Algo-Prep
```py
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
c = 0
res = []
while(l1 or l2):
s = 0 + c
if l1:
s += int(l1.val)
l1 = l1.next
if l2:
s += int(l2.val)
l2 = l2.next
print(s)
resa = s % 10
res.append(resa)
c = s // 10
if(c!=0):
res.append(c)
l3 = ListNode(0)
head = l3
for i in range(0, len(res)):
lt = res[i]
l3.next = ListNode(lt)
l3 = l3.next
return head.nextclass Solution:
def findContentChildren(self, g: List[int], s: List[int]) -> int:
g = sorted(g)
s = sorted(s)
content = 0
while s and g:
if s[-1] >= g[-1]:
s.pop()
content += 1
g.pop()
return content# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def convertBST(self, root: TreeNode) -> TreeNode:
self.ans = 0
def add(node):
if not node:
return
add(node.right)
self.ans += node.val
node.val = self.ans
add(node.left)
add(root)
return root## Recursive Solution
# ------------------------------------------
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
io = []
if(root==None):
return []
def inorder(x):
if(x.left!=None):
inorder(x.left)
io.append(int(x.val))
if(x.right!=None):
inorder(x.right)
inorder(root)
return io
# ------------------------------------------
## Iterative Solution using Stack
# ------------------------------------------
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
if(root==None):
return []
stack = []
io = []
c = root
while(c!=None or len(stack)!=0):
while(c!=None):
stack.append(c)
c = c.left
c = stack.pop()
io.append(c.val)
c = c.right
return io
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# ------------------------------------------
## Recursive Solution
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
if (root==None):
return []
po = []
def postorder(x):
if not x:
return
postorder(x.left)
postorder(x.right)
po.append(x.val)
postorder(root)
return po
# ------------------------------------------
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# ------------------------------------------
## Recursive Solution
class Solution:
def preorderTraversal(self, root: TreeNode) -> List[int]:
if(root==None):
return []
po = []
def preorder(x):
if x:
po.append(x.val)
preorder(x.left)
preorder(x.right)
preorder(root)
return po
# ------------------------------------------
## Iterative Solution
class Solution:
def preorderTraversal(self, root: TreeNode) -> List[int]:
if(root==None):
return []
stack = []
po = []
c = root
while(c!=None or len(stack)!=0):
while(c!=None):
stack.append(c)
po.append(c.val)
c = c.left
c = stack.pop()
c = c.right
return po
class Solution:
def backspaceCompare(self, S: str, T: str) -> bool:
def deleteBackSpace(X):
stack = []
for i in X:
if not i=='#':
stack.append(i)
elif(len(stack)==0):
continue
else:
stack.pop()
return stack
return deleteBackSpace(S)==deleteBackSpace(T)# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def balanceBST(self, root: TreeNode) -> TreeNode:
result = []
def inorder(node):
if node:
if node.left!=None:
inorder(node.left)
result.append(int(node.val))
if node.right!=None:
inorder(node.right)
def constructBalancedTree(arr):
if not arr:
return None
mid = len(arr)//2
root = TreeNode(arr[mid])
root.left = constructBalancedTree(arr[:mid])
root.right = constructBalancedTree(arr[mid+1:])
return root
inorder(root)
# result = [int(x.val) for x in result]
return constructBalancedTree(result)class Solution:
def maxProfit(self, prices: List[int]) -> int:
profit = 0
for i in range(0, len(prices)-1):
if(prices[i+1] > prices[i]):
profit += prices[i+1] - prices[i]
return profitimport math
class Solution:
def rangeBitwiseAnd(self, m: int, n: int) -> int:
ans = m
if not m==0:
x = math.log2(m)
x = int(x)+1
x = 2**x
else:
return 0
if(n>=x):
return 0
for i in range(m+1, n+1):
ans = ans & i
return ansfrom collections import Counter
class Solution:
def count(self, d1, d2):
s = 0
for i in 'abcdefghijklmnopqrstuvwxyz':
s += d1[i] - d2[i]
if s < 0:
return False
return True
def checkIfCanBreak(self, s1: str, s2: str) -> bool:
d1 = Counter(s1)
d2 = Counter(s2)
return self.count(d1, d2) | self.count(d2, d1)class Solution:
def maxArea(self, height: List[int]) -> int:
i = 0
j = len(height)-1
maxarea = 0
while(i!=j):
a = (j-i)*(min(height[i], height[j]))
if(a>maxarea):
maxarea = a
if(height[i]>height[j]):
j -= 1
else:
i += 1
return maxareaclass Solution:
def containsDuplicate(self, nums: List[int]) -> bool:
return not len(nums) == len(set(nums))class Solution:
def findMaxLength(self, nums: List[int]) -> int:
dic = { 0:-1 }
ps = 0
max_length = 0
for idx, number in enumerate(nums):
if number:
ps += 1
else:
ps -= 1
if ps in dic:
max_length = max(max_length, idx-dic[ps])
else:
dic[ps] = idx
return max_length# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def countNodes(self, root: TreeNode) -> int:
if not root:
return 0
self.c = 0
def count(node):
if node:
if node.left:
count(node.left)
