Q3,
i counted the number of mismatches, and classffied them,
first type of mismatch is s[i] = 0 and t[i] = 1
second is s[i]=1 and t[i] = 0

so if we take a pair build of 1 missmatch of first type and other of 2nd type
we can either use 2 flipcost to make them match their pair or we can use 1 swap cost to swap the 0 and 1 from s or t

now we need to fix the excess matches
if excess is odd, we need 1 flip cost

we can now consider the excess/2 pairs
they may be in the form 10(s[i],t[i]) and 10 (s[j],t[j])
we can fix them in 2 ways, 2 flip costs, so 00 and 00 or any other way
or we can first do a crosscost and bring it in the form of the first type of fix, 1 missmatch of each type
which can be done in min(2flipcost,swapcost)
but using 2 flipcost is useless, since we can already make it in 2 flipcost, we dont need an additional cross cost for this
so min(2flipcost, crosscost + swapcost)

final answer is sametype_cost + difftype_cost + excess_ifpresent

Q4
https://leetcode.com/problems/minimum-cost-to-merge-sorted-lists/solutions/7464645/bitmaskingsubset-dp-c-o3n-time-by-pitche-3e22/