Given a permutation P = p1, p2, …., pn of first n natural numbers (1 ≤ n ≤ 10). One can swap any two consecutive elements pi and pi + 1 (1 ≤ i < n). The task is to find the minimum number of swaps to change P to another permutation P' = p'1, p’2, …., p’n.
Input: P = “213”, P' = “321”
213 <-> 231 <-> 321
Input: P = “1234”, P' = “4123”
Approach: This problem can be solved using Dijkstra’s Shortest Path Algorithm. Seem like there is nothing related to a graph in the statement. But assume one permutation is one vertex, then every swap of a permutation’s elements is an edge which connects this vertex with another vertex. So finding the minimum number of swaps now becomes a simple BFS/shortest path problem.
Now let’s analyze time complexity. We have n! vertices, each vertex has n – 1 adjacent vertices. We also have to store vertices visited state by map because their representations are hard to be stored by normal arrays. So total time complexity is O(N log(N!) * N!). Meet In The Middle technique can be used to make the solution faster.
Meet In The Middle solution is similar to Dijkstra’s solution with some modifications.
- Let P be the start vertex and P' be the finish Vertex.
- Let both start and finish be roots. We start BFS from both the roots, start and finish at the same time but using only one queue.
- Push start and finish into queue’s back, visitedstart = visitedfinish = true.
- Let srcu be the root of vertex u in BFS progression. So, srcstart = start and srcfinish = finish.
- Let Du be the shortest distance from vertex u to it’s tree’s root. So Dstart = Dfinish = 0.
- While queue is not empty, pop queue’s front which is vertex u then push all vertices v which are adjacent with u and haven’t been visited yet (visitedv = false) into queue’s back, then let Dv = Du + 1, srcv = srcu and visitedv = true. Especially, if v was visited and srcv != to srcu then we can immediately return Du + Du + 1.
Below is the implementation of the above approach:
- Meet in the middle
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