Print a given matrix in spiral form

Last Updated : 21 Dec, 2024

Given a matrix mat[][] of size m x n, the task is to print all elements of the matrix in spiral form.

Examples:

Input: mat[][] = [[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]]
Output: [ 1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10 ]
Example of matrix in spiral form
Input: mat[][]= [[1, 2, 3, 4, 5, 6],
[7, 8, 9, 10, 11, 12],
[13, 14, 15, 16, 17, 18]]
Output: [ 1, 2, 3, 4, 5, 6, 12, 18, 17, 16, 15, 14, 13, 7, 8, 9, 10, 11 ]

Table of Content

[Naive Approach] Using Simulation by Visited Matrix - O(mn) Time and O(mn) Space

The idea is to simulate the spiral traversal by marking the cells that have already been visited. We can use a direction array to keep track of the movement (right, down, left, up) and change direction when we hit the boundary or a visited cell.
Initialize a 2D array vis[][] to keep track of visited cells.
Use direction arrays dr and dc to represent right, down, left, and up movements.
Start from the top-left cell and follow the direction arrays to visit each cell.
Change direction when encountering a boundary or a visited cell.
Continue until all cells are visited.

C++

//C++ program to perform spiral order 
// traversal of a matrix
#include <iostream>
#include <vector>
using namespace std;

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

    // Arrays to represent the changes in row and column indices: 
  	// move right(0, +1), move down(+1, 0), 
  	// move left(0, -1), move  up(-1, 0)
  
    // Change in row index for each direction
    vector<int> dr = { 0, 1, 0, -1 };

    // Change in column index for each direction
    vector<int> dc = { 1, 0, -1, 0 };

    // Initial position in the matrix
    int r = 0, c = 0;

    // Initial direction index (0 corresponds to 'right')
    int idx = 0;

    for (int i = 0; i < m * n; ++i) {
        res.push_back(mat[r][c]);
        vis[r][c] = true;

        // Calculate the next cell coordinates based on
        // current direction
        int newR = r + dr[idx];
        int newC = c + dc[idx];

        // Check if the next cell is within bounds and not
        // visited
        if (0 <= newR && newR < m && 0 <= newC && newC < n
            && !vis[newR][newC]) {

            // Move to the next row
            r = newR;

            // Move to the next column
            c = newC;
        }
        else {

            // Change direction (turn clockwise)
            idx = (idx + 1) % 4;

            // Move to the next row according to new
            // direction
            r += dr[idx];

            // Move to the next column according to new
            // direction
            c += dc[idx];
        }
    }
  
    return res;
}


int main() {
    vector<vector<int>> mat = { { 1, 2, 3, 4 },
                                    { 5, 6, 7, 8 },
                                    { 9, 10, 11, 12 },
                                    { 13, 14, 15, 16 } };

    vector<int> res = spirallyTraverse(mat);

    for (int num : res) {
        cout << num << " ";
    }
    return 0;
}

Java

// Java program to perform spiral order 
// traversal of a matrix
import java.util.ArrayList;
import java.util.List;

class GfG {
	static ArrayList<Integer> spirallyTraverse(int[][] mat) {
        int m = mat.length;
        int n = mat[0].length;

        ArrayList<Integer> res = new ArrayList<>();
        boolean[][] vis = new boolean[m][n];

        // Change in row index for each direction
        int[] dr = { 0, 1, 0, -1 };

        // Change in column index for each direction
        int[] dc = { 1, 0, -1, 0 };

        // Initial position in the matrix
        int r = 0, c = 0;

        // Initial direction index (0 corresponds to 'right')
        int idx = 0;

        for (int i = 0; i < m * n; ++i) {

            // Add current element to result list
            res.add(mat[r][c]);

            // Mark current cell as visited
            vis[r][c] = true;

            // Calculate the next cell coordinates based on
            // current direction
            int newR = r + dr[idx];
            int newC = c + dc[idx];

            // Check if the next cell is within bounds and not
            // visited
            if (0 <= newR && newR < m && 0 <= newC && newC < n
                    && !vis[newR][newC]) {

                // Move to the next row
                r = newR;

                // Move to the next column
                c = newC;
            } else {

                // Change direction (turn clockwise)
                idx = (idx + 1) % 4;