if node.right:
count(node.right)
self.c += 1
return self.c
count(root)
return self.cclass Solution:
def countBits(self, num: int) -> List[int]:
ans = [0]
offset = 1
for i in range(1, num+1):
if(offset*2 == i):
offset = i
ans.append(ans[i-offset]+1)
return ansclass Solution:
def countElements(self, arr: List[int]) -> int:
s = set()
s = set(arr)
c = 0
for i in range(len(arr)):
if(arr[i]+1 in s):
c += 1
return cfrom collections import deque
class Solution:
def canFinish(self, numCourses, prerequisites) -> bool:
adjList = [[] for _ in range(numCourses)]
inDegree = [0 for _ in range(numCourses)]
queue = deque()
visited = 0
for i in range(len(prerequisites)):
adjList[prerequisites[i][0]].append(prerequisites[i][1])
for i in range(numCourses):
for j in adjList[i]:
inDegree[j] += 1
for i in range(len(inDegree)):
if inDegree[i] == 0:
queue.append(i)
while queue:
el = queue.popleft()
for i in adjList[el]:
inDegree[i] -= 1
if inDegree[i] == 0:
queue.append(i)
visited += 1
if visited == numCourses:
return True
else:
return Falseclass Solution:
def minDeletionSize(self, A: List[str]) -> int:
s = 0
for col in zip(*A):
if any(col[i] > col[i+1] for i in range(len(col)-1)):
s += 1
return sclass Solution:
def deleteNode(self, root, key):
if not root:
return
if key > root.val:
root.right = self.deleteNode(root.right, key)
elif key < root.val:
root.left = self.deleteNode(root.left, key)
else:
if not root.left:
return root.right
else:
temp = root.left
while temp.right:
temp = temp.right
root.val = temp.val
root.left = self.deleteNode(root.left, temp.val)
return root# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def deleteNode(self, node):
node.val = node.next.val
node.next = node.next.nextclass MyCircularDeque:
def __init__(self, k: int):
"""
Initialize your data structure here. Set the size of the deque to be k.
"""
self.maxsize = k
self.size = 0
self.decq = [0]*k
self.front = self.rear = -1
def insertFront(self, value: int) -> bool:
"""
Adds an item at the front of Deque. Return true if the operation is successful.
"""
if self.size == self.maxsize:
return False
else:
if self.front == -1:
self.front = self.rear = 0
else:
self.front = (self.front-1)%self.maxsize
self.decq[self.front] = value
self.size += 1
return True
def insertLast(self, value: int) -> bool:
"""
Adds an item at the rear of Deque. Return true if the operation is successful.
"""
if self.size == self.maxsize:
return False
else:
if self.rear == -1:
self.front = self.rear = 0
else:
self.rear = (self.rear+1)%self.maxsize
self.decq[self.rear] = value
self.size += 1
return True
def deleteFront(self) -> bool:
"""
Deletes an item from the front of Deque. Return true if the operation is successful.
"""
if self.size == 0:
return False
else:
if self.front == self.rear:
self.front = self.rear = -1
else:
self.decq[self.front] = 0
self.front = (self.front+1)%self.maxsize
self.size -= 1
return True
def deleteLast(self) -> bool:
"""
Deletes an item from the rear of Deque. Return true if the operation is successful.
"""
if self.size == 0:
return False
else:
if self.front == self.rear:
self.front = self.rear = -1
else:
self.decq[self.rear] = 0
self.rear = (self.rear-1)%self.maxsize
self.size -= 1
return True
def getFront(self) -> int:
"""
Get the front item from the deque.
"""
return self.decq[self.front] if self.size != 0 else -1
def getRear(self) -> int:
"""
Get the last item from the deque.
"""
return self.decq[self.rear] if self.size != 0 else -1
def isEmpty(self) -> bool:
"""
Checks whether the circular deque is empty or not.
"""
return self.size == 0
def isFull(self) -> bool:
"""
Checks whether the circular deque is full or not.
"""
return self.size == self.maxsize
# ------------------------------------------
# Your MyCircularDeque object will be instantiated and called as such:
# obj = MyCircularDeque(k)
# param_1 = obj.insertFront(value)
# param_2 = obj.insertLast(value)
# param_3 = obj.deleteFront()
# param_4 = obj.deleteLast()
# param_5 = obj.getFront()
# param_6 = obj.getRear()
# param_7 = obj.isEmpty()
# param_8 = obj.isFull()
# ------------------------------------------
# Another Optimized Solution
class MyCircularDeque:
def __init__(self, k: int):
"""
Initialize your data structure here. Set the size of the deque to be k.
"""
self.decq = []
self.maxsize = k
def insertFront(self, value: int) -> bool:
"""
Adds an item at the front of Deque. Return true if the operation is successful.
"""
if len(self.decq) < self.maxsize:
self.decq.append(value)
return True
def insertLast(self, value: int) -> bool:
"""
Adds an item at the rear of Deque. Return true if the operation is successful.
"""
if len(self.decq) < self.maxsize:
self.decq.insert(0, value)
return True
def deleteFront(self) -> bool:
"""
Deletes an item from the front of Deque. Return true if the operation is successful.
