
if s <= e:
return [s,e]
else:
return [-1,-1]Calulate power
# (a ^ b) % p.
d = pow(a, b, p)Division with remainder
divmod(8, 3) # (2, 2)
divmod(3, 8) # (0, 3)Evaluates an expression
x = 1
print(eval('x + 1'))Creates iterator from container object such as list, tuple, dictionary and set
mytuple = ("apple", "banana", "cherry")
myit = iter(mytuple)
print(next(myit)) # apple
print(next(myit)) # bananamap(func, *iterables)
my_pets = ['alfred', 'tabitha', 'william', 'arla']
uppered_pets = list(map(str.upper, my_pets)) # ['ALFRED', 'TABITHA', 'WILLIAM', 'ARLA']
my_strings = ['a', 'b', 'c', 'd', 'e']
my_numbers = [1,2,3,4,5]
results = list(map(lambda x, y: (x, y), my_strings, my_numbers)) # [('a', 1), ('b', 2), ('c', 3), ('d', 4), ('e', 5)]A1 = [1, 4, 9]
''.join(map(str, A1))filter(func, iterable)
scores = [66, 90, 68, 59, 76, 60, 88, 74, 81, 65]
over_75 = list(filter(lambda x: x>75, scores)) # [90, 76, 88, 81]scores = [66, 90, 68, 59, 76, 60, 88, 74, 81, 65]
def is_A_student(score):
return score > 75
over_75 = list(filter(is_A_student, scores)) # [90, 76, 88, 81]dromes = ("demigod", "rewire", "madam", "freer", "anutforajaroftuna", "kiosk")
palindromes = list(filter(lambda word: word == word[::-1], dromes)) # ['madam', 'anutforajaroftuna']Get degrees == 0 from list
stk = list(filter(lambda x: degree[x]==0, degree.keys()))reduce(func, iterable[, initial]) where initial is optional
numbers = [3, 4, 6, 9, 34, 12]
result = reduce(lambda x, y: x+y, numbers) # 68
result = reduce(lambda x, y: x+y, numbers, 10) #78itertools.accumulate(iterable[, func]) –> accumulate object
import itertools
data = [3, 4, 6, 2, 1, 9, 0, 7, 5, 8]
list(itertools.accumulate(data)) # [3, 7, 13, 15, 16, 25, 25, 32, 37, 45]
list(accumulate(data, max)) # [3, 4, 6, 6, 6, 9, 9, 9, 9, 9]
cashflows = [1000, -90, -90, -90, -90] # Amortize a 5% loan of 1000 with 4 annual payments of 90
list(itertools.accumulate(cashflows, lambda bal, pmt: bal*1.05 + pmt)) [1000, 960.0, 918.0, 873.9000000000001, 827.5950000000001]
for k,v in groupby("aabbbc") # group by common letter
print(k) # a,b,c
print(list(v)) # [a,a],[b,b,b],[c,c]RE module allows regular expressions in python
def removeVowels(self, S: str) -> str:
return re.sub('a|e|i|o|u', '', S)from typing import List, Set, Dict, Tuple, Optional cheat sheet
Useful helpful function
R = len(grid)
C = len(grid[0])
def neighbors(r, c):
for nr, nc in ((r,c-1), (r,c+1), (r-1, c), (r+1,c)):
if 0<=nr<R and 0<=nc<C:
yield nr, nc
def dfs(r,c, index):
area = 0
grid[r][c] = index
for x,y in neighbors(r,c):
if grid[x][y] == 1:
area += dfs(x,y, index)
return area + 1Stack with appendleft() and popleft()
from collections import deque
deq = deque([1, 2, 3])
deq.appendleft(5)
deq.append(6)
deq
deque([5, 1, 2, 3, 6])
deq.popleft()
5
deq.pop()
6
deq
deque([1, 2, 3])from collections import Counter
count = Counter("hello") # Counter({'h': 1, 'e': 1, 'l': 2, 'o': 1})
count['l'] # 2
count['l'] += 1
count['l'] # 3Get counter k most common in list of tuples
# [1,1,1,2,2,3]
# Counter [(1, 3), (2, 2)]
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
if len(nums) == k:
return nums
return [n[0] for n in Counter(nums).most_common(k)] # [1,2]elements() lets you walk through each number in the Counter
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
c1 = collections.Counter(nums1) # [1,2,2,1]
c2 = collections.Counter(nums2) # [2,2]
dif = c1 & c2 # {2:2}
return list(dif.elements()) # [2,2]operators work on Counter
c = Counter(a=3, b=1)
d = Counter(a=1, b=2)
c + d # {'a': 4, 'b': 3}
c - d # {'a': 2}
c & d # {'a': 1, 'b': 1}
c | d # {'a': 3, 'b': 2}
c = Counter(a=2, b=-4)
