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▎678. Valid Parenthesis String

💻 Difficulty: Medium

Given a string s containing only three types of characters: '(', ')', and '*', return true if s is valid.

Rules for a valid string:

Any left parenthesis '(' must have a corresponding right parenthesis ')'.
Any right parenthesis ')' must have a corresponding left parenthesis '('.
Left parenthesis '(' must go before the corresponding right parenthesis ')'.
'*' can be treated as '(', ')', or an empty string "".

Examples

Example 1:
Input: s = "()"
Output: true

Example 2:
Input: s = "(*)"
Output: true

Example 3:
Input: s = "(*))"
Output: true


Constraints:

1 <= s.length <= 100

s[i] is '(', ')', or '*'.

---

Solution (Greedy Approach)

def checkValidString(s: str) -> bool:
low = high = 0 # low: min open count, high: max open count

for char in s:
if char == '(':
low += 1
high += 1
elif char == ')':
low = max(0, low - 1) # Decrease open count but not below zero
high -= 1
else: # '*'
low = max(0, low - 1) # Treat '*' as ')'
high += 1 # Treat '*' as '('

if high < 0: # Too many ')', invalid
return False

return low == 0 # If low is zero, valid parentheses

# Test cases
print(checkValidString("()")) # True
print(checkValidString("(*)")) # True
print(checkValidString("(*))")) # True



▎Explanation

We keep track of the possible range of open parentheses using low and high:

• low (minimum open count) decreases when ')' appears but never goes below 0.

• high (maximum open count) increases when '(' or '*' (treated as '(') appears.

• If high becomes negative at any point, the string is invalid.

• Finally, if low == 0, the string is valid.

🚀 Time Complexity: O(n)
🔹 Space Complexity: O(1)
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let me share one conversation with GPT
if i want to apply to FAANG what should i expect to know
ChatGPT said:
If you want to apply to FAANG (Facebook/Meta, Amazon, Apple, Netflix, Google), you need to be well-prepared in multiple technical and non-technical areas. Here’s a breakdown of what you should know:

1. Data Structures & Algorithms (DSA) – High Priority
You should be able to solve LeetCode Hard problems efficiently. Focus on:
Arrays, Strings, Linked Lists
Trees (Binary, BST, Trie)
Graphs (BFS, DFS, Dijkstra’s)
Dynamic Programming
Backtracking
Bit Manipulation
Heaps & Priority Queues
Two Pointers, Sliding Window, Prefix Sum

🔥 Recommended Resources:

LeetCode (Top 150 FAANG problems)
Codeforces (Div2 & Div1 contests)
NeetCode, Striver’s SDE Sheet
Cracking the Coding Interview by Gayle Laakmann McDowell
2. System Design (For Senior Roles)
FAANG interviews often test system design for mid/senior positions. You should know:
Scalability (Load Balancing, Sharding, Caching)
Microservices & Distributed Systems
Databases (SQL & NoSQL)
API Design, Rate Limiting, Consistency Models
Message Queues (Kafka, RabbitMQ)
CAP Theorem

🔥 Resources:

"Designing Data-Intensive Applications" by Martin Kleppmann
System Design Primer (GitHub)
Grokking the System Design Interview
3. Object-Oriented Design (OOD) & Design Patterns
SOLID Principles
Factory, Singleton, Observer, Decorator, Strategy Patterns

🔥 Resources:

Head First Design Patterns
Grokking the Object-Oriented Design Interview
4. Low-Level Programming & Optimization (For Performance-Critical Roles)
Memory Management (Heap vs Stack)
Threading & Concurrency
Cache Optimization
Assembly & Compiler Optimizations

5. Core CS Concepts
Operating Systems (Processes, Threads, Deadlocks, Virtual Memory)
Networking (TCP/IP, HTTP, Load Balancing, CDN)
Databases (Indexes, ACID, Transactions, Query Optimization)

🔥 Resources:

Computer Networking: A Top-Down Approach
Database System Concepts
6. Behavioral Interviews – Don't Ignore This!
FAANG uses STAR (Situation, Task, Action, Result) format for behavioral questions.
Leadership & Ownership
Conflict Resolution
Past Projects & Impact

🔥 Resources:

"The Amazon Leadership Principles"
"Decode & Conquer" by Lewis Lin
7. Programming Languages & Tech Stack
You should be fluent in at least one high-level language, such as:

Java, Python, C++, JavaScript, Go (most used in FAANG)
Knowledge of React, Node.js, Spring Boot, Django is a plus
8. Mock Interviews & Time Management
Start practicing mock interviews with peers or platforms like:
Pramp
Interviewing.io
Leetcode Contest + Codeforces
👍41
Forwarded from Codeforces Official
Codeforces Round #1010 (Div. 1, Div. 2) will take place on the 15th of March at 05:35 UTC.
Please, join by the link https://codeforces.com/contests/2081,2082?locale=en
330. Patching Array
Difficulty: Hard

Problem Statement:

Given a sorted integer array nums and an integer n, add/patch elements to the array such that any number in the range [1, n] (inclusive) can be formed by the sum of some elements in the array.

