My first instinct was to think about it like a sorting algorithm and count swaps, which was wrong.
The minimum number of moves to sort an array by moving elements is equal to the length of the array minus the length of the longest non-decreasing subsequence (LNDS). This is because elements in the LNDS are already in correct relative order and can remain in place, while all other elements must be moved. So, compute the LNDS using dynamic programming or patience sorting, then subtract its length from n.
Pro tip: Clarify that 'move' means removing an element and inserting it anywhere else; if the operation were a swap, the answer would differ. Also, mention that duplicates are allowed and the LNDS handles them correctly by using non-decreasing order.
Confirm that a move operation takes one element and inserts it at any other position, shifting other elements. The goal is to minimize the number of such moves to achieve a sorted (non-decreasing) array.
Elements that are already in correct relative order (i.e., form a non-decreasing subsequence) do not need to be moved. The largest such subsequence can stay, minimizing moves.
The minimum number of moves equals n minus the length of the longest non-decreasing subsequence (LNDS). This is because all elements not in the LNDS must be moved, and we can always insert them in the correct positions.
Use dynamic programming O(n^2) for simplicity, or patience sorting with binary search O(n log n) for optimal performance. For the example [5,2,2,4,3], the LNDS length is 3 (e.g., 2,2,4 or 2,2,3), so moves = 5 - 3 = 2.
Show that moving 5 to the end and 3 before 4 (or similar) sorts the array in 2 moves. Confirm that 1 move is impossible because no single move can sort it.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.