+c # {'a': 2}
-c # {'b': 4}d={}
print(d['Grapes'])# This gives Key Error
from collections import defaultdict
d = defaultdict(int) # set default
print(d['Grapes']) # 0, no key error
d = collections.defaultdict(lambda: 1)
print(d['Grapes']) # 1, no key errorfrom collections import defaultdict
dd = defaultdict(list)
dd['key'].append(1) # defaultdict(<class 'list'>, {'key': [1]})
dd['key'].append(2) # defaultdict(<class 'list'>, {'key': [1, 2]})- Get all info
- Debug example, is it a special case?
- Brute Force
- Get to brute-force solution as soon as possible. State runtime and then optimize, don't code yet
- Optimize
- Look for unused info
- Solve it manually on example, then reverse engineer thought process
- Space vs time, hashing
- BUDS (Bottlenecks, Unnecessary work, Duplication)
- Walk through approach
- Code
- Test
- Start small
- Hit edge cases
def firstBadVersion(self, n):
l, r = 0, n
while l < r:
m = l + (r-l) // 2
if isBadVersion(m):
r = m
else:
l = m + 1
return l"""
12345678
FFTTTTTT
"""
def mySqrt(self, x: int) -> int:
def condition(value, x) -> bool:
return value * value > x
if x == 1:
return 1
left, right = 1, x
while left < right:
mid = left + (right-left) // 2
if condition(mid, x):
right = mid
else:
left = mid + 1
return left - 1Use values to detect if number is missing
def isCompleteTree(self, root: TreeNode) -> bool:
self.total = 0
self.mx = float('-inf')
def dfs(node, cnt):
if node:
self.total += 1
self.mx = max(self.mx, cnt)
dfs(node.left, (cnt*2))
dfs(node.right, (cnt*2)+1)
dfs(root, 1)
return self.total == self.mxGet a range sum of values
def rangeSumBST(self, root: TreeNode, L: int, R: int) -> int:
self.total = 0
def helper(node):
if node is None:
return 0
if L <= node.val <= R:
self.total += node.val
if node.val > L:
left = helper(node.left)
if node.val < R:
right = helper(node.right)
helper(root)
return self.totalCheck if valid
def isValidBST(self, root: TreeNode) -> bool:
if not root:
return True
stk = [(root, float(-inf), float(inf))]
while stk:
node, floor, ceil = stk.pop()
if node:
if node.val >= ceil or node.val <= floor:
return False
stk.append((node.right, node.val, ceil))
stk.append((node.left, floor, node.val))
return TrueKahn's algorithm, detects cycles through degrees and needs all the nodes represented to work
- Initialize vertices as unvisited
- Pick vertex with zero indegree, append to result, decrease indegree of neighbors
- Now repeat for neighbors, resulting list is sorted by source -> dest
If cycle, then degree of nodes in cycle will not be 0 since there is no origin
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
# Kahns algorithm, topological sort
adj = collections.defaultdict(list)
degree = collections.Counter()
for dest, orig in prerequisites:
adj[orig].append(dest)
degree[dest] += 1
bfs = [c for c in range(numCourses) if degree[c] == 0]
for o in bfs:
for d in adj[o]:
degree[d] -= 1
if degree[d] == 0:
bfs.append(d)
return len(bfs) == numCoursesdef alienOrder(self, words: List[str]) -> str:
nodes = set("".join(words))
adj = collections.defaultdict(list)
degree = collections.Counter(nodes)
for w1, w2 in zip(words, words[1:]):
for c1, c2 in zip(w1, w2):
if c1 != c2:
adj[c1].append(c2)
degree[c2] += 1
break
else:
if len(w1) > len(w2):
return ""
stk = list(filter(lambda x: degree[x]==1, degree.keys()))
ans = []
while stk:
node = stk.pop()
ans.append(node)
for nei in adj[node]:
degree[nei] -= 1
if degree[nei] == 1:
stk.append(nei)
return "".join(ans) * (set(ans) == nodes)- Have a counter or hash-map to count specific array input and keep on increasing the window toward right using outer loop.
