SiliconANGLE & theCUBE’s Post

Fortanix CEO and co-founder Anand Kashyap joined David Vellante as part of SiliconANGLE & theCUBE’s coverage of RSA 2024. The two discussed all things Fortanix, diving into the company's history, solutions and the future of protecting data. Anand shares what makes Fortanix’s data protection solutions stand out and how having a confidential computing-based platform that runs on AI gives the company confidentiality, integrity and security. Listen to the full conversation below. #Dataprotection #AI #RSA

Day 20 of #geekstreak2024 challenge powered by Deutsche Bank and GeeksforGeeks! 𝐏𝐎𝐓𝐃: Total count You are given an array arr[] of positive integers and a threshold value k. For each element in the array, divide it into the minimum number of small integers such that each divided integer is less than or equal to k. Compute the total number of these integer across all elements of the array. Examples: Input: k = 3, arr[] = [5, 8, 10, 13] Output: 14 Explanation: Each number can be expressed as sum of different numbers less than or equal to k as 5 (3 2), 8 (3 3 2), 10 (3 3 3 1), 13 (3 3 3 3 1). So, the sum of count of each element is (2 3 4 5)=14. Expected Time Complexity: O(n) Expected Auxiliary Space: O(1) 𝐀𝐥𝐠𝐨𝐫𝐢𝐭𝐡𝐦: 1. Initialization: -Start with a variable total_count initialized to 0. This variable will keep track of the total number of divided integers across all elements in the array. 2. Iterate through the Array: - For each element in the array: - Calculate how many parts this element can be divided into. This can be computed using: count= element\k - This can also be derived using integer division: count=element k−1/k - This formula gives the correct number of parts because it effectively rounds up the result of the division. 3. Accumulate Counts: Add the count for the current element to total_count. 4. Return Result: After processing all elements, return total_count. 𝐒𝐨𝐮𝐫𝐜𝐞 𝐜𝐨𝐝𝐞: class Solution {  public:   int totalCount(int k, vector<int>& arr) {     int count=0;     for(int i=0;i<arr.size();i ){       count =arr[i]/k;       if(arr[i]%k!=0) count ;     }     return count;   } }; #geekstreak2024 #deutschebank #GeeksforGeeks

Day 3 of the #geekstreak2024 challenge supported by Deutsche Bank! Another Problem of the Day tackled on GeeksforGeeks! Each day presents an opportunity to enhance my problem-solving skills. Staying consistent and eager to evolve on this journey! 💻🏅 #PROBLEM_STATEMENT: Utilizing the "two-pointer technique" - the slow and fast pointer approach - to crack challenges effectively. Here's how it operates: #Steps: - Initialize both slow and fast pointers at the head of the list. - Traverse the list: Move slow by one step and fast by two steps iteratively. #End_of_Traversal: Once the fast pointer reaches the end, the slow pointer will be at the middle node. #Approach: #Two_Pointers: #Slow_pointer (slow): Moves one step at a time. #Fast_pointer (fast): Moves two steps at a time. #How_it_Works: Both pointers initiate at the head, with the slow moving one node ahead of the fast in each loop iteration. #Key_Insight: For an odd node count, the slow pointer aligns with the exact middle. In the case of an even count, the slow pointer naturally points to the second middle node when the fast pointer completes traversal. #Edge_Case: If the list is empty (head is None), the return value is -1. #DeutscheBank #codingchallenge #ProblemOfTheDay #geeksforgeeks #deutschebank GeeksforGeeks Deutsche Bank Sandeep Jain Sir

✨ Day 13 of #GeekStreak2024 with GeeksforGeeks and Deutsche Bank ✨ Approach: 1. Frequency Count Arrays: Use two arrays of size 128 to store character frequencies, one for the target string `p` (`pCount[]`) and another for the current window in `s` (`windowCount[]`). 2. Sliding Window Expand the window by moving the `right` pointer, adding characters to `windowCount[]`. Count valid characters only when their frequency doesn't exceed that in `pCount[]`. 3. Contract Window: Once all characters are matched, contract the window from the left by moving `left`, updating `windowCount[]`, and tracking the smallest valid window. 4. Result: If a valid window is found, return the substring; otherwise, return `"-1"`. #CodingChallenge #GeeksforGeeks #GeekStreak2024 #DeutscheBank

