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The direct address table approach is simple and efficient for this problem since the maximum key size is limited to 106. We can use a boolean array of size 106+1 where each index represents a key. The value stored at a given index indicates whether the key is present in the HashSet. This allows for O(1) time complexity for the 'add', 'remove', and 'contains' operations.
Time Complexity: O(1) for all operations (add, remove, contains).
Space Complexity: O(MAX), which is O(10^6).
1public class MyHashSet {
2 private boolean[] container;
3
4 public MyHashSet() {
5 container = new boolean[1000001];
6 }
7
8 public void add(int key) {
9 container[key] = true;
10 }
11
12 public void remove(int key) {
13 container[key] = false;
14 }
15
16 public boolean contains(int key) {
17 return container[key];
18 }
19}
20In Java, we create a boolean array to serve as our direct address table. The add and remove operations toggle the boolean value, and contains checks it.
This approach simulates a hash set using a hash function to assign keys to buckets, dealing with potential collisions using chaining. We create an array of lists, where each list contains keys that hash to the same index. This way, operations require traversing these short lists upon collisions.
Time Complexity: O(1) average, O(n) worst-case for add, remove, and contains due to collision chains.
Space Complexity: O(n), where n is the number of keys actually added.
1
In Python, we use a class 'Bucket' holding keys as lists and another class 'MyHashSet' to manage these buckets. Each bucket corresponds to keys by modulus operation, and operations use list membership test and updates to handle keys in buckets.