- Have a while loop inside to reduce the window side by sliding toward right. Movement will be based on constraints of problem.
- Store the current maximum window size or minimum window size or number of windows based on problem requirement.
- Get min/max/number of satisfied sub arrays
- Return length of the subarray with max sum/product
- Return max/min length/number of subarrays whose sum/product equals K
Can require 2 or 3 pointers to solve
def slidingWindowTemplate(self, s: str):
#init a collection or int value to save the result according the question.
rtn = []
# create a hashmap to save the Characters of the target substring.
# (K, V) = (Character, Frequence of the Characters)
hm = {}
# maintain a counter to check whether match the target string as needed
cnt = collections.Counter(s)
# Two Pointers: begin - left pointer of the window; end - right pointer of the window if needed
l = r = 0
# loop at the begining of the source string
for r, c in enumerate(s):
if c in hm:
l = max(hm[c]+1, l) # +/- 1 or set l to index, max = never move l left
# update hm
hm[c] = r
# increase l pointer to make it invalid/valid again
while cnt == 0: # counter condition
cnt[c] += 1 # modify counter if needed
# Save result / update min/max after loop is valid
rtn = max(rtn, r-l+1)
return rtndef fruits_into_baskets(fruits):
maxCount, j = 0, 0
ht = {}
for i, c in enumerate(fruits):
if c in ht:
ht[c] += 1
else:
ht[c] = 1
if len(ht) <= 2:
maxCount = max(maxCount, i-j+1)
else:
jc = fruits[j]
ht[jc] -= 1
if ht[jc] <= 0:
del ht[jc]
j += 1
return maxCountMake the optimal choice at each step.
Increasing Triplet Subsequence, true if i < j < k
def increasingTriplet(self, nums: List[int]) -> bool:
l = m = float('inf')
for n in nums:
if n <= l:
l = n
elif n <= m:
m = n
else:
return True
return FalseBottom up solution with arguments for min, max
def maxAncestorDiff(self, root: TreeNode) -> int:
if not root:
return 0
self.ans = 0
def dfs(node, minval, maxval):
if not node:
self.ans = max(self.ans, abs(maxval - minval))
return
dfs(node.left, min(node.val, minval), max(node.val, maxval))
dfs(node.right, min(node.val, minval), max(node.val, maxval))
dfs(root, float('inf'), float('-inf'))
return self.ansBuilding a path through a tree
def binaryTreePaths(self, root: TreeNode) -> List[str]:
rtn = []
if root is None: return []
stk = [(root, str(root.val))]
while stk:
node, path = stk.pop()
if node.left is None and node.right is None:
rtn.append(path)
if node.left:
stk.append((node.left, path + "->" + str(node.left.val)))
if node.right:
stk.append((node.right, path + "->" + str(node.right.val)))
return rtnUsing return value to sum
def diameterOfBinaryTree(self, root: TreeNode) -> int:
self.mx = 0
def dfs(node):
if node:
l = dfs(node.left)
r = dfs(node.right)
total = l + r
self.mx = max(self.mx, total)
return max(l, r) + 1
else:
return 0
dfs(root)
return self.mxChange Tree to Graph
def distanceK(self, root: TreeNode, target: TreeNode, K: int) -> List[int]:
