Merge pull request #1507 from VaishnaviMankala19/lcsol-3028

added 3028 lc solution

Author: ajay-dhangar

dsa-solutions/lc-solutions/3000-3099/3028-Ant-on-the-Boundary.md

@@ -0,0 +1,138 @@
+---
+id: ant-on-the-boundary
+title: Ant on the Boundary (LeetCode)
+sidebar_label: 3028-AntOnTheBoundary
+tags:
+  - Array
+  - Simulation
+  - Boundary Conditions
+description: Determine how many times an ant returns to the boundary after reading an array of non-zero integers and moving according to their values.
+sidebar_position: 3028
+---
+
+## Problem Description
+
+| Problem Statement | Solution Link | LeetCode Profile |
+| :---------------- | :------------ | :--------------- |
+| [Ant on the Boundary](https://leetcode.com/problems/ant-on-the-boundary/) | [Ant on the Boundary Solution on LeetCode](https://leetcode.com/problems/ant-on-the-boundary/solutions/) | [vaishu_1904](https://leetcode.com/u/vaishu_1904/) |
+
+## Problem Description
+
+An ant is on a boundary. It sometimes goes left and sometimes right.
+
+You are given an array of non-zero integers `nums`. The ant starts reading `nums` from the first element of it to its end. At each step, it moves according to the value of the current element:
+
+- If `nums[i] < 0`, it moves left by `-nums[i]` units.
+- If `nums[i] > 0`, it moves right by `nums[i]` units.
+
+Return the number of times the ant returns to the boundary.
+
+### Notes:
+- There is infinite space on both sides of the boundary.
+- We check whether the ant is on the boundary only after it has moved `|nums[i]|` units. In other words, if the ant crosses the boundary during its movement, it does not count.
+
+### Example 1
+- **Input:** `nums = [2,3,-5]`
+- **Output:** `1`
+- **Explanation:** 
+  - After the first step, the ant is 2 steps to the right of the boundary.
+  - After the second step, the ant is 5 steps to the right of the boundary.
+  - After the third step, the ant is on the boundary.
+  - So the answer is 1.
+
+### Example 2
+- **Input:** `nums = [3,2,-3,-4]`
+- **Output:** `0`
+- **Explanation:** 
+  - After the first step, the ant is 3 steps to the right of the boundary.
+  - After the second step, the ant is 5 steps to the right of the boundary.
+  - After the third step, the ant is 2 steps to the right of the boundary.
+  - After the fourth step, the ant is 2 steps to the left of the boundary.
+  - The ant never returned to the boundary, so the answer is 0.
+
+### Constraints
+- `1 <= nums.length <= 100`
+- `-10 <= nums[i] <= 10`
+- `nums[i] != 0`
+
+## Approach
+
+To determine how many times the ant returns to the boundary, we can use a simple simulation approach. Here's the approach:
+
+1. Initialize a variable `position` to keep track of the ant's current position relative to the boundary.
+2. Initialize a counter `boundary_count` to keep track of the number of times the ant returns to the boundary.
+3. Iterate through each element in `nums`:
+   - Update the `position` based on the value of the current element.
+   - Check if the `position` is zero (i.e., the ant is back at the boundary).
+   - If the ant is back at the boundary, increment the `boundary_count`.
+4. Return the `boundary_count`.
+
+### Solution Code
+
+#### Python
+
+```python
+class Solution:
+    def returnToBoundaryCount(self, nums: List[int]) -> int:
+        position = 0
+        boundary_count = 0
+        
+        for num in nums:
+            position += num
+            if position == 0:
+                boundary_count += 1
+        
+        return boundary_count
+```
+
+#### C++
+```c++
+#include <vector>
+using namespace std;
+
+class Solution {
+public:
+    int returnToBoundaryCount(vector<int>& nums) {
+        int position = 0;
+        int boundary_count = 0;
+        
+        for (int num : nums) {
+            position += num;
+            if (position == 0) {
+                boundary_count += 1;
+            }
+        }
+        
+        return boundary_count;
+    }
+};
+
+```
+
+#### Java
+```java
+class Solution {
+    public int returnToBoundaryCount(int[] nums) {
+        int position = 0;
+        int boundary_count = 0;
+        
+        for (int num : nums) {
+            position += num;
+            if (position == 0) {
+                boundary_count += 1;
+            }
+        }
+        
+        return boundary_count;
+    }
+}
+
+```
+
+### Conclusion
+The problem of determining how many times an ant returns to the boundary can be effectively solved 
+using a straightforward simulation approach. By keeping track of the ant's position as it moves 
+according to the values in the array, and counting each time the position returns to zero, we can 
+determine the desired result. The provided solutions in Python, C++, and Java demonstrate this 
+approach, ensuring the solution is efficient and easy to understand across different programming 
+languages.