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Leetcode with dani

Leetcode with dani

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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

it is started

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

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)

Did you understand Floyd’s Cycle-Finding Algorithm for Linked List? It’s pretty cool and clever!

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 p1

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.next

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.next

https://leetcode.com/problems/remove-linked-list-elements/ Easy question for beginner to linked list

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

🚀 A2SV Internship Program – Level Up Your Tech Skills! 🚀 Gain hands-on experience in Frontend, Backend, Mobile Development, UI/UX, and Product Management while working on real projects with industry experts. Perfect for students looking to grow and launch their tech careers! 📌 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! 🚀

A2SV __ Linked List Lecture II No code.pdf3.59 MB

Linked List 1 -with out code.pdf2.76 MB

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 maxm

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.

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

Linked List Questions.pdf1.12 KB

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