adj = collections.defaultdict(list)
def dfsa(node):
if node.left:
adj[node].append(node.left)
adj[node.left].append(node)
dfsa(node.left)
if node.right:
adj[node].append(node.right)
adj[node.right].append(node)
dfsa(node.right)
dfsa(root)
def dfs(node, prev, d):
if node:
if d == K:
rtn.append(node.val)
else:
for nei in adj[node]:
if nei != prev:
dfs(nei, node, d+1)
rtn = []
dfs(target, None, 0)
return rtnSubsection of sliding window, solve with Counter Dict
i.e. abc = bca != eba 111 111 111
def isAnagram(self, s: str, t: str) -> bool:
sc = collections.Counter(s)
st = collections.Counter(t)
if sc != st:
return False
return TrueSliding Window version (substring)
def findAnagrams(self, s: str, p: str) -> List[int]:
cntP = collections.Counter(p)
cntS = collections.Counter()
P = len(p)
S = len(s)
if P > S:
return []
ans = []
for i, c in enumerate(s):
cntS[c] += 1
if i >= P:
if cntS[s[i-P]] > 1:
cntS[s[i-P]] -= 1
else:
del cntS[s[i-P]]
if cntS == cntP:
ans.append(i-(P-1))
return ansdef coinChange(self, coins: List[int], amount: int) -> int:
MAX = float('inf')
dp = [MAX] * (amount + 1)
dp[0] = 0
for c in coins:
for a in range(c, amount+1):
dp[a] = min(dp[a], dp[a-c]+1)
return dp[amount] if dp[amount] != MAX else -1Classic DP grid, longest common subsequence
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
Y = len(text2)+1
X = len(text1)+1
dp = [[0] * Y for _ in range(X)]
# [[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]]
for i, c in enumerate(text1):
for j, d in enumerate(text2):
if c == d:
dp[i + 1][j + 1] = 1 + dp[i][j]
else:
dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j])
return dp[-1][-1]
# [[0,0,0,0],[0,1,1,1],[0,1,1,1],[0,1,2,2],[0,1,2,2],[0,1,2,3]]
# abcde
# "ace"- Useful algo when sorting in place
# if my number is equal to my index, i+1
# if my number is equal to this other number, i+1 (dups)
# else swap
def cyclic_sort(nums):
i = 0
while i < len(nums):
j = nums[i] - 1
if nums[i] != nums[j]:
nums[i], nums[j] = nums[j], nums[i]
else:
i += 1
return nums- Can be modified for divide in conquer problems
def quickSort(array):
def sort(arr, l, r):
if l < r:
p = part(arr, l, r)
sort(arr, l, p-1)
sort(arr, p+1, r)
def part(arr, l, r):
pivot = arr[r]
a = l
for i in range(l,r):
if arr[i] < pivot:
arr[i], arr[a] = arr[a], arr[i]
a += 1
arr[r], arr[a] = arr[a], arr[r]
return a
sort(array, 0, len(array)-1)
return arrayfrom collections import deque
def mergeSort(array):
def sortArray(nums):
if len(nums) > 1:
mid = len(nums)//2
l1 = sortArray(nums[:mid])
l2 = sortArray(nums[mid:])
nums = sort(l1,l2)
return nums
def sort(l1,l2):
result = []
l1 = deque(l1)
l2 = deque(l2)
while l1 and l2:
if l1[0] <= l2[0]:
result.append(l1.popleft())
else:
result.append(l2.popleft())
result.extend(l1 or l2)
return result
return sortArray(array)Merge K sorted Arrays with a heap
def mergeSortedArrays(self, arrays):
return list(heapq.merge(*arrays))Or manually with heappush/heappop.
