saltandwater - 1 month ago 9

Java Question

I read a few articles which said that, in a heap, only the root element can be deleted. However, why can't we delete elements, using the below approach?

- Find the index of the key element you want to delete
- Swap this element with the last element, so the key becomes the last element
- Re-heapify starting at the original index of the key(which you have now swapped with the last element)
- Reduce size of heap by 1

So, the code would look like

`static void delete(int[] a, int key)`

{

int i = 0;

for(i = 0;i<a.length;i++)

{

if(a[i] == key)

break;

}

swap(a, i, a.length-1);

max_heapify_forheapsort(a, i, a.length-2);

}

static void max_heapify_forheapsort(int[] a, int i, int limit)

{

int left = (2*i)+1;

int right = (2*i) + 2;

int next = i;

if(left <= limit && a[left] > a[i])

next = left;

if(right <= limit && a[right] > a[next] )

next = right;

if(next!=i)

{

swap(a, next, i);

max_heapify_forheapsort(a, next, limit);

}

}

Answer

It's certainly possible to delete elements from a heap with sift-up or sift-down operations as you propose. (It's similarly possible to change the key of any item in the heap and use sift-up or -down operations to restore the heap property.)

The tricky bit is what you stated quickly at the outset: "Find the index." A naive implementation requires O(n) to do this, and that kills heap efficiency. What people mean when they say "you can't delete" is that with a naive implementation you can't delete efficiently.

Fortunately it's not hard to make finding the index O(1). Just maintain a "reverse map" in parallel with the heap array, e.g. a hash map from objects to heap array indices. It's easy to update this map without adding asymptotic complexity to any of the heap operations, and with it - as you point out - delete has complexity O(log n). The sift up/down dominates the map lookup to find the index.

If you're interested in a tested implementation, here's one where the heap contains indices into another array of objects. Therefore the reverse map is implemented as an array `q->locs[k]`

, which is the index of the heap element that contains `k`

(which is an index into the object table).

Many problems in computer science can be solved by adding a level of indirection!

**Edit**

Since OP asked about why a sift up might be required to delete, consider the min heap

```
1
20 2
30 40 3 4
```

We want to delete 30, so move 4 (the last array element) into its position:

```
1
20 2
4 40 3
```

Clearly a sift up is required.