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The core insight is modeling it as a graph where each index points to where its value belongs in the sorted array, then counting cycles.
First, clarify that the problem is about finding the minimum swaps to transform the array into any non-decreasing order, not necessarily a stable sort. Then, explain that the optimal strategy involves sorting the array and analyzing cycles in the permutation mapping from original to sorted positions, while handling duplicates by grouping identical values. Finally, describe how to compute the minimum swaps as the sum over each distinct value of (cycle length - 1) for cycles formed by positions of that value.
Pro tip: Mention that duplicates can be handled by treating all occurrences of the same value as interchangeable, so cycles should be formed only among positions of the same value; this avoids overcounting swaps and is a common pitfall. Also, note that the answer is not simply the number of inversions, as swaps can fix multiple inversions at once.
Confirm that the goal is to sort the array in non-decreasing order using the minimum number of swaps of any two elements, and that duplicates are allowed. State that the sorted order is unique up to permutation of equal elements.
Create a sorted copy of the array. For each distinct value, collect the list of indices where it appears in the original array and the list of indices where it should appear in the sorted array. Pair these indices to define a permutation for each value.
For each distinct value, build a graph where each original index points to its target index in the sorted array. Find all cycles in this permutation. The number of swaps needed for that value is the sum of (cycle length - 1) over all cycles.
Add the swap counts from all distinct values to get the total minimum number of swaps. Explain that swaps never need to cross different values because equal elements are interchangeable.
State that the algorithm runs in O(n log n) time due to sorting, and O(n) extra space for the mappings and cycle detection. Mention that this is optimal for comparison-based sorting.
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