class Solution:
def mergeSortedArrays(self, arrays):
pq = []
for i, arr in enumerate(arrays):
pq.append((arr[0], i, 0))
heapify(pq)
res = []
while pq:
num, i, j = heappop(pq)
res.append(num)
if j + 1 < len(arrays[i]):
heappush(pq, (arrays[i][j + 1], i, j + 1))
return resMerging K Sorted Lists
def mergeKLists(self, lists: List[ListNode]) -> ListNode:
prehead = ListNode()
heap = []
for i in range(len(lists)):
node = lists[i]
while node:
heapq.heappush(heap, node.val)
node = node.next
node = prehead
while len(heap) > 0:
val = heapq.heappop(heap)
node.next = ListNode()
node = node.next
node.val = val
return prehead.next- Solutions typically require 3 pointers: current, previous and next
- Solutions are usually made simplier with a prehead or dummy head node you create and then add to. Then return dummy.next
Reverse:
def reverseLinkedList(head):
prev, node = None, head
while node:
node.next, prev, node = prev, node, node.next
return prevReversing is easier if you can modify the values of the list
def reverse(head):
node = head
stk = []
while node:
if node.data % 2 == 0:
stk.append(node)
if node.data % 2 == 1 or node.next is None:
while len(stk) > 1:
stk[-1].data, stk[0].data = stk[0].data, stk[-1].data
stk.pop(0)
stk.pop(-1)
stk.clear()
node = node.next
return headMerge:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
dummy = ListNode(-1)
prev = dummy
while l1 and l2:
if l1.val < l2.val:
prev.next = l1
l1 = l1.next
else:
prev.next = l2
l2 = l2.next
prev = prev.next
prev.next = l1 if l1 is not None else l2
return dummy.next- Typically two steps. A digit modulo step and a integer division step by the next base then reverse the result or use a deque()
Base 10 to 16, or any base by changing '16' and index
def toHex(self, num: int) -> str:
rtn = []
index = "0123456789abcdef"
if num == 0: return '0'
if num < 0: num += 2 ** 32
while num > 0:
digit = num % 16
rtn.append(index[digit])
num = num // 16
return "".join(rtn[::-1])- Count can be used if simple case, otherwise stack. Basic Calculator is an extension of this algo
def isValid(self, s) -> bool:
cnt = 0
for c in s:
if c == '(':
cnt += 1
elif c == ')':
cnt -= 1
if cnt < 0:
return False
return cnt == 0Stack can be used if more complex
def isValid(self, s: str) -> bool:
stk = []
mp = {")":"(", "}":"{", "]":"["}
for c in s:
if c in mp.values():
stk.append(c)
elif c in mp.keys():
test = stk.pop() if stk else '#'
if mp[c] != test:
return False
return len(stk) == 0Or must store parenthesis index for further modification
def minRemoveToMakeValid(self, s: str) -> str:
rtn = list(s)
stk = []
for i, c in enumerate(s):
if c == '(':
stk.append(i)
elif c == ')':
if len(stk) > 0:
stk.pop()
else:
rtn[i] = ''
while stk:
rtn[stk.pop()] = ''
return "".join(rtn)Infinite Transactions, base formula
def maxProfit(self, prices: List[int]) -> int:
t0, t1 = 0, float('-inf')
for p in prices:
t0old = t0
t0 = max(t0, t1 + p)
t1 = max(t1, t0old - p)
return t0Single Transaction, t0 (k-1) = 0
def maxProfit(self, prices: List[int]) -> int:
t0, t1 = 0, float('-inf')
for p in prices:
t0 = max(t0, t1 + p)
t1 = max(t1, - p)
return t0K Transactions
t0 = [0] * (k+1)
t1 = [float(-inf)] * (k+1)
for p in prices:
for i in range(k, 0, -1):
t0[i] = max(t0[i], t1[i] + p)
t1[i] = max(t1[i], t0[i-1] - p)