Explore the insights from our co-founders, Thanh-Long Huynh and Ghizlaine Amrani, in Moody's recent feature! Learn about the transformative power of alternative data, its real-time applications, and how it complements traditional financial analysis to provide a deeper understanding of economic and corporate performance. Read the full feature below. #AlternativeData #FinancialInsights #QuantCube

🚀 Day 11 of the 21 Days #GeekStreak2024 🚀 Today's Problem: Clone a Linked List with Next and Random Pointer Problem Statement: You are given a special linked list where each node has a next pointer pointing to its next node, and a random pointer which can point to any node in the list. The task is to create a deep copy of this linked list. The copied list should replicate the same structure as the original, where both next and random pointers point to corresponding new nodes. My Approach: I solve this problem, First traverse the whole list, For each node in the original list, create a new node and insert it next to the original node. In Second Pass: Set the random pointer for the new nodes using the original list’s random pointer. In Third Pass: Separate the original and copied lists to restore the original list and return the head of the new list. Time Complexity: O(n) Space Complexity: O(1) GeeksforGeeks Deutsche Bank

Data-driven modernization is not just a buzzword. Thought Machine's latest IDC report provides a comprehensive analysis of data migration challenges and strategies, helping you build a resilient, flexible data infrastructure that drives business growth. Download now: https://bit.ly/3UTFjav #fintech #datamigration #Innovation

✨ Day 20 of the #geekstreak2024 Challenge – Powered by Deutsche Bank 💡Problem of the Day: Total Count 📊 Today’s problem was about breaking down large numbers into smaller integers while keeping each one under a specified threshold. The task was to find out the total number of smaller integers needed to achieve this across all elements of the array. 🧠 Problem Details: Given an array of positive integers and a threshold value k, for each element, we had to divide it into the minimum number of small integers such that each one is less than or equal to k. The goal was to compute the total count of these divided integers for the entire array. Example 1: Input: k = 3, arr[] = [5, 8, 10, 13] Output: 14 Explanation: 5 → (3 2) → 2 integers 8 → (3 3 2) → 3 integers 10 → (3 3 3 1) → 4 integers 13 → (3 3 3 3 1) → 5 integers Total count: 2 3 4 5 = 14 Key Challenge: The challenge lay in figuring out how to divide each number efficiently while ensuring that the number of divisions was minimized and that each part stayed within the given threshold. What I Learned: 1. Gained more insight into breaking down larger numbers into smaller parts, focusing on minimizing steps while adhering to the constraints. 2. Improved my understanding of greedy algorithms and how to approach similar real-world problems with division and resource allocation. 3. Practiced implementing the solution with an O(n) time complexity, ensuring that it performs well even for larger datasets. Big thanks to GeeksforGeeks and Sandeep Jain sir for these exciting problem-solving opportunities, and to Deutsche Bank for powering this incredible challenge! 🚀 #geekstreak2024 #geeksforGeeks #deutschebank #techJourney #problemSolvers #codeChallenge

In the dynamic world of finance, reliable software is the foundation of success. Metoda faced challenges in modernizing software due to technical debt. Aionys' FinTech expertise delivered 🏦: - Modernization: New technologies & architecture for a future-proof platform. - Scalability: Microservices architecture enables scalability, agility, and easy updates. - Innovation: Powerful new features to drive market expansion. At Aionys, we understand the significance of financial software and the importance of reliable solutions. Our team is committed to delivering innovative and secure customized solutions to meet our customers' specific needs. We strive to combine our expertise with innovation, empowering financial institutions to stay ahead. 🎯 #Aionys, #Fintech, #SoftwareDevelopment, #TechTransformation, #FintechSolutions, #ReducedCosts.

🚀 Day 9 of #GeekStreak2024, and yesterday’s challenge was "Facing the Sun" on GeeksforGeeks. 🌅🏙️ The task was to find out how many buildings can see the sunrise, which boiled down to comparing building heights from left to right. A straightforward problem, but it reinforced the importance of scanning and keeping track of maximum values efficiently. 💡 Big thanks to Deutsche Bank for supporting this journey of constant growth. Onward to the next challenge! 💪 #DeutscheBank #GeekStreak2024 #FacingTheSun #CompetitiveProgramming #GeeksforGeeks #CodingJourney