to add to my question, suppose there is a 2*2 matrix, when we pop (0,0) the queue will have (1, 0), (-1, 0), (0, 1), (0, -1) and a length of 4

Yes, this is due to the constant factor.

You can assume the space complexity will be something like: O( C * min( M, N) ) where 'C' is a constant.

Since we ignore constants for asymptotic analysis, that's why the space complexity is O( min(M, N)).

to be honest you should write out the entire solution before reducing and simplifying. You do it for the sliding window solution where O(26) is approximated as O(1). Took me a while to write out a 2x2 matrix to understand this.

I still don't understand why if someone could share an explanation / visualization or an intuition I would appreciate

https://imgur.com/gallery/M58OKvB

For those who were initially confused like myself,

min( m, n) denotes the smaller of the two numbers m and n (or their common value if m = n). For example, the rank of a 3 x 5 matrix can be no more than 3, and the rank of a 4 x 2 matrix can be no more than 2

This help...

Thanks Alexander Yang