return t0[k]Arrays can be shifted right by reversing the whole string, and then reversing 0,k-1 and k,len(str)
def rotate(self, nums: List[int], k: int) -> None:
def reverse(l, r, nums):
while l < r:
nums[l], nums[r] = nums[r], nums[l]
l += 1
r -= 1
if len(nums) <= 1: return
k = k % len(nums)
reverse(0, len(nums)-1, nums)
reverse(0, k-1, nums)
reverse(k, len(nums)-1, nums)The total number of continuous subarrays with sum k can be found by hashing the continuous sum per value and adding the count of continuous sum - k
def subarraySum(self, nums: List[int], k: int) -> int:
mp = {0: 1}
rtn, total = 0, 0
for n in nums:
total += n
rtn += mp.get(total - k, 0)
mp[total] = mp.get(total, 0) + 1
return rtnEvents pattern can be applied when to many interval problems such as 'Find employee free time between meetings' and 'find peak population' when individual start/ends are irrelavent and sum start/end times are more important
def employeeFreeTime(self, schedule: '[[Interval]]') -> '[Interval]':
events = []
for e in schedule:
for m in e:
events.append((m.start, 1))
events.append((m.end, -1))
events.sort()
itv = []
prev = None
bal = 0
for t, c in events:
if bal == 0 and prev is not None and t != prev:
itv.append(Interval(prev, t))
bal += c
prev = t
return itvMerging a new meeting into a list
def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
bisect.insort(intervals, newInterval)
merged = [intervals[0]]
for i in intervals:
ms, me = merged[-1]
s, e = i
if me >= s:
merged[-1] = (ms, max(me, e))
else:
merged.append(i)
return mergedGood for autocomplete, spell checker, IP routing (match longest prefix), predictive text, solving word games
class Trie:
def __init__(self):
self.root = {}
def addWord(self, s: str):
tmp = self.root
for c in s:
if c not in tmp:
tmp[c] = {}
tmp = tmp[c]
tmp['#'] = s # Store full word at '#' to simplify
def matchPrefix(self, s: str, tmp=None):
if not tmp: tmp = self.root
for c in s:
if c not in tmp:
return []
tmp = tmp[c]
rtn = []
for k in tmp:
if k == '#':
rtn.append(tmp[k])
else:
rtn += self.matchPrefix('', tmp[k])
return rtn
def hasWord(self, s: str):
tmp = self.root
for c in s:
if c in tmp:
tmp = tmp[c]
else:
return False
return TrueSearch example with . for wildcards
def search(self, word: str) -> bool:
def searchNode(word, node):
for i,c in enumerate(word):
if c in node:
node = node[c]
elif c == '.':
return any(searchNode(word[i+1:], node[cn]) for cn in node if cn != '$' )
else:
return False
return '$' in node
return searchNode(word, self.trie)local_maxiumum[i] = max(A[i], A[i] + local_maximum[i-1]) Explanation Determine max subarray sum
# input: [-2,1,-3,4,-1,2,1,-5,4]
def maxSubArray(self, nums: List[int]) -> int:
for i in range(1, len(nums)):
if nums[i-1] > 0:
nums[i] += nums[i-1]
return max(nums) # max([-2,1,-2,4,3,5,6,1,5]) = 6Union Find is a useful algorithm for graph
DSU for integers
class DSU:
def __init__(self, N):
self.par = list(range(N))
def find(self, x): # Find Parent
if self.par[x] != x:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
xr, yr = self.find(x), self.find(y)
if xr == yr: # If parents are equal, return False
return False
self.par[yr] = xr # Give y node parent of x
return True # return True if union occuredDSU for strings
class DSU:
def __init__(self):
self.par = {}
def find(self, x):
if x != self.par.setdefault(x, x):
