Walk a Matrix in Spiral Order

medium matrix simulation

Problem

Given an m × n matrix, return its values visited in clockwise spiral order starting from the top-left. Track four shrinking boundaries — top, bottom, left, right — and walk a side at a time: rightward across the top row, downward along the right column, leftward across the bottom row, upward along the left column. After each side, pull that boundary inward; stop the moment any pair crosses.

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

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

Outer ring 1→2→3→6→9→8→7→4, then the inner cell 5.

matrix (rows separated by ';', values by ',')

def spiral_order(matrix):
    out = []
    top, bottom = 0, len(matrix) - 1
    left, right = 0, len(matrix[0]) - 1
    while top <= bottom and left <= right:
        for c in range(left, right + 1):
            out.append(matrix[top][c])
        top += 1
        for r in range(top, bottom + 1):
            out.append(matrix[r][right])
        right -= 1
        if top <= bottom:
            for c in range(right, left - 1, -1):
                out.append(matrix[bottom][c])
            bottom -= 1
        if left <= right:
            for r in range(bottom, top - 1, -1):
                out.append(matrix[r][left])
            left += 1
    return out

function spiralOrder(matrix) {
  const out = [];
  let top = 0, bottom = matrix.length - 1;
  let left = 0, right = matrix[0].length - 1;
  while (top <= bottom && left <= right) {
    for (let c = left; c <= right; c++) out.push(matrix[top][c]);
    top++;
    for (let r = top; r <= bottom; r++) out.push(matrix[r][right]);
    right--;
    if (top <= bottom) {
      for (let c = right; c >= left; c--) out.push(matrix[bottom][c]);
      bottom--;
    }
    if (left <= right) {
      for (let r = bottom; r >= top; r--) out.push(matrix[r][left]);
      left++;
    }
  }
  return out;
}

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> out = new ArrayList<>();
        int top = 0, bottom = matrix.length - 1;
        int left = 0, right = matrix[0].length - 1;
        while (top <= bottom && left <= right) {
            for (int c = left; c <= right; c++) out.add(matrix[top][c]);
            top++;
            for (int r = top; r <= bottom; r++) out.add(matrix[r][right]);
            right--;
            if (top <= bottom) {
                for (int c = right; c >= left; c--) out.add(matrix[bottom][c]);
                bottom--;
            }
            if (left <= right) {
                for (int r = bottom; r >= top; r--) out.add(matrix[r][left]);
                left++;
            }
        }
        return out;
    }
}

vector<int> spiralOrder(vector<vector<int>>& matrix) {
    vector<int> out;
    int top = 0, bottom = (int)matrix.size() - 1;
    int left = 0, right = (int)matrix[0].size() - 1;
    while (top <= bottom && left <= right) {
        for (int c = left; c <= right; ++c) out.push_back(matrix[top][c]);
        ++top;
        for (int r = top; r <= bottom; ++r) out.push_back(matrix[r][right]);
        --right;
        if (top <= bottom) {
            for (int c = right; c >= left; --c) out.push_back(matrix[bottom][c]);
            --bottom;
        }
        if (left <= right) {
            for (int r = bottom; r >= top; --r) out.push_back(matrix[r][left]);
            ++left;
        }
    }
    return out;
}

Time: O(m · n) Space: O(1) extra