                // Move to the next row according to new
                // direction
                r += dr[idx];

                // Move to the next column according to new
                // direction
                c += dc[idx];
            }
        }

        return res;
    }
  
    public static void main(String[] args) {
        int[][] mat = { { 1, 2, 3, 4 },
                        { 5, 6, 7, 8 },
                        { 9, 10, 11, 12 },
                        { 13, 14, 15, 16 } };
      
        ArrayList<Integer> res = spirallyTraverse(mat);

        for (int num : res) {
            System.out.print(num + " ");
        }
    }
}

Python

# Python program to perform spiral order
# traversal of a matrix

def spirallyTraverse(mat):
    m = len(mat)
    n = len(mat[0])

    res = []
    vis = [[False] * n for _ in range(m)]

    # Change in row index for each direction
    dr = [0, 1, 0, -1]

    # Change in column index for each direction
    dc = [1, 0, -1, 0]

    # Initial position in the matrix
    r, c = 0, 0

    # Initial direction index (0 corresponds to 'right')
    idx = 0

    for _ in range(m * n):
        res.append(mat[r][c])
        vis[r][c] = True

        # Calculate the next cell coordinates based on
        # current direction
        newR, newC = r + dr[idx], c + dc[idx]

        # Check if the next cell is within bounds and not
        # visited
        if 0 <= newR < m and 0 <= newC < n and not vis[newR][newC]:

            # Move to the next row
            r, c = newR, newC
        else:

            # Change direction (turn clockwise)
            idx = (idx + 1) % 4

            # Move to the next row according to new
            # direction
            r += dr[idx]

            # Move to the next column according to new
            # direction
            c += dc[idx]

    return res


if __name__ == "__main__":
    mat = [[1, 2, 3, 4],
           [5, 6, 7, 8],
           [9, 10, 11, 12],
           [13, 14, 15, 16]]

    res = spirallyTraverse(mat)

    print(" ".join(map(str, res)))

// C# program to perform spiral order traversal of a matrix
using System;
using System.Collections.Generic;

class GfG {
    static List<int> spirallyTraverse(int[][] mat) {
        int m = mat.Length;
        int n = mat[0].Length;

        List<int> res = new List<int>();
        bool[, ] vis = new bool[m, n];

        // Arrays to represent the changes in row and column
        // indices: turn right(0, +1), turn down(+1, 0),
        // turn left(0, -1), turn up(-1, 0)
        int[] dr = { 0, 1, 0, -1 };
        int[] dc = { 1, 0, -1, 0 };

        // Initial position in the matrix
        int r = 0, c = 0;

        // Initial direction index (0 corresponds to
        // 'right')
        int idx = 0;

        for (int i = 0; i < m * n; ++i) {
            res.Add(mat[r][c]);
            vis[r, c] = true;

            // Calculate the next cell coordinates based on
            // current direction
            int newR = r + dr[idx];
            int newC = c + dc[idx];

            // Check if the next cell is within bounds and
            // not visited
            if (newR >= 0 && newR < m && newC >= 0
                && newC < n && !vis[newR, newC]) {
                r = newR;
                c = newC;
            }
            else {
              
                // Change direction (turn clockwise)
                idx = (idx + 1) % 4;
                r += dr[idx];
                c += dc[idx];
            }
        }

        return res;
    }

    static void Main() {
        int[][] mat = { new int[] { 1, 2, 3, 4 },
                        new int[] { 5, 6, 7, 8 },
                        new int[] { 9, 10, 11, 12 },
                        new int[] { 13, 14, 15, 16 } };

        List<int> res = spirallyTraverse(mat);

        foreach(int num in res)
            Console.Write(num + " ");
    }
}

JavaScript

// JavaScript program to perform spiral order traversal of a
// matrix
function spirallyTraverse(mat) {
    const m = mat.length;
    const n = mat[0].length;
    
    const res = [];
    const vis = Array.from({length : m},
                           () => Array(n).fill(false));

    // Arrays to represent the changes in row and column
    // indices: turn right(0, +1), turn down(+1, 0), turn
    // left(0, -1), turn up(-1, 0)
    const dr = [ 0, 1, 0, -1 ];
    const dc = [ 1, 0, -1, 0 ];

