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Print the matrix in spiral manner

Problem Statement:​

Given a Matrix, print the given matrix in spiral order.

  • Example:

    Example 1:
    Input: Matrix[][] = { { 1, 2, 3, 4 },
    		      { 5, 6, 7, 8 },
    		      { 9, 10, 11, 12 },
    	              { 13, 14, 15, 16 } }

    Outhput: 1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10.
    Explanation: The output of matrix in spiral form.

    Example 2:
    Input: Matrix[][] = { { 1, 2, 3 },
    	              { 4, 5, 6 },
    		      { 7, 8, 9 } }
    			    
    Output: 1, 2, 3, 6, 9, 8, 7, 4, 5.
    Explanation: The output of matrix in spiral form.

Solution:​

vector<int> spirallyTraverse(vector<vector<int>>& mat) {
    int n = mat.size();
    int m = mat[0].size();
    vector<int> res;

    int top = 0, bottom = n - 1;
    int left = 0, right = m - 1;

    while (top <= bottom && left <= right) {
        // Left to right
        for (int i = left; i <= right; i++) {
            res.push_back(mat[top][i]);
        }
        top++;

        // Top to bottom
        for (int i = top; i <= bottom; i++) {
            res.push_back(mat[i][right]);
        }
        right--;

        // Right to left
        if (top <= bottom) {
            for (int i = right; i >= left; i--) {
                res.push_back(mat[bottom][i]);
            }
            bottom--;
        }

        // Bottom to top
        if (left <= right) {
            for (int i = bottom; i >= top; i--) {
                res.push_back(mat[i][left]);
            }
            left++;
        }
    }

    return res;
}

🧠 How it Works​

  • Use four boundaries: top, bottom, left, right
  • Traverse layer by layer:
    • ➡️ Left to Right on top row
    • ⬇️ Top to Bottom on right column
    • ⬅️ Right to Left on bottom row
    • ⬆️ Bottom to Top on left column
  • Shrink boundaries after each direction

⚠️ Edge Cases​

  • Single row or column → handled via boundary checks
  • Matrix with 1x1 or empty → returns as expected
  • Always check bounds (top <= bottom, left <= right) before inner loops

📉 Time & Space Complexity​

MetricValue
TimeO(m × n) (all elements visited once)
SpaceO(1) extra (excluding result vector)

💡 Other Possible Solutions​

  • No better time complexity — this is optimal
  • Can be done recursively (for fun or interviews), but not recommended for large inputs

🔁 Related Problems​

  • LC 54 – Spiral Matrix (exact match)
  • LC 59 – Spiral Matrix II (construct spiral)
  • Boundary traversal
  • Zigzag traversal

📚 Concepts Used​

  • 2D Boundary Control
  • Layered Traversal
  • Four Directional Pointers
💬

Discussion & Doubts