Spiral Matrix - Leetcode 54 - Arrays & Strings (Python)

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[@GregHogg]

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[@ngneerin]

Simpler solution
l, r, t, b, res = 0, len(matrix[0]), 0, len(matrix), []
while l < r and t < b:
for j in range(l, r):
res.append(matrix[t][j])
t += 1
for i in range(t, b):
res.append(matrix[i][r - 1])
r -= 1
if not (l < r and t < b): break
for j in range(r - 1, l - 1, -1):
res.append(matrix[b - 1][j])
b -= 1
for i in range(b - 1, t - 1, -1):
res.append(matrix[i][l])
l += 1
return res

[@carnagen]

Another solution would be to cut off the first row of the matrix and then rotate it to the left. Repeat until only one element remains. It's not faster, however.

[@chandlerkenworthy3185]

Here is my solution to this problem. It utilises the fact you can just take the outward (perimeter) spiral each time and then select the inner (m-2) x (n-2) elements each time (the inner central matrix) until you run out of elements.
m, n = len(matrix), len(matrix[0])
answer = []
def getPerimeter(mat):
if len(mat) == 1:
return mat[0]
if len(mat[0]) == 1:
return [i[0] for i in mat]
return mat[0] + [i[-1] for i in mat[1:-1]] + mat[-1][::-1] + [i[0] for i in mat[1:-1][::-1]]
while len(answer) != (m * n):
answer += getPerimeter(matrix)
matrix = [row[1:-1] for row in matrix[1:-1]]
return answer

[@JoeTan-nq4fq]

My solution - Add the popped elements (from matrix) to result array over one loop, Recur until matrix is empty
Step 1 - Pop the first row and add to the result array
Within the try / catch block
Step 2 - Pop the elements from rightmost column and add to the result array
Step 3 - Pop the last row in reverse and add to the result array
Step 4 - Pop the elements from leftmost column and add to the result array
Step 5 - Recur until matix is empty - return result + spiralOrder(matrix)
Catch IndexError (which means matrix is empty) and return result

[@vasu_1014]

Hlo greg what are your thoughts on my code using java
static List spiralOrder(int[][] m) {
List list = new ArrayList();
int size = 0;
if(m.length %2 == 0)
size = m.length/2;
else
size = m.length/2 + 1;
for(int i=0; i i)
j--;
else
k--;
}
}
return list;
}

[@floccinau263]

can you solve this as well?
981. Time Based Key-Value Store
Please🙏🏻