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Design A Number Container System - Leetcode Solution

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LeetCode:  Design A Number Container System Leetcode Solution

Difficulty: Medium

Topics: design heap-priority-queue hash-table  

Unfortunately, as an AI language model, I am not able to access the problem description or solutions of LeetCode as it requires a login. However, I can provide you with a general solution for designing a number container system.

Designing a number container system involves creating a data structure that can efficiently store and search for numerical values. Here, we can consider using a Hash Table or a Binary Search Tree (BST) as our data structure.

  1. Hash Table Approach: We can use a hash table to store the values. In this approach, we use a hash function to convert the values into array indices. The hash function takes a key (numeric value) as input and returns an index value. We can use a simple modulo operation to convert the key into an array index.

Once we have the index, we can store the key-value pair in the respective index. In case of collision, we can use techniques like chaining or probing. Chaining involves creating a linked list at each array index where multiple keys hash to the same index. Probing involves searching for the next available empty index in the array.

The advantage of using a hash table is that it has a constant average time complexity of O(1) for insertion, deletion and search operations. However, it may have a worst-case time complexity of O(n) in case of collisions.

  1. Binary Search Tree Approach: We can also use a Binary Search Tree (BST) as our data structure. In this approach, we insert the keys in a binary tree such that for any node, all the keys in its left subtree are less than its key, and all the keys in its right subtree are greater or equal to its key.

We can use the principles of binary search to perform efficient search operations. For insertion, we can traverse the tree starting from the root to the leaf node where we want to insert the new key. For deletion, we can also traverse the tree to find the node to be deleted and then replace it with its successor (the smallest value in the right subtree) or its predecessor (the largest value in the left subtree).

The advantage of using a BST is that it maintains the key order, and it has a worst-case time complexity of O(logn) for insertion, deletion and search operations. However, the balance of the tree affects its efficiency, and it may have a worst-case time complexity of O(n) if the tree is severely unbalanced.

Overall, a hash table is a better choice if we want a faster average time complexity, while a BST is a better choice if we want to maintain the key order and have a better worst-case time complexity.

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