GATE CS 2015 Set 1 — Question 37
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Programming & Data Structures → Heaps → Heap Insert & Extract
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Question
Consider a max heap, represented by the array: 40, 30, 20, 10, 15, 16, 17, 8, 4.
Now consider that a value 35 is inserted into this heap. After insertion, the new heap is
| Array Index | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|
| Value | 40 | 30 | 20 | 10 | 15 | 16 | 17 | 8 | 4 |
Correct answer
(B) 40, 35, 20, 10, 30, 16, 17, 8, 4, 15
Solution
The initial max heap array is:
This corresponds to the following heap structure:
When a new value (35) is inserted into a max heap, it is first added to the end of the array, and then 'heapified up' to maintain the max heap property.
The new element 35 is at index 10.
Array:
The value 35 is now at index 5.
Array:
The value 35 is now at index 2.
[40, 30, 20, 10, 15, 16, 17, 8, 4].This corresponds to the following heap structure:
40
/ \
30 20
/ \ / \
10 15 16 17
/ \
8 4
When a new value (35) is inserted into a max heap, it is first added to the end of the array, and then 'heapified up' to maintain the max heap property.
1.Add 35 to the end:
The array becomes [40, 30, 20, 10, 15, 16, 17, 8, 4, 35].The new element 35 is at index 10.
2.Heapify up (percolate up):
- Compare 35 (index 10) with its parent (index ):
Array:
[40, 30, 20, 10, 35, 16, 17, 8, 4, 15].The value 35 is now at index 5.
- Compare 35 (index 5) with its new parent (index ):
Array:
[40, 35, 20, 10, 30, 16, 17, 8, 4, 15].The value 35 is now at index 2.
- Compare 35 (index 2) with its new parent (index ):
[40, 35, 20, 10, 30, 16, 17, 8, 4, 15].This matches option (B).The final heap structure: 40
/ \
35 20
/ \ / \
10 30 16 17
/ \
8 4
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