The problem Given an array nums of size n, return the majority element. The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array. Example 1: Input: nums = [3,2,3] Output: 3 Example 2: Input: nums = [2,2,1,1,1,2,2] Output: 2 Constraints: n == nums.length 1 <= n <= 5 * 10^4 -10^9 <= nums[i] <= 10^9 The input is generated such that a majority element will exist in the array. ...
The problem Design a HashMap without using any built-in hash table libraries. Implement the MyHashMap class: MyHashMap() initializes the object with an empty map. void put(int key, int value) inserts a (key, value) pair into the HashMap. If the key already exists in the map, update the corresponding value. int get(int key) returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key. ...
The problem Design a HashSet without using any built-in hash table libraries. Implement MyHashSet class: void add(key) Inserts the value key into the HashSet. bool contains(key) Returns whether the value key exists in the HashSet. void remove(key) Removes the value key from the HashSet. If key does not exist in the HashSet, do nothing. Example 1: Input ["MyHashSet", "add", "add", "contains", "contains", "add", "contains", "remove", "contains"] [[], [1], [2], [1], [3], [2], [2], [2], [2]] Output [null, null, null, true, false, null, true, null, false] Explanation MyHashSet myHashSet = new MyHashSet(); myHashSet.add(1); // set = [1] myHashSet.add(2); // set = [1, 2] myHashSet.contains(1); // return True myHashSet.contains(3); // return False, (not found) myHashSet.add(2); // set = [1, 2] myHashSet.contains(2); // return True myHashSet.remove(2); // set = [1] myHashSet.contains(2); // return False, (already removed) Constraints: ...