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Repost from Codeforces Official
Codeforces Round 1007 (Div. 2) will take place on the 28th of February at 14:35 UTC.
Please, join by the link https://codeforces.com/contests/2071?locale=en
1 258
Repost from Codeforces Official
Educational Codeforces Round 175
(rated for Div. 2) starts in ~2 hours.
Please, join by the link https://codeforces.com/contests/2070
1 258
Top 10 Non-Technical Interview Questions for FAANG Companies: Insights and Preparation Tips
▎🧠 Behavioral Leadership
1. "Tell me about a time you failed and what you learned."
Tests self-awareness and growth mindset.
2. "Describe a team conflict you resolved."
Assesses emotional intelligence and mediation skills.
3. "Share an example of showing leadership without authority."
Evaluates initiative and influence.
▎🎯 Career Motivation
4. "Why do you want to work here specifically?"
Probes company research and cultural fit.
5. "Where do you see yourself in 3-5 years?"
Checks alignment with company growth paths.
▎🌟 Strengths Values
6. "What unique value would you bring to this team?"
Reveals self-assessment accuracy.
7. "Describe your most innovative professional contribution."
Assesses creativity and business impact.
▎🛠 Work Approach
8. "How do you prioritize when facing multiple deadlines?"
Tests organizational and decision-making skills.
9. "Tell me about a project requiring deep analysis."
Examines problem-solving methodology.
▎🤝 Cultural Fit
10. "What does ideal team collaboration look like to you?"
Matches working style with company culture.
---
▎Key Preparation Tips:
• Use the STAR (Situation-Task-Action-Result) format for behavioral answers.
• Align responses with FAANG leadership principles.
• Practice concise storytelling (90-120 seconds per answer).
• Research specific company values (e.g., Amazon's 16 Leadership Principles).
For 53 additional common questions and answer frameworks, refer to:
FAANG Behavioral Guide (https://igotanoffer.com/blogs/tech/faang-interview-questions)
Non-Technical Question Strategies (https://www.indeed.com/career-advice/interviewing/non-tech-interview-questions)
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Did you understand Floyd’s Cycle-Finding Algorithm for Linked List? It’s pretty cool and clever!
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class Solution:
def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
p1 = head
p2 = head
while p2 and p2.next:
p1=p1.next
p2 = p2.next.next
return p11 258
class Solution:
def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
dummy = ListNode(-1,head)
p1 = dummy
p2 = dummy.next
count = 0
while p2:
if count <n:
p2 = p2.next
count+=1
else:
p1 = p1.next
p2 = p2.next
if p1:
p1.next = p1.next.next
return dummy.next1 258
class Solution:
def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
dummy = ListNode(0,head)
p1 = dummy
p2 = dummy.next
while p2:
if p2.val==val:
p1.next = p2.next
p2 = p2.next
else:
p1 = p1.next
p2 = p2.next
return dummy.next1 258
https://leetcode.com/problems/remove-linked-list-elements/
Easy question for beginner to linked list
1 258
Repost from Codeforces Official
Educational Codeforces Round 175
(rated for Div. 2) starts on the 27th of February at 14:35 UTC.
Please, join by the link https://codeforces.com/contests/2070
1 258
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📌 Requirements: Must be a continuing student in their internship phase and have completed or be enrolled in A2SV’s G6 program.
📍 Location: Abrehot Library (In-person)
⏰ Hours: 9 AM - 12 PM, 2 PM - 6 PM (G6: 2 PM - 4 PM)
Apply now and start your journey with A2SV! 🚀
1 258
class Solution:
def findMaxLength(self, nums: List[int]) -> int:
s = {0:-1}
t = 0
maxm = 0
c1 = 0
c0 = 0
for i in range(len(nums)):
if nums[i]:
c1+=1
else:
c0+=1
if c1-c0 not in s:
s[c1-c0] = i
else:
maxm = max(maxm,i-s[c1-c0])
return maxm1 258
▎LeetCode Problem 525: Contiguous Array
Difficulty: Medium
Topics: Array, Hashing, Prefix Sum
Companies: Many top tech companies have featured similar problems in interviews
Link: Contiguous Array on LeetCode
---
▎Problem Statement
Given a binary array
nums, return the maximum length of a contiguous subarray with an equal number of 0 and 1.
A contiguous subarray is one where the elements are consecutive, and an equal number of 0s and 1s means that the sum (after a transformation) is 0.
Note:
• You can transform the array by replacing each 0 with -1.
• Then, the problem reduces to finding the largest subarray whose sum is 0.
---
▎Examples
Example 1:
Input: nums = [0, 1]
Output: 2
Explanation: The entire array [0, 1] has one 0 and one 1.
Example 2:
Input: nums = [0, 1, 0]
Output: 2
Explanation: Either subarray [0, 1] or [1, 0] has an equal number of 0s and 1s.
---
▎Approach Explanation
1. Transform the Array:
Replace every 0 with -1. This way, a subarray with an equal number of 0s and 1s will have a total sum of 0.
2. Prefix Sum Hash Map:
• Prefix Sum: Compute the cumulative sum while iterating through the transformed array.
• Hash Map: Use a dictionary to store the first occurrence of each cumulative sum.
• Finding a Subarray: If the same cumulative sum is seen again, the subarray between these two indices has a sum of 0 (equal number of 0s and 1s).
• Update Maximum Length: Calculate the length of the subarray and update the maximum length if this subarray is longer.
3. Efficiency:
This approach works in O(n) time and O(n) space, making it efficient for large inputs.
---
▎ Solution
def findMaxLength(nums):
# Replace 0 with -1 for transformation
for i in range(len(nums)):
if nums[i] == 0:
nums[i] = -1
cumulative_sum = 0
max_length = 0
sum_index_map = {} # To store the first occurrence of each cumulative sum
for i, num in enumerate(nums):
cumulative_sum += num
# Check if cumulative_sum is 0, which means subarray from index 0 to i is valid
if cumulative_sum == 0:
max_length = i + 1
# If cumulative_sum has been seen before, update max_length accordingly
if cumulative_sum in sum_index_map:
max_length = max(max_length, i - sum_index_map[cumulative_sum])
else:
sum_index_map[cumulative_sum] = i
return max_length
# Testing the function with provided examples
print(findMaxLength([0, 1])) # Output: 2
print(findMaxLength([0, 1, 0])) # Output: 2
▎Final Recap
This solution transforms the binary array into one that uses -1 instead of 0, so that we can use the cumulative sum to detect subarrays with equal numbers of 0s and 1s. By maintaining a hash map of the first occurrence of each cumulative sum, we can efficiently calculate the maximum length of any subarray with a sum of 0. This is a neat and efficient way to tackle the problem!
For more details, visit the problem link: Contiguous Array on LeetCode.1 258
Repost from Codeforces Official
Codeforces Round 1006 (Div. 3) will take place on the 25th of February at 14:35 UTC.
Please, join by the link https://codeforces.com/contests/2072?locale=en
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