"""
if self.decq:
self.decq.pop()
return True
def deleteLast(self) -> bool:
"""
Deletes an item from the rear of Deque. Return true if the operation is successful.
"""
if self.decq:
del self.decq[0]
return True
def getFront(self) -> int:
"""
Get the front item from the deque.
"""
return self.decq[-1] if self.decq else -1
def getRear(self) -> int:
"""
Get the last item from the deque.
"""
return self.decq[0] if self.decq else -1
def isEmpty(self) -> bool:
"""
Checks whether the circular deque is empty or not.
"""
return len(self.decq)==0
def isFull(self) -> bool:
"""
Checks whether the circular deque is full or not.
"""
return len(self.decq)==self.maxsize
# ------------------------------------------
# Your MyCircularDeque object will be instantiated and called as such:
# obj = MyCircularDeque(k)
# param_1 = obj.insertFront(value)
# param_2 = obj.insertLast(value)
# param_3 = obj.deleteFront()
# param_4 = obj.deleteLast()
# param_5 = obj.getFront()
# param_6 = obj.getRear()
# param_7 = obj.isEmpty()
# param_8 = obj.isFull()class MyCircularQueue:
def __init__(self, k: int):
self.size = 0
self.maxsize = k
self.cq = [0]*k
self.front = self.rear = -1
def enQueue(self, value: int) -> bool:
if self.size == self.maxsize:
return False
else:
if self.rear == -1:
self.rear = self.front = 0
else:
self.rear = (self.rear+1)%self.maxsize
self.cq[self.rear] = value
self.size += 1
return True
def deQueue(self) -> bool:
if self.size == 0:
return False
if self.front == self.rear:
self.front = self.rear = -1
else:
self.front = (self.front+1)%self.maxsize
self.size -= 1
return True
def Front(self) -> int:
return self.cq[self.front] if self.size!=0 else -1
def Rear(self) -> int:
return self.cq[self.rear] if self.size!=0 else -1
def isEmpty(self) -> bool:
return self.size == 0
def isFull(self) -> bool:
return self.size == self.maxsize
# ------------------------------------------
# Your MyCircularQueue object will be instantiated and called as such:
# obj = MyCircularQueue(k)
# param_1 = obj.enQueue(value)
# param_2 = obj.deQueue()
# param_3 = obj.Front()
# param_4 = obj.Rear()
# param_5 = obj.isEmpty()
# param_6 = obj.isFull()# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def diameterOfBinaryTree(self, root: TreeNode) -> int:
self.depth = 1
def findDepth(first):
if not first:
return 0
ld = findDepth(first.left)
rd = findDepth(first.right)
self.depth = max(self.depth, ld+rd+1)
return max(ld,rd) + 1
findDepth(root)
return self.depth - 1class Solution:
def trailingZeroes(self, n: int) -> int:
count = 0
m = 5
while (n/m >= 1):
count += int(n/m)
m *= 5
return count # Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def findBottomLeftValue(self, root: TreeNode) -> int:
queue = deque([root])
visited = set()
while queue:
size = len(queue)
leftmost = math.inf
for i in range(size):
node = queue.popleft()
if leftmost == math.inf:
leftmost = node.val
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
if not queue:
return leftmost
return 0# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def findDuplicateSubtrees(self, root: TreeNode) -> List[TreeNode]:
tree = collections.defaultdict()
tree.default_factory = tree.__len__
c = collections.Counter()
anslist = []
def find(node):
if node:
tid = tree[node.val, find(node.left), find(node.right)]
c[tid] += 1
if c[tid] == 2:
anslist.append(node)
return tid
find(root)
return anslistfrom collections import Counter
class Solution:
def firstUniqChar(self, s: str) -> int:
c = Counter(s)
for i in range(len(s)):
if c[s[i]] == 1:
return i
return -1class Solution:
def fizzBuzz(self, n: int) -> List[str]:
res = []
for i in range(1, n+1):
if i % 3 == 0 and i % 5 == 0:
res.append("FizzBuzz")
elif i % 3 == 0:
res.append("Fizz")
elif i % 5 == 0:
res.append("Buzz")
else:
res.append(str(i))
return resimport collections
class Solution:
def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
word = collections.defaultdict(list)
for s in strs:
word[tuple(sorted(s))].append(s)
return word.values()class MyQueue:
def __init__(self):
"""
Initialize your data structure here.
"""
self.s1 = []
self.s2 = []
def push(self, x: int) -> None:
"""
Push element x to the back of queue.
"""
self.s1.append(x)
def pop(self) -> int:
"""
Removes the element from in front of queue and returns that element.
"""
while len(self.s1) > 1:
self.s2.append(self.s1.pop())
temp = self.s1.pop()
while len(self.s2):
self.s1.append(self.s2.pop())
return temp
def peek(self) -> int:
"""
Get the front element.
"""
return self.s1[0]
def empty(self) -> bool:
"""
Returns whether the queue is empty.
"""
return False if len(self.s1) else True
# ------------------------------------------
# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()## Implementation Using queue.Queue()
## Faster than 67%
from queue import Queue
class MyStack:
def __init__(self):
"""
Initialize your data structure here.
"""
self.q1 = Queue(maxsize=0)
self.q2 = Queue(maxsize=0)
def push(self, x: int) -> None:
"""
Push element x onto stack.
