There are a number of clues which may provide intuition for the full solution, including:

• When comparing integers, higher bit positions are more significant than lower bit positions.
• If some element has a bit set at the highest bit position, then every element to its right must also have this highest bit set.

This motivates the following observation:

If the bit positions are labelled from ~0~ to ~B~, where ~0~ represents the lowest bit position and ~B~ represents the highest bit position, then an array is sorted if and only if:

• There is a (potentially empty) prefix of the elements which do not have the bit at position ~B~ set, and the remaining (potentially empty) suffix has the bit at position ~B~ set, and
• Both the prefix and suffix are sorted themselves, when only considering the bit positions up to ~B-1~.

We can now formulate a dynamic programming recurrence.

Define ~dp[L][R][B]~ as the minimum number of moves to sort the subarray indexed from ~L~ to ~R~ (inclusive), when we only consider the bits up to position ~B~. If ~ones[L][R][B]~ is the number of elements in the subarray ~[L, R]~ with the bit at position ~B~ set, and ~zeroes[L][R][B]~ is the number of elements in the subarray ~[L, R]~ with the bit at position ~B~ unset, then

$$\displaystyle dp[L][R][B] = \min_{i=L-1}^R (dp[L][i][B-1] + dp[i+1][R][B-1] + ones[L][i][B] + zeroes[i+1][R][B])$$

Time complexity: ~\mathcal{O}(N^3 \log(\max(a_i)))~