Easy & Efficient code Largest 1-Bordered Square 
IntuitionThe problem requires finding the largest square with all 1s on its border in a binary grid. To efficiently check for valid squares, we need to precompu
# Intuition
The problem requires finding the largest square with all `1`s on its border in a binary grid. To efficiently check for valid squares, we need to precompute how many `1`s extend **rightward** and **downward** from each cell.

Instead of checking every possible square from scratch (which would be inefficient), we leverage **precomputed prefix values** to verify the existence of valid bordered squares in **constant time**.

---

# Approach
1. **Precompute Right and Down Matrices**  
   - `right[i][j]`: Number of consecutive `1`s to the right of `grid[i][j]`.
   - `down[i][j]`: Number of consecutive `1`s downward from `grid[i][j]`.

2. **Iterate Through Each Cell**  
   - Determine the maximum possible side length (`maxside`) for a square starting at `(i, j)`.
   - Try to form a square of side `maxside` by checking:
     - If the **bottom row** has at least `side` consecutive `1`s (`right[i + side - 1][j]`).
     - If the **rightmost column** has at least `side` consecutive `1`s (`down[i][j + side - 1]`).
   - **Update `maxsize`** accordingly.

3. **Return `maxsize * maxsize`** as the area of the largest valid square.

---

# Complexity
- **Precomputing `right` and `down` arrays:**  
  - Each cell is processed once: $$O(m 	imes n)$$  
- **Checking for the largest square:**  
  - In the worst case, we check each cell and iterate over possible side lengths: $$O(m 	imes n)$$  
- **Overall Complexity:** $$O(m 	imes n)$$  
  - Efficient for large grids.

- **Space Complexity:** $$O(m 	imes n)$$  
  - We store two additional matrices (`right` and `down`), making it $$O(m 	imes n)$$ extra space.

---

# Code
```cpp
class Solution {
public:
    int largest1BorderedSquare(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> right(n, vector<int>(m, 0));
        vector<vector<int>> down(n, vector<int>(m, 0));

        // Precompute right and down arrays
        for (int i = n - 1; i >= 0; i--) {
            for (int j = m - 1; j >= 0; j--) {
                if (grid[i][j] == 1) {
                    right[i][j] = (j == m - 1) ? 1 : 1 + right[i][j + 1];
                    down[i][j] = (i == n - 1) ? 1 : 1 + down[i + 1][j];
                }
            }
        }

        int maxsize = 0;

        // Iterate to find the largest square
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int maxside = min(right[i][j], down[i][j]);

                for (int side = maxside; side > maxsize; side--) { // Early stopping
                    if (right[i + side - 1][j] >= side && down[i][j + side - 1] >= side) {
                        maxsize = side;
                        break;  // No need to check smaller sizes
                    }
                }
            }
        }
        return maxsize * maxsize;
    }
};
```

---

# Key Optimizations
✅ **Handles Rectangular Grids** (supports `m × n`, not just `n × n`)  
✅ **Early Stopping** (Avoids unnecessary iterations by breaking early when a valid square is found)  
✅ **Optimized Side-Length Search** (Only checks sizes greater than the current `maxsize`)  

This solution efficiently finds the largest 1-bordered square in any binary grid. 🚀

['C++']