"""
self.q1.put(x)
def pop(self) -> int:
"""
Removes the element on top of the stack and returns that element.
"""
while(self.q1.qsize()>1):
self.q2.put(self.q1.get())
temp = self.q1.get()
while(self.q2.qsize()>0):
self.q1.put(self.q2.get())
return temp
def top(self) -> int:
"""
Get the top element.
"""
while(self.q1.qsize()>1):
self.q2.put(self.q1.get())
temp = self.q1.get()
while(self.q2.qsize()>0):
self.q1.put(self.q2.get())
self.q1.put(temp)
return temp
def empty(self) -> bool:
"""
Returns whether the stack is empty.
"""
return self.q1.empty()
# ------------------------------------------
## Implementation using Deque
## Faster than 100%
from collections import deque
class MyStack:
def __init__(self):
"""
Initialize your data structure here.
"""
self.q = deque()
def push(self, x: int) -> None:
"""
Push element x onto stack.
"""
self.q.append(x)
def pop(self) -> int:
"""
Removes the element on top of the stack and returns that element.
"""
return self.q.pop()
def top(self) -> int:
"""
Get the top element.
"""
temp = self.q.pop()
self.q.append(temp)
return temp
def empty(self) -> bool:
"""
Returns whether the stack is empty.
"""
return False if len(self.q) else True
# ------------------------------------------
# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()class CustomStack:
def __init__(self, maxSize: int):
self.stack = []
self.maxSize = maxSize
def push(self, x: int) -> None:
if(len(self.stack) < self.maxSize):
self.stack.append(x)
def pop(self) -> int:
if(len(self.stack)!=0):
return self.stack.pop()
else:
return -1
def increment(self, k: int, val: int) -> None:
for i in range(min(k, len(self.stack))):
self.stack[i] += val
# ------------------------------------------
# Your CustomStack object will be instantiated and called as such:
# obj = CustomStack(maxSize)
# obj.push(x)
# param_2 = obj.pop()
# obj.increment(k,val)import math
class Solution:
def integerReplacement(self, n: int) -> int:
s = 0
while(n!=1):
if(n%2==0):
n //= 2
s += 1
continue
if(n==3):
return s+2
else:
if(math.ceil(math.log2(n-1))==math.floor(math.log2(n-1))):
n -= 1
elif((math.ceil(math.log2(n+1))==math.floor(math.log2(n+1))) or ((n+1)%4==0)):
n += 1
else:
n -= 1
s += 1
return sclass Solution:
def intToRoman(self, num: int) -> str:
dic = { 1000:'M', 900:'CM', 500:'D', 400:'CD', 100:'C', 90:'XC', 50:'L', 40:'XL', 10:'X', 9:'IX', 5:'V', 4:'IV', 1:'I' }
ans = ''
for i in dic:
while num>=i:
ans += dic[i]
num -= i
return ansclass Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
if len(nums1) > len(nums2):
i = 0
while i < len(nums2):
if nums2[i] in set(nums1):
nums1.remove(nums2[i])
i += 1
else:
nums2.remove(nums2[i])
return nums2
else:
i = 0
while i < len(nums1):
if nums1[i] in set(nums2):
nums2.remove(nums1[i])
i += 1
else:
nums1.remove(nums1[i])
return nums1
# ------------------------------------------
# Alternate Approach
from collections import Counter
class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
return list((Counter(nums1)&Counter(nums2)).elements())# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
def invert(node):
if node.left:
invert(node.left)
if node.right:
invert(node.right)
temp = node.left
node.left = node.right
node.right = temp
if root:
invert(root)
return rootclass Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(s) == 0:
return True
if len(t) == 0:
return False
sp = 0
for tc in t:
if s[sp] == tc:
sp += 1
if sp == len(s):
return True
return Falseclass Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
while(len(stones)!=1 and len(stones)!=0):
stones = sorted(stones)
t = abs(stones[-1]-stones[-2])
if(t!=0):
stones.pop()
stones[-1] = t
else:
stones.pop()
stones.pop()
if(stones):
return stones[0]
return 0class Solution:
def lemonadeChange(self, bills: List[int]) -> bool:
denom = {
5: 0,
10: 0,
20: 0
}
for i in range(len(bills)):
denom[bills[i]] += 1
if bills[i] > 5:
bal = bills[i] - 5
if bal % 5 == 0 and bal % 10 != 0:
if denom[5] > 0:
denom[5] -= 1
bal -= 5
if bal == 0:
continue
if bal % 10 == 0 and bal % 20 != 0:
if denom[10] > 0:
denom[10] -= 1
bal -= 10
if bal == 0:
continue
if denom[5] > 1:
denom[5] -= 2
bal -= 10
if bal == 0:
continue
if bal > 0:
return False
return Trueclass Solution:
def longestCommonPrefix(self, strs: List[str]) -> str:
x = 0
for i in zip(*strs):
r = all(a == i[0] for a in i)
if r:
x += 1
else:
break
return strs[0][0:x] if x else ''class Solution:
def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
A = sorted(A)
for i in range(len(A)):
if A[i] < 0:
