# LeetCode - Spiral Matrix

## Problem statement

Given an m x n matrix, return all elements of the matrix in spiral order.

Problem statement taken from: https://leetcode.com/problems/spiral-matrix

Example 1:

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

Example 2:

``````Input: matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]
Output: [1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]
``````

Constraints:

``````- m == matrix.length
- n == matrix[i].length
- 1 <= m, n <= 10
- -100 <= matrix[i][j] <= 100
``````

### Explanation

#### Clockwise(Spiral) matrix traversal

As per the above examples, we can see the elements of the outer loop are printed first in a clockwise manner then the elements of the inner loop are printed. So the problem can be solved by dividing the matrix into boundaries. We need to use four loops that print the array element in the spiral clockwise form.

Let's check the algorithm:

``````- set k = 0, l = 0
set m = matrix.size(), n = matrix[0].size()
initialize result array and i

/*
k - starting row index
m - ending row index
l - starting column index
n - ending column index
i - iterator
*/
- loop while k < m && l < n
- loop for i = l; i < n; i++
- result.push(matrix[k][i])
- k++

- loop for i = k; i < m; i++
- result.push(matrix[i][n - 1])
- n--

- if k < m
- loop for i = n - 1; i >= l; i--
- result.push(matrix[m - 1][i])
- m--

- if l < n
- loop for i = m - 1; i >= k; i--
- result.push(matrix[i][l])
- l++

- return result
``````

#### C++ solution

``````class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
int k = 0, l = 0;
int i;
int m = matrix.size();
int n = matrix[0].size();
vector<int> result;

while(k < m && l < n) {
for(i = l; i < n; i++) {
result.push_back(matrix[k][i]);
}
k++;

for(i = k; i < m; i++) {
result.push_back(matrix[i][n - 1]);
}
n--;

if(k < m) {
for(i = n - 1; i >= l; i--) {
result.push_back(matrix[m - 1][i]);
}
m--;
}

if(l < n) {
for(i = m - 1; i >= k; i--) {
result.push_back(matrix[i][l]);
}
l++;
}
}

return result;
}
};
``````

#### Golang solution

``````func spiralOrder(matrix [][]int) []int {
m := len(matrix)
n := len(matrix[0])
result := make([]int, m * n)
counter := 0
k , l := 0, 0
var i int

for k < m && l < n {
for i = l; i < n; i++ {
result[counter] = matrix[k][i]
counter++
}
k++

for i = k; i < m; i++ {
result[counter] = matrix[i][n - 1]
counter++
}
n--

if k < m {
for i = n - 1; i >= l; i-- {
result[counter] = matrix[m - 1][i]
counter++
}
m--
}

if l < n {
for i = m - 1; i >= k; i-- {
result[counter] = matrix[i][l]
counter++
}
l++
}
}

return result
}``````

#### Javascript solution

``````var spiralOrder = function(matrix) {
let m = matrix.length, n = matrix[0].length;
let result = [];
let k = 0, l = 0, i;

while(k < m && l < n) {
for(i = l; i < n; i++) {
result.push(matrix[k][i]);
}
k++;

for(i = k; i < m; i++) {
result.push(matrix[i][n - 1]);
}
n--;

if(k < m) {
for(i = n - 1; i >= l; i--) {
result.push(matrix[m - 1][i]);
}
m--;
}

if(l < n) {
for(i = m - 1; i >= k; i--) {
result.push(matrix[i][l]);
}
l++;
}
}

return result;
};
``````

#### Dry Run

Let's dry-run our algorithm to see how the solution works.

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

Step 1: k = 0, l = 0, i
m = matrix.size()
= 3
n = matrix[0].size()
= 3
initialize vector<int> result

Step 2: loop while k < m && l < n
0 < 3 && 0 < 3
true

loop for i = l; i < n; i++
result.push_back(matrix[k][i])

// the for loop iterates for i = 0 to 2
// we fetch matrix[0][0], matrix[0][1] and matrix[0][2]
result = [1, 2, 3]
k++
k = 1

loop for i = k; i < m; i++
result.push_back(matrix[i][n - 1])

// the for loop iterates for i = 1 to 2
// we fetch matrix[1][2] and matrix[2][2]
result = [1, 2, 3, 6, 9]
n--
n = 2

if k < m
1 < 3
true

loop for i = n - 1; i >= l; i--
result.push_back(matrix[m - 1][i])

// the for loop iterates for i = 2 to 0
// we fetch matrix[2][1] and matrix[2][0]
result = [1, 2, 3, 6, 9, 8, 7]
m--
m = 2

if l < n
0 < 2
true

loop for i = m - 1; i >= k; i--
result.push_back(matrix[i][l])

// the for loop iterates for i = 1 to 1
// we fetch matrix[1][0]
result = [1, 2, 3, 6, 9, 8, 7, 4]
l++
l = 1

Step 3: loop while k < m && l < n
1 < 2 && 1 < 2
true

loop for i = l; i < n; i++
result.push_back(matrix[k][i])

// the for loop iterates for i = 1 to 1
// we fetch matrix[1][1]
result = [1, 2, 3, 6, 9, 8, 7, 4, 5]
k++
k = 2

loop for i = k; i < m; i++
result.push_back(matrix[i][n - 1])

// no iteration as k is 2 and m is 2
// i = k; i = 2 and 2 < 2 false
n--
n = 1

if k < m
2 < 3
true

loop for i = n - 1; i >= l; i--
result.push_back(matrix[m - 1][i])

// no iteration as n is 1 and l is 1
// i = n - 1; i = 0 and 0 >= 1 false
m--
m = 1

if l < n
1 < 1
false

l++
l = 2

Step 4: loop while k < m && l < n
2 < 1 && 2 < 1
false

Step 5: return result

So we return the answer as [1, 2, 3, 6, 9, 8, 7, 4, 5].
``````