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Solution 1: Orbital Navigation (Spiral Matrix)

Approach Explanation

We maintain four boundaries: top, bottom, left, and right. While the boundaries haven't crossed each other, we traverse the matrix in a clockwise spiral.

Step-by-Step Logic

  1. top = 0, bottom = m - 1, left = 0, right = n - 1.
  2. While top <= bottom and left <= right:
    • Traverse from left to right along top. Increment top.
    • Traverse from top to bottom along right. Decrement right.
    • If top <= bottom:
      • Traverse from right to left along bottom. Decrement bottom.
    • If left <= right:
      • Traverse from bottom to top along left. Increment left.
  3. Return the result list.

Complexity

  • Time Complexity: O(M * N).
  • Space Complexity: O(1) (ignoring output array).

Code

def spiral_order(matrix):
    if not matrix: return []
    
    m, n = len(matrix), len(matrix[0])
    top, bottom, left, right = 0, m - 1, 0, n - 1
    res = []
    
    while len(res) < m * n:
        # 1. Left to Right
        for j in range(left, right + 1):
            res.append(matrix[top][j])
        top += 1
        
        # 2. Top to Bottom
        if len(res) < m * n:
            for i in range(top, bottom + 1):
                res.append(matrix[i][right])
            right -= 1
            
        # 3. Right to Left
        if len(res) < m * n:
            for j in range(right, left - 1, -1):
                res.append(matrix[bottom][j])
            bottom -= 1
            
        # 4. Bottom to Top
        if len(res) < m * n:
            for i in range(bottom, top - 1, -1):
                res.append(matrix[i][left])
            left += 1
            
    return res