# How To Go Around A Square Matrix In Spiral Order

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## Problem statement

Given a positive integer `n`, generate an `n x n` matrix filled with elements from `1` to `n^2` in spiral order.

### Example 1

``````Input: n = 3
Output: [[1,2,3],[8,9,4],[7,6,5]]
``````

### Example 2

``````Input: n = 1
Output: [[1]]
``````

### Constraints

• `1 <= n <= 20`.

## Solution

1. Starting from the top left of the matrix.

2. Going along the spiral direction.

3. Put the value to the matrix, starting from `1`.

### Code

``````#include <vector>
#include <iostream>
using namespace std;
enum Direction {RIGHT, DOWN, LEFT, UP};
vector<vector<int>> generateMatrix(int n) {
vector<vector<int>> m(n, vector<int>(n));
int bottom = n - 1;
int right = n - 1;
int top = 0;
int left = 0;
int row = 0;
int col = 0;
Direction d = RIGHT;
int a = 1;
while (top <= bottom && left <= right) {
m[row][col] = a++;
switch (d) {
case RIGHT: if (col == right) {
top++;
d = DOWN;
row++;
} else {
col++;
}
break;
case DOWN:  if (row == bottom) {
right--;
d = LEFT;
col--;
} else {
row++;
}
break;
case LEFT:  if (col == left) {
bottom--;
d = UP;
row--;
} else {
col--;
}
break;
case UP:    if (row == top) {
left++;
d = RIGHT;
col++;
} else {
row--;
}
break;
}
}
return m;
}
void printResult(vector<vector<int>>& m) {
cout << "[";
for (auto& r : m) {
cout << "[";
for (int a : r) {
cout << a << ",";
}
cout << "]";
}
cout << "]\n";
}
int main() {
auto m = generateMatrix(3);
printResult(m);
m = generateMatrix(1);
printResult(m);
}
``````
``````Output:
[[1,2,3,][8,9,4,][7,6,5,]]
[[1,]]
``````

### Code explanation

1. The code creates a 2D vector `m` of size `n x n` to represent the square matrix. It initializes it with all zeros.

2. It defines variables `bottom`, `right`, `top`, and `left` to keep track of the boundaries of the matrix. Initially, they are set to the indexes of the outermost rows and columns of the matrix.

3. The code initializes `row` and `col` to 0, representing the current position in the matrix.

4. It creates an {index}`enumDirection` to represent the current direction of movement (`RIGHT`, `DOWN`, `LEFT`, `UP`). It initializes `d` to `RIGHT` to start moving to the right.

5. The code initializes an integer `a` to 1, which will be used to fill the matrix with incrementing values.

6. It uses a while loop to traverse the matrix and fills it with values as long as the `top` boundary is less than or equal to the `bottom` boundary and the `left` boundary is less than or equal to the `right` boundary.

7. Inside the loop, it fills the current position `(row, col)` in the matrix with the value `a` and increment `a` by 1.

8. The code uses a {index}`switch` statement based on the current direction `d` to determine the next move:

• If `d` is RIGHT, it checks if `col` has reached the `right` boundary. If it has, it moves to the next direction (`DOWN`) and update `row` accordingly; otherwise, increment `col`.

• Similar logic is applied for the `DOWN`, `LEFT`, and `UP` directions, adjusting `row` and `col` accordingly and changing direction when needed.

9. Once the loop is complete, the matrix is filled in a spiral order, and the function returns the filled matrix.

### Complexity

• Runtime: `O(n^2)`, where `n x n` is the size of the matrix.

• Extra space: `O(1)`.