706. Design HashMap LeetCode Solution

Last updated: February 19, 2025

In this guide, you will get 706. Design HashMap LeetCode Solution with the best time and space complexity. The solution to Design HashMap problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Design HashMap solution in C++
Design HashMap solution in Java
Design HashMap solution in Python
Additional Resources

706. Design HashMap LeetCode Solution image

Problem Statement of Design HashMap

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.
void remove(key) removes the key and its corresponding value if the map contains the mapping for the key.

Example 1:

Input
[“MyHashMap”, “put”, “put”, “get”, “get”, “put”, “get”, “remove”, “get”]
[[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
Output
[null, null, null, 1, -1, null, 1, null, -1]

Explanation
MyHashMap myHashMap = new MyHashMap();
myHashMap.put(1, 1); // The map is now [[1,1]]
myHashMap.put(2, 2); // The map is now [[1,1], [2,2]]
myHashMap.get(1); // return 1, The map is now [[1,1], [2,2]]
myHashMap.get(3); // return -1 (i.e., not found), The map is now [[1,1], [2,2]]
myHashMap.put(2, 1); // The map is now [[1,1], [2,1]] (i.e., update the existing value)
myHashMap.get(2); // return 1, The map is now [[1,1], [2,1]]
myHashMap.remove(2); // remove the mapping for 2, The map is now [[1,1]]
myHashMap.get(2); // return -1 (i.e., not found), The map is now [[1,1]]

Constraints:

0 <= key, value <= 106
At most 104 calls will be made to put, get, and remove.

Complexity Analysis

Time Complexity: O(1)
Space Complexity: O(n)

706. Design HashMap LeetCode Solution in C++

class MyHashMap {
 public:
  MyHashMap() : lists(kSize) {}

  void put(int key, int value) {
    auto& pairs = lists[key % kSize];
    for (auto& [k, v] : pairs)
      if (k == key) {
        v = value;
        return;
      }
    pairs.emplace_back(key, value);
  }

  int get(int key) {
    const list<pair<int, int>>& pairs = lists[key % kSize];
    for (const auto& [k, v] : pairs)
      if (k == key)
        return v;
    return -1;
  }

  void remove(int key) {
    auto& pairs = lists[key % kSize];
    for (auto it = pairs.begin(); it != pairs.end(); ++it)
      if (it->first == key) {
        pairs.erase(it);
        return;
      }
  }

 private:
  static constexpr int kSize = 10000;
  // Each slot stores the (key, value) list.
  vector<list<pair<int, int>>> lists;
};
/* code provided by PROGIEZ */

706. Design HashMap LeetCode Solution in Java

class MyHashMap {
  public MyHashMap() {
    lists = new List[kSize];
    for (int i = 0; i < kSize; ++i)
      lists[i] = new ArrayList<>();
  }

  public void put(int key, int value) {
    for (int[] pair : lists[key % kSize])
      if (pair[0] == key) {
        pair[1] = value;
        return;
      }
    lists[key % kSize].add(new int[] {key, value});
  }

  public int get(int key) {
    for (int[] pair : lists[key % kSize])
      if (pair[0] == key)
        return pair[1];
    return -1;
  }

  public void remove(int key) {
    for (int i = 0; i < lists[key % kSize].size(); ++i)
      if (lists[key % kSize].get(i)[0] == key) {
        lists[key % kSize].remove(i);
        return;
      }
  }

  private static final int kSize = 10000;
  List<int[]>[] lists; // Each slot stores the (key, value) list.
}
// code provided by PROGIEZ

706. Design HashMap LeetCode Solution in Python

N/A
 # code by PROGIEZ

Additional Resources

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