    // Initial position in the matrix
    let r = 0, c = 0;

    // Initial direction index (0 corresponds to 'right')
    let idx = 0;

    for (let i = 0; i < m * n; ++i) {
    
        // Add current element to result array
        res.push(mat[r][c]);

        // Mark current cell as visited
        vis[r][c] = true;

        // Calculate the next cell coordinates based on
        // current direction
        const newR = r + dr[idx];
        const newC = c + dc[idx];

        // Check if the next cell is within bounds and not
        // visited
        if (newR >= 0 && newR < m && newC >= 0 && newC < n
            && !vis[newR][newC]) {
            r = newR;
            c = newC;
        }
        else {
        
            // Change direction (turn clockwise)
            idx = (idx + 1) % 4;
            r += dr[idx];
            c += dc[idx];
        }
    }

    return res;
}

// Driver Code
const mat = [
    [ 1, 2, 3, 4 ], 
    [ 5, 6, 7, 8 ], 
    [ 9, 10, 11, 12 ],
    [ 13, 14, 15, 16 ]
];

const res = spirallyTraverse(mat);
console.log(res.join(" "));

Output

1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10

Time Complexity: O(m*n), where m and n are the number of rows and columns of the given matrix respectively.
Auxiliary Space: O(m*n), for the visited matrix and the result vector.

[Expected Approach] Using Boundary Traversal - O(m*n) Time and O(1) Space

We can print the matrix in a spiral order by dividing it into loops or boundaries. We print the elements of the outer boundary first, then move inward to print the elements of the inner boundaries.
Define the boundaries of the matrix with variables top, bottom, left, and right.
Print the top row from left to right and increment top.
Print the right column from top to bottom and decrement right.
Check if boundaries have crossed; if not, print the bottom row from right to left and decrement bottom.
Print the left column from bottom to top and increment left.
Repeat until all boundaries are crossed.

Illustration:

C++

// C++ program to perform spiral order
// traversal of a matrix
#include <iostream>
#include <vector>
using namespace std;

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

    vector<int> res;

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

    // Iterate until all elements are printed
    while (top <= bottom && left <= right) {

        // Print top row from left to right
        for (int i = left; i <= right; ++i) {
            res.push_back(mat[top][i]);
        }
        top++;

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

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

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

    return res;
}

int main() {

    vector<vector<int>> mat = {{1, 2, 3, 4}, 
                               {5, 6, 7, 8}, 
                               {9, 10, 11, 12}, 
                               {13, 14, 15, 16}};

    vector<int> res = spirallyTraverse(mat);
    for (int ele : res)
        cout << ele << " ";
    return 0;
}

Java

// Java program to perform spiral order
// traversal of a matrix
import java.util.ArrayList;
import java.util.List;

class GfG {
    static List<Integer> spirallyTraverse(int[][] mat) {
        int m = mat.length;
        int n = mat[0].length;

        ArrayList<Integer> res = new ArrayList<>();

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

        // Iterate until all elements are printed
        while (top <= bottom && left <= right) {

            // Print top row from left to right
            for (int i = left; i <= right; ++i) {
                res.add(mat[top][i]);
            }
            top++;

            // Print right column from top to bottom
            for (int i = top; i <= bottom; ++i) {
                res.add(mat[i][right]);
            }
            right--;

            // Print bottom row from right to left (if exists)
            if (top <= bottom) {
                for (int i = right; i >= left; --i) {
                    res.add(mat[bottom][i]);
                }
                bottom--;
            }

            // Print left column from bottom to top (if exists)
            if (left <= right) {
                for (int i = bottom; i >= top; --i) {
                    res.add(mat[i][left]);
                }
                left++;
            }
        }

        return res;
    }

    public static void main(String[] args) {
        int[][] mat = {{1, 2, 3, 4}, 
                       {5, 6, 7, 8}, 
                       {9, 10, 11, 12}, 
                       {13, 14, 15, 16}};

        List<Integer> res = spirallyTraverse(mat);
        for (int ele : res) {
            System.out.print(ele + " ");
        }
    }
}

Python

# Python program to perform spiral order
# traversal of a matrix

def spirallyTraverse(mat):
    m, n = len(mat), len(mat[0])

    res = []