A[i] = -A[i]
K -= 1
elif A[i] >= 0:
if K % 2 == 0:
break
else:
A[A.index(min(A))] = -A[A.index(min(A))]
break
if K == 0:
break
return sum(A) # Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
dummy = l3 = ListNode(0)
while l1 and l2:
if l1.val > l2.val:
l3.next = ListNode(l2.val)
l3 = l3.next
l2 = l2.next
else:
l3.next = ListNode(l1.val)
l3 = l3.next
l1 = l1.next
while l1:
l3.next = ListNode(l1.val)
l3 = l3.next
l1 = l1.next
while l2:
l3.next = ListNode(l2.val)
l3 = l3.next
l2 = l2.next
return dummy.next
# ------------------------------------------
# Slightly More Optimized
# ------------------------------------------
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
head = l3 = ListNode()
while l1 and l2:
if l1.val < l2.val:
l3.next = ListNode(l1.val)
l1 = l1.next
l3 = l3.next
else:
l3.next = ListNode(l2.val)
l2 = l2.next
l3 = l3.next
l3.next = l1 or l2
return head.next# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def middleNode(self, head: ListNode) -> ListNode:
A = [head]
while A[-1].next:
A.append(A[-1].next)
return A[len(A)//2]
# ------------------------------------------
# Solution using Slow and Fast Pointers
class Solution:
def middleNode(self, head: ListNode) -> ListNode:
slowPointer = head
fastPointer = head
while(fastPointer and fastPointer.next):
slowPointer = slowPointer.next
fastPointer = fastPointer.next.next
return slowPointer import math
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
self.min = math.inf
def push(self, x: int) -> None:
self.x = x
self.stack.append(x)
if(x < self.min):
self.min = x
def pop(self) -> None:
t = self.stack.pop()
if(t==self.min and len(self.stack)):
self.min = min(self.stack)
elif(t==self.min and not len(self.stack)):
self.min = math.inf
def top(self) -> int:
return self.stack[-1]
def getMin(self) -> int:
return self.min
# ------------------------------------------
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()class Solution:
def minSubsequence(self, nums: List[int]) -> List[int]:
nums = sorted(nums)[::-1]
x = sum(nums)
s = 0
if len(nums) == 1:
return nums
for i in range(len(nums)+1):
s += nums[i]
if s > x-s:
return nums[0:i+1]class Solution:
def isMonotonic(self, A: List[int]) -> bool:
inc = True
dec = True
for i in range(0, len(A)-1):
if(A[i] > A[i+1]):
inc = False
if(A[i] < A[i+1]):
dec = False
return inc or dec
class Solution:
def moveZeroes(self, nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
t = nums.count(0)
nzpos = 0
if(t==0):
return nums
for i in range(0, len(nums)):
if(nums[i]!=0):
nums[nzpos] = nums[i]
nzpos = nzpos+1
for i in range(len(nums)-t, len(nums)):
nums[i] = 0 from collections import deque
class Solution:
def numIslands(self, grid) -> int:
if not grid:
return 0
r = len(grid)
c = len(grid[0])
queue = deque()
islands = 0
for i in range(r):
for j in range(c):
if grid[i][j] == '1':
islands += 1
grid[i][j] = '0'
queue.append([i, j])
while queue:
el = queue.popleft()
rs = el[0]
cs = el[1]
if rs - 1 >= 0 and grid[rs-1][cs] == '1':
queue.append([rs-1, cs])
grid[rs-1][cs] = '0'
if rs + 1 < r and grid[rs+1][cs] == '1':
queue.append([rs+1, cs])
grid[rs+1][cs] = '0'
if cs - 1 >= 0 and grid[rs][cs-1] == '1':
queue.append([rs, cs-1])
grid[rs][cs-1] = '0'
if cs + 1 < c and grid[rs][cs+1] == '1':
queue.append([rs, cs+1])
grid[rs][cs+1] = '0'
return islandsclass RecentCounter:
def __init__(self):
self.q = collections.deque()
def ping(self, t: int) -> int:
self.q.append(t)
while self.q[0] < t-3000:
self.q.popleft()
return len(self.q)class Solution:
def hammingWeight(self, n: int) -> int:
return (bin(n).count('1'))# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution1:
def isPalindrome(self, head: ListNode) -> bool:
temp = head
stack = []
l = 0
while temp:
l += 1
temp = temp.next
temp = head
for i in range(0, l//2):
stack.append(temp.val)
temp = temp.next
if l % 2 != 0:
temp = temp.next
for i in range(0, l//2):
if temp.val == stack.pop():
continue
return False
return True
# ------------------------------------------
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution2:
def isPalindrome(self, head: ListNode) -> bool:
el = []
while head:
el.append(head.val)
head = head.next
for i in range(0, len(el)//2):
if not el[i] == el[-i-1]:
return False
return Trueclass Solution:
def isPalindrome(self, x: int) -> bool:
a = []
x = str(x)
x = list(x)
a = x[::-1]
if (str(a)==str(x)):
return True
else:
return False"""
This first solution uses the interval splitting method
This is achieved by using a dp table.
