MEXOR

Let $P(l,r)$ denote the parity of the number of zeros in $A_l,\ldots,A_r$. Splitting every element into a singleton yields the XOR value $P(1,N)$.

If we replace the singleton segments inside an interval $[l,r]$ with one segment, the XOR changes by $\mathrm{MEX}(l,r)\oplus P(l,r)$. Thus, we may select disjoint intervals and XOR their changes into the baseline.

An interval with MEX $0$ has zero change. If the MEX is $1$, the interval contains no $1$. An odd number of zeros also yields zero change. If it contains an even number of zeros, we may replace it with the interval between a pair of consecutive zero occurrences having no $1$ between them. This interval has change $1$.

Now consider an interval whose MEX is $m\ge2$. Inside it, select a shortest subinterval $[u,v]$ containing all values $0,1,\ldots,m-1$. Each endpoint must occur exactly once inside this subinterval and must have value less than . Call such intervals . Enumerate them by maintaining frequencies and MEX for every left endpoint.

If the removed part contains an even number of zeros, the essential interval produces the same change. Otherwise, extend it to the closest zero immediately to its left or right, choosing a side contained in the original interval. This adds exactly one zero. The MEX remains $m$ because the original interval contains no $m$. Hence, it suffices to consider every essential interval and its one-sided extensions to the nearest zero, provided the MEX does not change. The number of essential intervals, and therefore the number of candidates, is $O(N)$.

Let $dp[i]$ be the set of XOR changes obtainable from disjoint candidate intervals contained in the prefix $[1,i]$. We may skip position $i$, or select any candidate $[l,i]$ and XOR its change with every state of $$.

Let $G$ be the MEX of the entire array. Every segment MEX is at most $G$, so all possible XOR values are below $2^{\lceil\log_2(G+1)\rceil}$. Representing each DP row as a bitset yields time and space. The answer is the maximum of over all .