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
xr, yr = self.find(x), self.find(y)
if xr == yr: return
self.par[yr] = xrDSU with union by rank
class DSU:
def __init__(self, N):
self.par = list(range(N))
self.sz = [1] * N
def find(self, x):
if self.par[x] != x:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
xr, yr = self.find(x), self.find(y)
if xr == yr:
return False
if self.sz[xr] < self.sz[yr]:
xr, yr = yr, xr
self.par[yr] = xr
self.sz[xr] += self.sz[yr]
return TrueFast Power, or Exponential by squaring allows calculating squares in logn time (x^n)2 = x^(2n)
def myPow(self, x: float, n: int) -> float:
if n < 0:
n *= -1
x = 1/x
ans = 1
while n > 0:
if n % 2 == 1:
ans = ans * x
x *= x
n = n // 2
return ansFibonacci can be calulated with Golden Ratio
def fib(self, N: int) -> int:
golden_ratio = (1 + 5 ** 0.5) / 2
return int((golden_ratio ** N + 1) / 5 ** 0.5)A calculator can be simulated with stack
class Solution:
def calculate(self, s: str) -> int:
s += '$'
def helper(stk, i):
sign = '+'
num = 0
while i < len(s):
c = s[i]
if c == " ":
i += 1
continue
elif c.isdigit():
num = num * 10 + int(c)
i += 1
elif c == '(':
num, i = helper([], i+1)
else:
if sign == '+':
stk.append(num)
if sign == '-':
stk.append(-num)
if sign == '*':
stk.append(stk.pop() * num)
if sign == '/':
stk.append(int(stk.pop() / num))
i += 1
num = 0
if c == ')':
return sum(stk), i
sign = c
return sum(stk)
return helper([],0)class Solution:
def evalRPN(self, tokens: List[str]) -> int:
stk = []
while tokens:
c = tokens.pop(0)
if c not in '+-/*':
stk.append(int(c))
else:
a = stk.pop()
b = stk.pop()
if c == '+':
stk.append(a + b)
if c == '-':
stk.append(b-a)
if c == '*':
stk.append(a * b)
if c == '/':
stk.append(int(b / a))
return stk[0]Used to sample large unknown populations. Each new item added has a 1/count chance of being selected
def __init__(self, nums):
self.nums = nums
def pick(self, target):
res = None
count = 0
for i, x in enumerate(self.nums):
if x == target:
count += 1
chance = random.randint(1, count)
if chance == 1:
res = i
return resCan find the min number of subsequences of strings in some source through binary search and a dict of the indexes of the source array
def shortestWay(self, source: str, target: str) -> int:
ref = collections.defaultdict(list)
for i,c in enumerate(source):
ref[c].append(i)
ans = 1
i = -1
for c in target:
if c not in ref:
return -1
offset = ref[c]
j = bisect.bisect_left(offset, i)
if j == len(offset):
ans += 1
i = offset[0] + 1
else:
i = offset[j] + 1
return ansRemoving adjacent duplicates is much more effective with a stack
def removeDuplicates(self, s: str, k: int) -> str:
stk = []
for c in s:
if stk and stk[-1][0] == c:
stk[-1][1] += 1
if stk[-1][1] >= k:
stk.pop()
else:
stk.append([c, 1])
ans = []
for c in stk:
ans.extend([c[0]] * c[1])
return "".join(ans)Dutch National Flag Problem proposed by Edsger W. Dijkstra
def sortColors(self, nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
# for all idx < p0 : nums[idx < p0] = 0
# curr is an index of element under consideration
p0 = curr = 0
# for all idx > p2 : nums[idx > p2] = 2
p2 = len(nums) - 1
while curr <= p2:
if nums[curr] == 0:
nums[p0], nums[curr] = nums[curr], nums[p0]
p0 += 1
curr += 1
elif nums[curr] == 2:
nums[curr], nums[p2] = nums[p2], nums[curr]
p2 -= 1
else:
curr += 1