Time Complexity: O(n^1.5)
"""
class Solution1:
def numSquares(self, n) -> int:
if n <= 3:
return n
dp = [0 for _ in range(n+1)]
dp[1], dp[2], dp[3] = 1, 2, 3
for i in range(4, len(dp)):
dp[i] = i
j = 1
while j*j <= i:
dp[i] = min(dp[i], 1 + dp[i - j*j])
j += 1
return dp[-1]
"""
Lagrange's 4 square and 3 square theorem
Theorem: Every natural number can be represented as the sum of 4 integer squares.
N = a^2 + b^2 + c^2 + d^2
Theorem: A natural number can be represented as sum of 3 squares of integers.
N = a^2 + b^2 + c^2
if and only if the N is not of the form,
N = 4^a (8b + 7) -- (1)
LOGIC:
- if N is a perfect square, return 1
- if N is of form (1),
- keep dividing by 4
- divide by 8
- if rem == 7:
return 4
- check if N can be split into two perfect squares. If yes, return 2
- if all fails, return 3
"""
class Solution:
def numSquares(self, n: int) -> int:
if ceil(sqrt(n)) == floor(sqrt(n)):
return 1
while n % 4 == 0:
n /= 4
if n % 8 == 7:
return 4
j = 1
while j*j <= n:
if ceil(sqrt(n - j*j)) == floor(sqrt(n - j*j)):
return 2
j += 1
else:
return 3class Solution:
def stringShift(self, s: str, shift: List[List[int]]) -> str:
amount = 0
for i in range(len(shift)):
if(shift[i][0]==0):
amount += (-1)*shift[i][1]
else:
amount += 1*shift[i][1]
print(amount)
if amount == 0:
return s
elif amount < 0:
return s[(abs(amount)%len(s)):] + s[0:(abs(amount)%len(s))]
else:
return s[len(s)-(amount%len(s)):] + s[0:len(s)-(amount%len(s))]class Solution:
def plusOne(self, digits: List[int]) -> List[int]:
carry = 0
for i in range(len(digits)-1, -1, -1):
if digits[i] != 9:
digits[i] += 1
break
else:
digits[i] = 0
if i == 0:
digits.insert(0, 1)
return digitsclass Solution:
def isPowerOfThree(self, n: int) -> bool:
if n < 1:
return False
if n == 1:
return True
if sum(list(map(int, str(n)))) % 3 != 0:
return False
else:
while n > 1:
if n % 3 == 0:
n /= 3
else:
return False
if n != 1:
return False
else:
return True
# ------------------------------------------
# Alternate Approach
class Solution:
def isPowerOfThree(self, n: int) -> bool:
if n < 1:
return False
else:
return 1162261467 % n == 0# Classic Solution
class Solution:
def isPowerOfTwo(self, n: int) -> bool:
if n == 1:
return True
if n == 0:
return False
while n % 2 == 0:
n = n / 2
if n == 1:
return True
else:
return False
# ------------------------------------------
# Solution Using Bit Manipulation
class Solution:
def isPowerOfTwo(self, n: int) -> bool:
return n > 0 and bin(n).count('1') == 1class Solution:
def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]:
people.sort(key = lambda x: (-x[0], x[1]))
rec = []
for p in people:
rec.insert(p[1], p)
return recfrom random import choices
class Solution:
def __init__(self, w: List[int]):
self.w = w
def pickIndex(self) -> int:
return choices(range(len(self.w)), self.w)[0]
# ------------------------------------------
# Your Solution object will be instantiated and called as such:
# obj = Solution(w)
# param_1 = obj.pickIndex()class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
p = 0
while p < len(nums)-1:
if nums[p] == nums[p+1]:
nums.pop(p+1)
continue
p += 1
return len(nums)# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def removeElements(self, head: ListNode, val: int) -> ListNode:
pointer = ListNode(0)
pointer.next = head
tempnode = pointer
while tempnode.next != None:
if tempnode.next.val == val:
tempnode.next = tempnode.next.next
else:
tempnode = tempnode.next
return pointer.next# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
tail = head
length = 1
while tail.next:
length += 1
tail = tail.next
if(length==1):
return None
if(length==n):
return head.next
tempnode = head
for _ in range(0, length-n-1):
tempnode = tempnode.next
tempnode.next = tempnode.next.next
return head
# ------------------------------------------
# One Pass
# ------------------------------------------
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
dummy = fast = slow = ListNode()
dummy.next = head
if not head.next:
return None
for _ in range(n+1):
fast = fast.next
while fast:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return dummy.nextclass Solution:
def reorganizeString(self, S: str) -> str:
l = len(S)
A = []
for k, v in sorted((S.count(x), x) for x in set(S)):
if k > (l+1) / 2 :
return ""
A.extend(k * v)
# print(A)
ans = [None] * l
ans[::2], ans[1::2] = A[(l//2) : ], A[: (l//2)]
return ''.join(ans)class Solution:
def findRepeatedDnaSequences(self, s: str) -> List[str]:
dic = {}
ans = []
for i in range(0, len(s)-9):
if s[i:i+10] not in dic:
dic[s[i:i+10]] = 1
else:
dic[s[i:i+10]] += 1
for k, v in dic.items():
if v>1:
ans.append(k)
return ansclass Solution:
def reverseBits(self, n: int) -> int:
s = str(bin(n))
s = s[2:]
s = '0'*(32-len(s)) + s
s = int(s[::-1], 2)
return sclass Solution:
def reverse(self, x: int) -> int:
x = str(x)
if (x[0] == '-'):
a = x[1:2147483648:1]
a = a[::-1]
if (int(a)>2147483648):
return 0
return int("-"+a)
else:
a = x[::-1]
if (int(a)>2147483647):
return 0
return int(a)# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
