54. Spiral Matrix 🌐

Difficulty: Medium - Tags: Matrix, Simulation

LeetCode Problem Link

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Problem Statement 📜

Given an m x n matrix, return all elements of the matrix in spiral order.

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Examples 🌟

🔹 Example 1:

Input:

int[][] matrix = {
    {1, 2, 3},
    {4, 5, 6},
    {7, 8, 9}
};

Output:

[1, 2, 3, 6, 9, 8, 7, 4, 5]

🔹 Example 2:

Input:

int[][] matrix = {
    {1, 2, 3, 4},
    {5, 6, 7, 8},
    {9, 10, 11, 12}
};

Output:

[1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]

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Constraints ⚙️

• m == matrix.length

• n == matrix[i].length

• 1 <= m, n <= 10

• -100 <= matrix[i][j] <= 100

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Solution 💡

To traverse the matrix in spiral order, we use boundary pointers for rows and columns and move in the order: left-to-right, top-to-bottom, right-to-left, and bottom-to-top.

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Java Solution

import java.util.ArrayList;
import java.util.List;

public class SpiralMatrix {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> result = new ArrayList<>();

        if (matrix == null || matrix.length == 0) return result;

        int top = 0, bottom = matrix.length - 1;
        int left = 0, right = matrix[0].length - 1;

        while (top <= bottom && left <= right) {
            // Traverse top row
            for (int i = left; i <= right; i++) {
                result.add(matrix[top][i]);
            }
            top++;

            // Traverse right column
            for (int i = top; i <= bottom; i++) {
                result.add(matrix[i][right]);
            }
            right--;

            // Traverse bottom row (if still within bounds)
            if (top <= bottom) {
                for (int i = right; i >= left; i--) {
                    result.add(matrix[bottom][i]);
                }
                bottom--;
            }

            // Traverse left column (if still within bounds)
            if (left <= right) {
                for (int i = bottom; i >= top; i--) {
                    result.add(matrix[i][left]);
                }
                left++;
            }
        }

        return result;
    }
}

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Explanation of the Solution

1. Initialization:

   • Define four boundary variables: top, bottom, left, and right.

2. Traversal:

   • Left to Right: Traverse the current top row and increment top.

   • Top to Bottom: Traverse the current right column and decrement right.

   • Right to Left: Traverse the current bottom row (if valid) and decrement bottom.

   • Bottom to Top: Traverse the current left column (if valid) and increment left.

3. Result:

   • Stop traversal when top > bottom or left > right.

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Time Complexity ⏳

• O(m * n):

  • Each element of the matrix is visited exactly once.

Space Complexity 💾

• O(1):

  • No additional space is used apart from the result list (output space not counted).

You can find the full solution here.