I actually do it an entirely different way, but you can decide which is faster.

I do sum(L_{a}=L_{b})=dim(L_{a}). Here's the theory:

If the lists are equal, "L_{a}=L_{b}" should return {1,1,1,1…}, the whole list is full of ones. By seeing if the sum of this list is equal to the dimensions, you can correctly affirm whether or not the lists are the same. Say L_{1} is {4,3,1,2} and L_{2} is {4,1,3,3}. Testing L_{1}=L_{2} will return {1,0,0,0}. Find the sum( of {1,0,0,0} which is 1, and check to see if it's the same as the dimension of one of the lists, which is 4. As 1 is not equal to 4, the command returns 0. If the lists were the same, then L_{1}=L,,2, would return {1,1,1,1} and running the sum( of that would return 4, which is equal to the dimension which is still 4 and the command returns 1.

I honestly am not sure which way is faster although I have the feeling that using min( would save more code and probably takes less time. Just another look at the problem. :)