temp = head
prev = None
while(temp!=None):
next = temp.next
temp.next = prev
prev = temp
temp = next
return prevclass Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
head = 0
temp = ''
tail = len(s) - 1
while head < tail:
temp = s[head]
s[head] = s[tail]
s[tail] = temp
head += 1
tail -= 1class Solution:
def reverseParentheses(self, s: str) -> str:
stack = []
s = list(s)
for i in range(len(s)):
if s[i] == '(':
stack.append(i)
continue
if s[i] == ')':
idx = stack.pop()
s[idx+1 : i] = s[idx+1 : i][::-1]
ans = ""
for i in s:
if i=="(" or i==")":
continue
ans += i
return ansclass Solution:
def romanToInt(self, s: str) -> int:
ans = 0
prev = ''
for i in range(len(s)):
if(s[i]=='M'):
if(prev=='C'):
ans += 800
prev = 'M'
continue
ans += 1000
prev = 'M'
continue
if(s[i]=='D'):
if(prev=='C'):
ans += 300
prev = 'D'
continue
ans += 500
prev = 'D'
continue
if(s[i]=='C'):
if(prev=='X'):
ans += 80
prev = 'C'
continue
ans += 100
prev = 'C'
continue
if(s[i]=='L'):
if(prev=='X'):
ans += 30
prev = 'L'
continue
ans += 50
prev = 'L'
continue
if(s[i]=='X'):
if(prev=='I'):
ans += 8
prev = 'X'
continue
ans += 10
prev = 'X'
continue
if(s[i]=='V'):
if(prev=='I'):
ans += 3
prev = 'V'
continue
ans += 5
prev = 'V'
continue
if(s[i]=='I'):
ans += 1
prev = 'I'
return ansclass Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
x = len(nums)
A = [0]*x
for i in range(0, x):
A[(i+k)%x] = nums[i]
for i in range(0, x):
nums[i] = A[i]class Solution:
def rotate(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
n = len(matrix)
for x in zip(*matrix):
matrix.pop(0)
matrix.append(x[::-1])# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def rotateRight(self, head: ListNode, k: int) -> ListNode:
if not head:
return None
if not k==0:
tail = head
length = 1
while(tail.next):
length += 1
tail = tail.next
k = k % length
tail.next = head
tempnode = head
for _ in range(0, length-k-1):
tempnode = tempnode.next
a = tempnode.next
tempnode.next = None
return a
return headclass Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
def binary(nums, low, high, target):
if low <= high:
mid = low + (high - low) // 2
if nums[mid] == target:
return mid
elif nums[mid] > target:
return binary(nums, low, mid-1, target)
else:
return binary(nums, mid+1, high, target)
else:
return high+1
return binary(nums, 0, len(nums)-1, target)
# ------------------------------------------
# Iterative Binary Search
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
low = 0
high = len(nums) - 1
while low <= high:
mid = (low + high) // 2
if nums[mid] == target:
return mid
elif nums[mid] > target:
high = mid - 1
else:
low = mid + 1
else:
return high + 1class Solution:
def __init__(self, nums: List[int]):
self.nums = nums
shuf = self.nums.copy()
self.shuf = shuf
def reset(self) -> List[int]:
"""
Resets the array to its original configuration and return it.
"""
self.shuf = self.nums.copy()
return self.shuf
def shuffle(self) -> List[int]:
"""
Returns a random shuffling of the array.
"""
x = len(self.nums)
for i in range(x):
j = random.randrange(i, x)
self.shuf[i], self.shuf[j] = self.shuf[j], self.shuf[i]
return self.shuf
# ------------------------------------------
# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.reset()
# param_2 = obj.shuffle()# from collections import deque
class Solution:
def simplifyPath(self, path: str) -> str:
if(len(path)==0 or path==None or path==''):
return '/'
p = path.split('/')
stack = []
for item in p:
if (item=='..'):
if(stack):
stack.pop()
continue
if item=='.' or len(item)==0:
pass
else:
stack.append(item)
return '/'+'/'.join(stack)
class Solution:
def singleNumber(self, nums: List[int]) -> int:
return int(((sum(set(nums))*3) - sum(nums))/2)class Solution:
def singleNumber(self, nums: List[int]) -> List[int]:
ans = []
for i in set(nums):
if nums.count(i)==1:
ans.append(i)
if(len(ans)==2):
return ans
return ansfrom collections import Counter
class Solution:
def sortColors(self, nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
n = Counter(nums)
nums.clear()
for v in range(3):
for i in range(n[v]):
nums.append(v)class Solution:
def balancedStringSplit(self, s: str) -> int:
c = 0
rc = 0
lc = 0
for i in range(len(s)):
if s[i] == 'R':
rc += 1
if s[i] == 'L':
lc += 1
if rc == lc:
c += 1
rc = 0
lc = 0
return cclass Solution:
def myAtoi(self, str: str) -> int:
if len(str)==0:
return 0
str = list(str.strip())
if len(str)==0:
return 0
if(str[0]=='-'):
s = -1
else:
s = 1
if str[0] in ['-', '+']:
del str[0]
i = 0
exp = 0
while(i < len(str) and str[i].isdigit()):
exp = exp*10 + ord(str[i]) - ord('0')
i += 1
return max(-2**31, min(exp*s, 2**31-1))class Solution:
def leastInterval(self, tasks: List[str], n: int) -> int:
tasksDict = collections.Counter(tasks)
heap = []
c = 0
for k, v in tasksDict.items():
heappush(heap, (-v,k))
while heap:
i = 0
stack = []
while i<=n:
if len(heap)>0:
index, task = heappop(heap)
if index!=-1:
stack.append((index+1, task))
c += 1
if len(heap)==0 and len(stack)==0:
break
i += 1
for i in stack:
heappush(heap, i)
return c## Naive Solution that runs in O(n^2)
