▎678. Valid Parenthesis String
💻 Difficulty: Medium
Given a string
▎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:
Output:
Example 2:
Input:
Output:
Example 3:
Input:
Output:
▎Constraints:
•
•
---
▎Solution (Greedy Approach)
💻 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:
trueExample 2:
Input:
s = "(*)" Output:
trueExample 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)
LeetCode
Valid Parenthesis String - LeetCode
Can you solve this real interview question? Valid Parenthesis String - Given a string s containing only three types of characters: '(', ')' and '*', return true if s is valid.
The following rules define a valid string:
* Any left parenthesis '(' must have…
The following rules define a valid string:
* Any left parenthesis '(' must have…
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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
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
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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 :
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
👍2
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
i didnt know it had a simple ans; my ans is like thousand lines
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
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Educational Codeforces Round 176
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Please, join by the link https://codeforces.com/contests/2075
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Here is our very first leetcode Tree problem (BFS). Answer will be posted at 6:00 PM
LeetCode
Binary Tree Level Order Traversal - LeetCode
Can you solve this real interview question? Binary Tree Level Order Traversal - 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).
Example 1:
[https://assets.leetcode.c…
Example 1:
[https://assets.leetcode.c…
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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:
• Output:
Example 2:
• Input:
• Output:
Example 3:
• Input:
• Output:
▎Constraints
• The number of nodes in the tree is in the range
•
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
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
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#Tree second question Answer will be posted at 4 PM
LeetCode
Average of Levels in Binary Tree - LeetCode
Can you solve this real interview question? Average of Levels in Binary Tree - Given the root of a binary tree, return the average value of the nodes on each level in the form of an array. Answers within 10-5 of the actual answer will be accepted.
Example…
Example…
👍3
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
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Codeforces Round 1011 (Div. 2) will take place on the 22nd of March at 14:35 UTC.
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Forwarded from Codeforces Official
Codeforces Round #1012 (Div. 1, Div. 2) will take place on the 23rd of March at 05:35 UTC.
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