Assignment - State Design

$dp[\text{mask}]$ = minimum cost when the first popcount(mask) elements of nums1 have been assigned to the nums2 elements indicated by mask.

Base case: $dp[0] = 0$. No assignments made, zero cost. Transition: Let $k = \text{popcount}(\text{mask})$. For each unset bit $j$, try assigning nums1[$k$] to nums2[$j$]. $dp[\text{mask} | (1 << j)] = \min(dp[\text{mask} | (1 << j)], dp[\text{mask}] + \text{nums1}[k] \oplus \text{nums2}[j])$