Shortest Path using Meet In The Middle

Difficulty Level : Hard
Last Updated : 10 Dec, 2020

Given a permutation P = p₁, p₂, …., p_n of first n natural numbers (1 ≤ n ≤ 10). One can swap any two consecutive elements p_i and p_{i + 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, visited_start = visited_finish = true.
Let src_u be the root of vertex u in BFS progression. So, src_start = start and src_finish = finish.
Let D_u be the shortest distance from vertex u to it’s tree’s root. So D_start = D_finish = 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 (visited_v = false) into queue’s back, then let D_v = D_u + 1, src_v = src_u and visited_v = true. Especially, if v was visited and src_v != src_u then we can immediately return D_u + D_v + 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

Output:

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