Because the optimal substructure does not hold.

Consider $$$A=[100_2,111_2,111_2,111_2,10_2,100_2,1_2,10_2]$$$.

The correct answer is $$$110_2(1->5->6->8)$$$,but your method will give $$$111_2$$$($$$dp[5]=11_2$$$).

greedy approach
think about filling up the answer in bitwise

First of all, $$$answer$$$ will be atleast $$$A[1] | A[N]$$$. I define a boolean vector $$$killed$$$ (initialised to all $$$false$$$), where $$$killed[i] = true$$$ would represent that I am not allowed to use $$$i_{th}$$$ index in my path.

Now, start from MSB($$$i = 19$$$ for this problem), if $$$i_{th}$$$ bit is already set in $$$answer$$$, then just continue.

Else see if it is possible to make this bit as $$$0$$$ in our $$$answer$$$ : Iterate over all indices ignoring the ones which are already killed, set $$$killed[j] = true$$$ for all such $$$j$$$ such that $$$i_{th}$$$ bit is 1 in $$$A[j]$$$ and then, if it is possible to reach index $$$N$$$ from index $$$1$$$, then we can afford to lose those indices and can have $$$i_{th}$$$ bit in our $$$answer$$$ as 0!

Checking for $$$i_{th}$$$ bit will take $$$O(N)$$$ time, hence total time complexity would be $$$O(N*log(MAXA_i))$$$.

You can check out my submission Link.