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.
Examples:
Input: P = “213”, P’ = “321”
Output: 2
213 <-> 231 <-> 321
Input: P = “1234”, P’ = “4123”
Output: 3
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 != srcu then we can immediately return Du + Dv + 1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to find minimum number of // swaps to make another permutation int ShortestPath( int n, string start, string finish) { unordered_map<string, bool > visited; unordered_map<string, int > D; unordered_map<string, string> src; visited[start] = visited[finish] = true ; D[start] = D[finish] = 0; src[start] = start; src[finish] = finish; queue<string> q; q.push(start); q.push(finish); while (!q.empty()) { // Take top vertex of the queue string u = q.front(); q.pop(); // Generate n - 1 of it's permutations for ( int i = 1; i < n; i++) { // Generate next permutation string v = u; swap(v[i], v[i - 1]); // If v is not visited if (!visited[v]) { // Set it visited visited[v] = true ; // Make root of u and root of v equal src[v] = src[u]; // Increment it's distance by 1 D[v] = D[u] + 1; // Push this vertex into queue q.push(v); } // If it is already visited // and roots are different // then answer is found else if (src[u] != src[v]) return D[u] + D[v] + 1; } } } // Driver code int main() { string p1 = "1234" , p2 = "4123" ; int n = p1.length(); cout << ShortestPath(n, p1, p2); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG{ // Function to find minimum number of // swaps to make another permutation static int ShortestPath( int n, String start, String finish) { HashMap<String, Boolean> visited = new HashMap<String, Boolean>(); HashMap<String, Integer> D = new HashMap<String, Integer>(); HashMap<String, String> src = new HashMap<String, String>(); visited.put(start, true ); visited.put(finish, true ); D.put(start, 0 ); D.put(finish, 0 ); src.put(start, start); src.put(finish, finish); Queue<String> q = new LinkedList<>(); q.add(start); q.add(finish); while (q.size() != 0 ) { // Take top vertex of the queue String u = (String)q.peek(); q.remove(); // Generate n - 1 of it's permutations for ( int i = 1 ; i < n; i++) { // Generate next permutation StringBuilder tmp = new StringBuilder(u); char t = tmp.charAt(i); tmp.setCharAt(i, tmp.charAt(i - 1 )); tmp.setCharAt(i - 1 , t); String v = tmp.toString(); // If v is not visited if (!visited.getOrDefault(v, false )) { // Set it visited visited.put(v, true ); // Make root of u and root of v equal src.put(v, src.get(u)); // Increment it's distance by 1 D.put(v, D.get(u) + 1 ); // Push this vertex into queue q.add(v); } // If it is already visited // and roots are different // then answer is found else if (src.get(u) != src.get(v)) return D.get(u) + D.get(v) + 1 ; } } return 0 ; } // Driver Code public static void main(String[] args) { String p1 = "1234" , p2 = "4123" ; int n = p1.length(); System.out.println(ShortestPath(n, p1, p2)); } } // This code is contributed by pratham76 |
Python3
# Python3 implementation of the approach from collections import deque, defaultdict # Function to find minimum number of # swaps to make another permutation def shortestPath(n: int , start: str , finish: str ) - > int : visited, D, src = defaultdict( lambda : False ), defaultdict( lambda : 0 ), defaultdict( lambda : '') visited[start] = visited[finish] = True D[start] = D[finish] = 0 src[start], src[finish] = start, finish q = deque() q.append(start) q.append(finish) while q: # Take top vertex of the queue u = q[ 0 ] q.popleft() # Generate n - 1 of it's permutations for i in range (n): v = list (u) v[i], v[i - 1 ] = v[i - 1 ], v[i] v = ''.join(v) if not visited[v]: # Set it visited visited[v] = True # Make root of u and root of v equal src[v] = src[u] # Increment it's distance by 1 D[v] = D[u] + 1 # Push this vertex into queue q.append(v) # If it is already visited # and roots are different # then answer is found elif u in src and src[u] ! = src[v]: return D[u] + D[v] + 1 # Driver Code if __name__ = = "__main__" : p1 = "1234" p2 = "4123" n = len (p1) print (shortestPath(n, p1, p2)) # This code is contributed by # sanjeev2552 |
C#
// C# implementation of the approach using System; using System.Collections; using System.Text; using System.Collections.Generic; class GFG{ // Function to find minimum number of // swaps to make another permutation static int ShortestPath( int n, string start, string finish) { Dictionary< string , bool > visited = new Dictionary< string , bool >(); Dictionary< string , int > D = new Dictionary< string , int >(); Dictionary< string , string > src = new Dictionary< string , string >(); visited[start] = true ; visited[finish] = true ; D[start] = 0; D[finish] = 0; src[start] = start; src[finish] = finish; Queue q = new Queue(); q.Enqueue(start); q.Enqueue(finish); while (q.Count != 0) { // Take top vertex of the queue string u = ( string )q.Peek(); q.Dequeue(); // Generate n - 1 of it's permutations for ( int i = 1; i < n; i++) { // Generate next permutation StringBuilder tmp = new StringBuilder(u); char t = tmp[i]; tmp[i] = tmp[i - 1]; tmp[i - 1] = t; string v = tmp.ToString(); // If v is not visited if (!visited.GetValueOrDefault(v, false )) { // Set it visited visited[v] = true ; // Make root of u and root of v equal src[v] = src[u]; // Increment it's distance by 1 D[v] = D[u] + 1; // Push this vertex into queue q.Enqueue(v); } // If it is already visited // and roots are different // then answer is found else if (src[u] != src[v]) return D[u] + D[v] + 1; } } return 0; } // Driver Code public static void Main( string [] args) { string p1 = "1234" , p2 = "4123" ; int n = p1.Length; Console.Write(ShortestPath(n, p1, p2)); } } // This code is contributed by rutvik_56 |
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