Return the minimum number of patches required.

Example 1:

Input: nums = [1, 3], n = 6
Output: 1
Explanation:
The combinations of nums are [1], [3], [1, 3], which form possible sums of 1, 3, 4.
Now if we add/patch 2 to nums, the combinations become [1], [2], [3], [1, 3], [2, 3], [1, 2, 3], covering all sums 1, 2, 3, 4, 5, 6.
Thus, only 1 patch is needed.
Example 2:

Input: nums = [1, 5, 10], n = 20
Output: 2
Explanation:
The two patches can be [2, 4].
Example 3:

Input: nums = [1, 2, 2], n = 5
Output: 0
Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10^4
nums is sorted in ascending order.
1 <= n <= 2^31 - 1
Solution :

class Solution:
def minPatches(self, nums: List[int], n: int) -> int:
patches = 0
miss = 1
i = 0
while miss <= n:
if i < len(nums) and nums[i] <= miss:
miss += nums[i]
i += 1
else:
miss += miss
patches += 1
return patches
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135. Candy
Hard
There are n children standing in a line. Each child is assigned a rating value given in the integer array ratings.

You are giving candies to these children subjected to the following requirements:

Each child must have at least one candy.
Children with a higher rating get more candies than their neighbors.
Return the minimum number of candies you need to have to distribute the candies to the children.



Example 1:

Input: ratings = [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
Example 2:

Input: ratings = [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
The third child gets 1 candy because it satisfies the above two conditions.


Constraints:

n == ratings.length
1 <= n <= 2 * 104
0 <= ratings[i] <= 2 * 104
class Solution:
def candy(self, ratings: List[int]) -> int:
n = len(ratings)
candies = [1] * n # Give every child 1 candy initially

# Left-to-right pass: if a child has a higher rating than the left neighbor, give one more candy
for i in range(1, n):
if ratings[i] > ratings[i - 1]:
candies[i] = candies[i - 1] + 1

# Right-to-left pass: if a child has a higher rating than the right neighbor, ensure they have more candy
for i in range(n - 2, -1, -1):
if ratings[i] > ratings[i + 1]:
candies[i] = max(candies[i], candies[i + 1] + 1)

return sum(candies)

i didnt know it had a simple ans; my ans is like thousand lines
👍2
Forwarded from Codeforces Official
Unfortunately, due to technical reasons, the round is postponed by 8.5 hours (to the usual scheduled time). We apologize for any inconvenience caused.
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A big thank you to everyone at this channel and  Alpha ,for a great session! I really enjoyed it.
7
For Ethiopians only
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Leetcode with dani
Here is our very first leetcode Tree problem (BFS). Answer will be posted at 6:00 PM
102. Binary Tree Level Order Traversal

Difficulty: Medium

Problem Statement

Given the root of a binary tree, return the level order traversal of its nodes' values (i.e., from left to right, level by level).

Examples

Example 1:

Input: root = [3,9,20,null,null,15,7]

Output: [[3],[9,20],[15,7]]

Example 2:

Input: root = [1]

Output: [[1]]

Example 3:

Input: root = []

Output: []

Constraints

• The number of nodes in the tree is in the range [0, 2000].

-1000 <= Node.val <= 1000
jakes-resume.pdf
102.5 KB
When u create a CV take this as model
👍2
Leetcode with dani
Here is our very first leetcode Tree problem (BFS). Answer will be posted at 6:00 PM
class Solution:
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:


def helper(qu):
if not qu:
return

i = len(qu)
temp = []
nextsize = 0


while i > 0 :
poped = qu.popleft()
temp.append(poped.val)

if poped.left:
qu.append(poped.left)
nextsize += 1

if poped.right:
qu.append(poped.right)
nextsize += 1


i -= 1

ans.append(temp)
helper(qu)

if not root:
return []

ans =[]
q = deque()
q.append(root)
helper(q)

return ans
1
class Solution:
def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:

def helper(qu):
if not qu:
return

i = len(qu)
leng = len(qu)
total = 0

while i > 0:
poped = qu.popleft()
total += poped.val

if poped.left:
qu.append(poped.left)
if poped.right:
qu.append(poped.right)

i -= 1
ans.append(total/leng)
helper(qu)
qu = deque()
qu.append(root)
ans = []

helper(qu)

return ans
👍21
Forwarded from Codeforces Official
Codeforces Round #1012 (Div. 1, Div. 2) will take place on the 23rd of March at 05:35 UTC.
Please, join by the link https://codeforces.com/contests/2089,2090?locale=en