Spirally traversing a matrix || geeksforgeeks || problem of the day || matrix || javascript.

This solution implements the spiral traversal of a matrix using the boundary traversal technique. The method spirallyTraverse accepts a 2D matrix mat and returns an array of integers representing the spiral order of its elements. Four boundary pointers—left, right, up, and down—are initialized to the edges of the matrix. A loop iterates as long as the left pointer is less than or equal to right, and the up pointer is less than or equal to down.

During each iteration, the following steps are performed:

Traverse the up row from left to right and increment up.
Traverse the right column from up to down and decrement right.
If valid, traverse the down row from right to left and decrement down.
If valid, traverse the left column from down to up and increment left.
This process continues until all matrix elements are traversed, and the result is stored in the res array. The algorithm ensures a time complexity of
𝑂(𝑚×𝑛) , where 𝑚 and 𝑛 are the dimensions of the matrix, and uses 𝑂(1) auxiliary space, making it efficient.

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