    # Initialize boundaries
    top, bottom, left, right = 0, m - 1, 0, n - 1

    # Iterate until all elements are printed
    while top <= bottom and left <= right:

        # Print top row from left to right
        for i in range(left, right + 1):
            res.append(mat[top][i])
        top += 1

        # Print right column from top to bottom
        for i in range(top, bottom + 1):
            res.append(mat[i][right])
        right -= 1

        # Print bottom row from right to left (if exists)
        if top <= bottom:
            for i in range(right, left - 1, -1):
                res.append(mat[bottom][i])
            bottom -= 1

        # Print left column from bottom to top (if exists)
        if left <= right:
            for i in range(bottom, top - 1, -1):
                res.append(mat[i][left])
            left += 1

    return res


if __name__ == "__main__":
    mat = [[1, 2, 3, 4], 
           [5, 6, 7, 8], 
           [9, 10, 11, 12], 
           [13, 14, 15, 16]]

    res = spirallyTraverse(mat)
    print(" ".join(map(str, res)))

// C# program to perform spiral order
// traversal of a matrix
using System;
using System.Collections.Generic;

class GfG {
    static List<int> spirallyTraverse(int[,] mat) {
        int m = mat.GetLength(0);
        int n = mat.GetLength(1);

        List<int> res = new List<int>();

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

        // Iterate until all elements are printed
        while (top <= bottom && left <= right) {

            // Print top row from left to right
            for (int i = left; i <= right; ++i) {
                res.Add(mat[top, i]);
            }
            top++;

            // Print right column from top to bottom
            for (int i = top; i <= bottom; ++i) {
                res.Add(mat[i, right]);
            }
            right--;

            // Print bottom row from right to left (if exists)
            if (top <= bottom) {
                for (int i = right; i >= left; --i) {
                    res.Add(mat[bottom, i]);
                }
                bottom--;
            }

            // Print left column from bottom to top (if exists)
            if (left <= right) {
                for (int i = bottom; i >= top; --i) {
                    res.Add(mat[i, left]);
                }
                left++;
            }
        }

        return res;
    }

    static void Main(string[] args) {
        int[,] mat = {{1, 2, 3, 4}, 
                      {5, 6, 7, 8}, 
                      {9, 10, 11, 12}, 
                      {13, 14, 15, 16}};

        List<int> res = spirallyTraverse(mat);
        Console.WriteLine(string.Join(" ", res));
    }
}

JavaScript

// JavaScript program to perform spiral order
// traversal of a matrix

function spirallyTraverse(mat) {
    const m = mat.length;
    const n = mat[0].length;

    const res = [];

    // Initialize boundaries
    let top = 0, bottom = m - 1, left = 0, right = n - 1;

    // Iterate until all elements are printed
    while (top <= bottom && left <= right) {

        // Print top row from left to right
        for (let i = left; i <= right; ++i) {
            res.push(mat[top][i]);
        }
        top++;

        // Print right column from top to bottom
        for (let i = top; i <= bottom; ++i) {
            res.push(mat[i][right]);
        }
        right--;

        // Print bottom row from right to left (if exists)
        if (top <= bottom) {
            for (let i = right; i >= left; --i) {
                res.push(mat[bottom][i]);
            }
            bottom--;
        }

        // Print left column from bottom to top (if exists)
        if (left <= right) {
            for (let i = bottom; i >= top; --i) {
                res.push(mat[i][left]);
            }
            left++;
        }
    }

    return res;
}

// Driver Code
const mat = [[1, 2, 3, 4], 
             [5, 6, 7, 8], 
             [9, 10, 11, 12], 
             [13, 14, 15, 16]];

const res = spirallyTraverse(mat);
console.log(res.join(" "));

Output

1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10

Time Complexity: O(m*n), where m and n are the number of rows and columns of the given matrix respectively.
Auxiliary Space: O(1).

Find distinct elements common to all rows of a matrix

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Practice Tags :

Print a given matrix in spiral form

[Naive Approach] Using Simulation by Visited Matrix - O(m*n) Time and O(m*n) Space

[Expected Approach] Using Boundary Traversal - O(m*n) Time and O(1) Space

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[Naive Approach] Using Simulation by Visited Matrix - O(mn) Time and O(mn) Space