11.5: Fixing MyHashMap

The problem with MyHashMap is in size, which is inherited from MyBetterMap:

public int size() {
    int total = 0;
    for (MyLinearMap<K, V> map: maps) {
        total += map.size();
    }
    return total;
}

To add up the total size it has to iterate the sub-maps. Since we increase the number of sub-maps, k, as the number of entries, n, increases, k is proportional to n, so size is linear.

And that makes put linear, too, because it uses size:

public V put(K key, V value) {
    V oldValue = super.put(key, value);

    if (size() > maps.size() * FACTOR) {
        rehash();
    }
    return oldValue;
}

Everything we did to make put constant time is wasted if size is linear!

Fortunately, there is a simple solution, and we have seen it before: we have to keep the number of entries in an instance variable and update it whenever we call a method that changes it.

You’ll find my solution in the repository for this book, in MyFixedHashMap.java. Here’s the beginning of the class definition:

public class MyFixedHashMap<K, V> extends MyHashMap<K, V> implements Map<K, V> {

    private int size = 0;

    public void clear() {
        super.clear();
        size = 0;
    }

Rather than modify MyHashMap, I define a new class that extends it. It adds a new instance variable, size, which is initialized to zero.

Updating clear is straightforward; we invoke clear in the superclass (which clears the sub-maps), and then update size.

Updating remove and put is a little more difficult because when we invoke the method on the superclass, we can’t tell whether the size of the sub-map changed. Here’s how I worked around that:

public V remove(Object key) {
    MyLinearMap<K, V> map = chooseMap(key);
    size -= map.size();
    V oldValue = map.remove(key);
    size += map.size();
    return oldValue;
}

remove uses chooseMap to find the right sub-map, then subtracts away the size of the sub-map. It invokes remove on the sub-map, which may or may not change the size of the sub-map, depending on whether it finds the key. But either way, we add the new size of the sub-map back to size, so the final value of size is correct.

The rewritten version of put is similar:

public V put(K key, V value) {
    MyLinearMap<K, V> map = chooseMap(key);
    size -= map.size();
    V oldValue = map.put(key, value);
    size += map.size();

    if (size() > maps.size() * FACTOR) {
        size = 0;
        rehash();
    }
    return oldValue;
}

We have the same problem here: when we invoke put on the sub-map, we don’t know whether it added a new entry. So we use the same solution, subtracting off the old size and then adding in the new size.

Now the implementation of the size method is simple:

public int size() {
    return size;
}

And that’s pretty clearly constant time.

When I profiled this solution, I found that the total time for putting n keys is proportional to n, which means that each put is constant time, as it’s supposed to be.

---