103 lines
1.8 KiB
Markdown
Executable File
103 lines
1.8 KiB
Markdown
Executable File
# Consider a Min Heap with These Elements: [4, 5, 6, 7, 8, 9, 10]. Insert 3 into This Min Heap.
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In a **min heap**, the root node contains the smallest element, and every parent node is less than its children. When we insert a new element into a min heap, we follow these steps:
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1. **Insert the element at the end** of the heap (maintaining the complete binary tree property).
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2. **Bubble up (or sift up)** the inserted element to restore the heap property by comparing it with its parent and swapping if necessary.
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### Given Min Heap:
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The given min heap is represented as an array:
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```csharp
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[4, 5, 6, 7, 8, 9, 10]
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```
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This corresponds to the following binary tree:
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```
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4
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/ \
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5 6
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/ \ / \
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7 8 9 10
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```
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### **Step 1: Insert 3 into the Heap**
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First, insert 3 at the end of the heap:
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```csharp
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[4, 5, 6, 7, 8, 9, 10, 3]
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```
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This corresponds to the following binary tree:
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```
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4
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/ \
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5 6
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/ \ / \
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7 8 9 10
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/
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3
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```
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### **Step 2: Bubble Up**
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Now, we need to **bubble up** the element `3` to restore the heap property. We compare `3` with its parent and swap if necessary:
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- The parent of `3` is `7` (index 3). Since `3 < 7`, we swap them.
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After the swap, the heap becomes:
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```
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4
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/ \
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5 6
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/ \ / \
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3 8 9 10
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/
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7
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```
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Now, we continue the bubbling up process:
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- The parent of `3` is `5` (index 1). Since `3 < 5`, we swap them.
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After the swap, the heap becomes:
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```
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4
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/ \
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3 6
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/ \ / \
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5 8 9 10
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/
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7
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```
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Finally, we compare `3` with its parent `4` (index 0):
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- Since `3 < 4`, we swap them.
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After the swap, the heap becomes:
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```
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3
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/ \
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4 6
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/ \ / \
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5 8 9 10
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/
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7
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```
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### **Final Array Representation**:
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The final min heap as an array is:
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```csharp
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[3, 4, 6, 5, 8, 9, 10, 7]
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```
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