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+# Consider a min heap with these elements: [4, 5, 6, 7, 8, 9, 10]. Insert 3 into this min heap.
+
+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:
+
+1. **Insert the element at the end** of the heap (maintaining the complete binary tree property).
+2. **Bubble up (or sift up)** the inserted element to restore the heap property by comparing it with its parent and swapping if necessary.
+
+### Given Min Heap:
+
+The given min heap is represented as an array:
+```csharp
+[4, 5, 6, 7, 8, 9, 10]
+```
+This corresponds to the following binary tree:
+```
+        4
+      /   \
+     5     6
+    / \   / \
+   7   8 9  10
+```
+### **Step 1: Insert 3 into the Heap**
+
+First, insert 3 at the end of the heap:
+```csharp
+[4, 5, 6, 7, 8, 9, 10, 3]
+```
+This corresponds to the following binary tree:
+```
+        4
+      /   \
+     5     6
+    / \   / \
+   7   8 9  10
+  /
+ 3
+```
+### **Step 2: Bubble Up**
+
+Now, we need to **bubble up** the element `3` to restore the heap property. We compare `3` with its parent and swap if necessary:
+
+- The parent of `3` is `7` (index 3). Since `3 < 7`, we swap them.
+
+After the swap, the heap becomes:
+```
+        4
+      /   \
+     5     6
+    / \   / \
+   3   8 9  10
+  /
+ 7
+```
+
+Now, we continue the bubbling up process:
+
+- The parent of `3` is `5` (index 1). Since `3 < 5`, we swap them.
+
+After the swap, the heap becomes:
+```
+        4
+      /   \
+     3     6
+    / \   / \
+   5   8 9  10
+  /
+ 7
+```
+
+Finally, we compare `3` with its parent `4` (index 0):
+
+- Since `3 < 4`, we swap them.
+
+After the swap, the heap becomes:
+
+```
+        3
+      /   \
+     4     6
+    / \   / \
+   5   8 9  10
+  /
+ 7
+```
+
+### **Final Array Representation**:
+
+The final min heap as an array is:
+```csharp
+[3, 4, 6, 5, 8, 9, 10, 7]
+```