Design HashMap

Problem

Design a class MyHashMap with int keys (0..10000) and int values. Implement put(int, int), get(int) (returns -1 if absent), and remove(int).

Constraints

• 0 ≤ key, value ≤ 10⁶
• at most 10⁴ operations

Approach — Hashing & Counting

This is a Hashing & Counting problem. The idea: store what you've seen in a hash map for O(1) lookups, trading a little space to avoid a nested scan. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(n) time, O(n) space.

Solution code

Python

class MyHashMap:
    def __init__(self):
        self.size = 1000
        self.buckets = [[] for _ in range(self.size)]

    def put(self, key, value):
        b = self.buckets[key % self.size]
        for i, (k, _) in enumerate(b):
            if k == key:
                b[i] = (key, value); return
        b.append((key, value))

    def get(self, key):
        for k, v in self.buckets[key % self.size]:
            if k == key:
                return v
        return -1

    def remove(self, key):
        b = self.buckets[key % self.size]
        for i, (k, _) in enumerate(b):
            if k == key:
                b.pop(i); return

Java

import java.util.*;

class MyHashMap {
    private int[] data;
    public MyHashMap() {
        data = new int[1000001];
        Arrays.fill(data, -1);
    }
    public void put(int key, int value) { data[key] = value; }
    public int get(int key) { return data[key]; }
    public void remove(int key) { data[key] = -1; }
}

class Solution {}

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.