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# 54. Spiral Matrix

## Intitution

We need to simulate how a spiral moves through the matrix — from the top-left corner, moving **right → down → left → up**, and repeat while shrinking the boundaries.

To keep track of what’s been visited, we can shrink the four boundaries (`top`, `bottom`, `left`, `right`) after every full layer is traversed. This approach avoids using extra space for visited cells.

## Complexity

| Space Complexity | Time Complexity  |
| ---------------- | ---------------- |
| $$\text{O}(1)$$  | $$\text{O}(nm)$$ |

## Code

```java
public List<Integer> spiralOrder(int[][] matrix) {
    int totalElements = matrix.length * matrix[0].length;
    List<Integer> result = new ArrayList<>();

    int leftBound = 0, rightBound = matrix[0].length - 1;
    int topBound = 0, bottomBound = matrix.length - 1;

    while (totalElements > 0) {
        // Traverse from left to right
        for (int col = leftBound; col <= rightBound && totalElements > 0; col++) {
            result.add(matrix[topBound][col]);
            totalElements--;
        }
        topBound++;

        // Traverse from top to bottom
        for (int row = topBound; row <= bottomBound && totalElements > 0; row++) {
            result.add(matrix[row][rightBound]);
            totalElements--;
        }
        rightBound--;

        // Traverse from right to left
        for (int col = rightBound; col >= leftBound && totalElements > 0; col--) {
            result.add(matrix[bottomBound][col]);
            totalElements--;
        }
        bottomBound--;

        // Traverse from bottom to top
        for (int row = bottomBound; row >= topBound && totalElements > 0; row--) {
            result.add(matrix[row][leftBound]);
            totalElements--;
        }
        leftBound++;
    }

    return result;
}

```