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Welcome to the GeeksforGeeks Problem of the Day (PotD) Solutions repository! This repository aims to provide daily solutions to the coding challenges presented by GeeksforGeeks as part of their Problem of the Day series.
Dive into a treasure trove of daily challenges meticulously crafted to sharpen your problem-solving skills. With diverse topics, detailed explanations, and a supportive community, this repository is your gateway to mastering algorithms, data structures, and more.
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Jan 8, 2024 · The questions will cover different topics based on Data Structures and Algorithms and you will have 24 hours to channel your inner Geek and solve the challenge. Table of Content. What is Problem Of The Day (POTD)? Instructions for Problem Of The Day (POTD) Nomenclature for POTD (Geek Nomenclature)
Deletion and Reverse in Circular Linked List. Intuit. Solve Problem. Medium 37K 60.7 %. Previous Problems. 2024. October. 3 October.
Oct 2, 2024 · leetcode-potd Design your implementation of the circular double-ended queue (deque). Implement the MyCircularDeque class: MyCircularDeque(int k) Initializes the deque with a maximum size of k.
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Here you'll find answers & solutions to all past seasons. If your browser does not support the viewing of the embedded PDF, you can view it here . If you'd like to contribute a solution, or fix a typo, or learn how to otherwise contribute, you can find the repository here.
Explanation: First create a list of stock with (price, index): [ (10,1), (7,2), (19,3)]. Then sort the list, it becomes : [ (7,2), (10,1), (19,3)] Now, keep buying the stock until we ran out of money, here with k = 45, we can buy: 7 2 + 10 1 + 19*1 = 43. Hence, total 4 stocks we can buy with k = 45; Time and Auxiliary Space Complexity.
Solution- class Solution { public: int searchInsertK (vector<int>Arr, int N, int k) { // code here int low=0; int high=N-1; int mid; while (low<=high) { mid= (low+high)/2; if (Arr [mid]==k) return mid; else if (Arr [mid]>k) { high=mid-1; } else { low=mid+1; } } if (Arr [mid]>k) { return mid; } else return mid+1; } };