## Traverses the array, and finds the maximum left or right element.
## rainwater that can be stored in the column = min(left, right) - height[i]
class NaiveSolution:
def trap(self, height) -> int:
res = 0
n = len(height)
for i in range(1, n-1):
left = height[i]
for j in range(i):
left = max(left, height[j])
right = height[i]
for j in range(i+1, n):
right = max(right, height[j])
res += min(left, right) - height[i]
return res
# ------------------------------------------
## Optimized solution that runs in O(N)
## Stores the left and right maximum values in two dp arrays.
## Space Complexity - O(N)
class Solution:
def trap(self, height: List[int]) -> int:
if height == []:
return 0
n = len(height)
res = 0
left = [0 for _ in range(n)]
right = [0 for _ in range(n)]
left[0] = height[0]
right[n-1] = height[n-1]
for i in range(1, n):
left[i] = max(left[i-1], height[i])
for i in range(n-2, -1, -1):
right[i] = max(right[i+1], height[i])
for i in range(n):
res += min(left[i], right[i]) - height[i]
return res class Solution:
def minimumTotal(self, triangle: List[List[int]]) -> int:
dp = [0 for _ in range(len(triangle)+1)]
for r in triangle[::-1]:
for i in range(len(r)):
dp[i] = r[i] + min(dp[i], dp[i+1])
return dp[0]class Solution:
def twoCitySchedCost(self, costs: List[List[int]]) -> int:
costs = sorted(costs, key = lambda x: x[0] - x[1])
return sum(i[0] for i in costs[0:len(costs)//2]) + sum(j[1] for j in costs[len(costs)//2:len(costs)])class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
m = len(obstacleGrid)
n = len(obstacleGrid[0])
if obstacleGrid[0][0] == 1:
return 0
obstacleGrid[0][0] = 1
for i in range(1, m):
obstacleGrid[i][0] = int(obstacleGrid[i][0] == 0 and obstacleGrid[i-1][0] == 1)
for i in range(1, n):
obstacleGrid[0][i] = int(obstacleGrid[0][i] == 0 and obstacleGrid[0][i-1] == 1)
for i in range(1, m):
for j in range(1, n):
if obstacleGrid[i][j] == 0:
obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1]
else:
obstacleGrid[i][j] = 0
return obstacleGrid[-1][-1] from collections import Counter
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
return Counter(s) == Counter(t)class Solution:
def isPalindrome(self, s: str) -> bool:
s = ''.join(a for a in s if a.isalnum()).lower()
return s == s[::-1]class Solution:
def checkValidString(self, s: str) -> bool:
lb = 0
rb = 0
for i in s:
if(i=='(' or i=='*'):
lb += 1
else:
lb -= 1
if lb < 0:
return False
if(lb==0):
return True
for i in range(len(s)-1, -1, -1):
if(s[i]==')' or s[i]=='*'):
rb += 1
else:
rb -= 1
if rb < 0:
return False
return Trueclass Solution(object):
def robotSim(self, commands, obstacles):
dx = [0, 1, 0, -1]
dy = [1, 0, -1, 0]
x = y = di = 0
obstacleSet = set(map(tuple, obstacles))
ans = 0
for cmd in commands:
if cmd == -2: #left
di = (di - 1) % 4
elif cmd == -1: #right
di = (di + 1) % 4
else:
for k in range(cmd):
if (x+dx[di], y+dy[di]) not in obstacleSet:
x += dx[di]
y += dy[di]
ans = max(ans, x*x + y*y)
# ------------------------------------------
# Solution for https://leetcode.com/problems/two-sum/
# Language : Python3
# ------------------------------------------
# O(n^2) Solution
class Solution:
def twoSum(self, nums, target):
for i in range(len(nums)):
for j in range(i+1,len(nums)):
if(nums[i]+nums[j]==target):
return i, j
# O(n) Solution
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
dict = { v:k for k, v in enumerate(nums) }
for i in range(len(nums)):
if target - nums[i] in dict and nums.index(target - nums[i]) != i:
return i, nums.index(target-nums[i])
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