Minimum number of prefix reversals to sort permutation of first N numbers

Given N numbers which have a permutation of first N numbers. In a single operation, any prefix can be reversed. The task is to find the minimum number of such operations such that the numbers in the array are in increasing order.

Examples:

Input : a[] = {3, 1, 2} 
Output : 2 
Step1: Reverse the complete array a, a[] = {2, 1, 3} 
Step2: Reverse the prefix(0-1) in s, a[] = {1, 2, 3} 

Input : a[] = {1, 2, 4, 3} 
Output : 3 
Step1: Reverse the complete array a, a[] = {3, 4, 2, 1} 
Step2: Reverse the prefix(0-1) in s, a[] = {4, 3, 2, 1} 
Step3: Reverse the complete array a, a[] = {1, 2, 3, 4}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The approach to solve this problem is to use BFS.

Encode the given numbers in a string. Sort the array and encode it into a string destination.
Then do a BFS from the initial permutation. Each time, check all permutations induced by reversing a prefix of current permutation.
If it is not visited, put it into the queue with the count of reversals done.
When the permutation of the encoded string is same as the destination string, return the numbers of reversals required to reach here.
That is, all permutations of strings are done and the minimal of those is returned as the answer.

Below is the implementation of the above approach:

C++

// C++ program to find 
// minimum number of prefix reversals 
// to sort permutation of first N numbers 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return the minimum prefix reversals 
int minimumPrefixReversals(int a[], int n) 
{ 
    string start = ""; 
    string destination = "", t, r; 
    for (int i = 0; i < n; i++) { 
        // converts the number to a character 
        // and add  to string 
        start += to_string(a[i]); 
    } 
    sort(a, a + n); 
    for (int i = 0; i < n; i++) { 
        destination += to_string(a[i]); 
    } 
  
    // Queue to store the pairs 
    // of string and number of reversals 
    queue<pair<string, int> > qu; 
    pair<string, int> p; 
  
    // Initially push the original string 
    qu.push(make_pair(start, 0)); 
  
    // if original string is the destination string 
    if (start == destination) { 
        return 0; 
    } 
  
    // iterate till queue is empty 
    while (!qu.empty()) { 
  
        // pair at the top 
        p = qu.front(); 
  
        // string 
        t = p.first; 
  
        // pop the top-most element 
        qu.pop(); 
  
        // peform prefix reversals for all index and push 
        // in the queue and check for the minimal 
        for (int j = 2; j <= n; j++) { 
            r = t; 
  
            // reverse the string till prefix j 
            reverse(r.begin(), r.begin() + j); 
  
            // if after reversing the string from first i index 
            // it is the destination 
            if (r == destination) { 
                return p.second + 1; 
            } 
  
            // push the number of reversals for string r 
            qu.push(make_pair(r, p.second + 1)); 
        } 
    } 
} 
  
// Driver Code 
int main() 
{ 
  
    int a[] = { 1, 2, 4, 3 }; 
    int n = sizeof(a) / sizeof(a[0]); 
  
    // Calling function 
    cout << minimumPrefixReversals(a, n); 
  
    return 0; 
} 

Output:

Java

// Java program to find minimum  
// number of prefix reversals to 
// sort permutation of first N numbers 
import java.util.*; 
  
public class Main 
{ 
      
    // function to find minimum prefix reversal through BFS 
    public static int minimumPrefixReversals(int[] a) 
    { 
        // size of array  
        int n = a.length; 
          
        // string for initial and goal nodes 
        String start = "", destination = ""; 
          
        // string for manipulation in while loop 
        String original = "",modified = ""; 
          
        // node to store temporary values from front of queue 
        Node temp = null; 
          
        // create the starting string  
        for (int i = 0; i < n; i++) 
        start += a[i]; 
          
        // sort the array and prepare final destination string 
        Arrays.sort(a); 
        for (int i = 0; i < n; i++) 
            destination += a[i]; 
          
        // this queue will store all the BFS siblings 
        Queue<Node> q = new LinkedList<>(); 
          
        // place the starting node in queue 
        q.add(new Node(start, 0)); 
          
        //base case:- if array is already sorted 
        if (start == destination) 
            return 0; 
          
          
        // loop until the size of queue is empty  
        while (q.size() != 0) 
        { 
            // put front node of queue in temporary variable 
            temp = q.poll(); 
              
            // store the original string at this step 
            original = temp.string; 
              
            for (int j = 2; j <= n; j++) 
            { 
                // modified will be used to genrate all  
                // manipulation of original string 
                // like if original = 1342 
                // modified = 3142 , 4312 , 2431 
                  
                modified = original; 
                  
                // generate the permutation by reversing  
                modified = reverse (modified , j); 
                  
                if (modified.equals(destination)) 
                { 
                    // if string match then return  
                    // the height of the current node 
                    return temp.steps + 1; 
                } 
                  
                // else put this node into queue 
                q.add(new Node(modified,temp.steps + 1)); 
            } 
              
        } 
          
    // if no case match then default value 
    return Integer.MIN_VALUE; 
    } 
      
    // function to reverse the string upto an index  
    public static String reverse (String s , int index) 
    { 
        char temp []= s.toCharArray(); 
        int i = 0; 
        while (i < index) 
        { 
            char c = temp[i]; 
            temp[i] = temp[index-1]; 
            temp[index-1] = c; 
            i++;index--; 
        } 
        return String.valueOf(temp); 
    } 
  
    // Driver code 
    public static void main(String []args) 
    { 
        int a[] = new int []{1,2,4,3}; 
        System.out.println(minimumPrefixReversals(a)); 
    } 
      
    // Node class to store a combined set of values 
    static class Node 
    { 
        public String string ; 
        public int steps; 
          
        public Node(String string,int steps) 
        { 
            this.string = string; 
            this.steps= steps; 
        } 
    } 
} 
  
// This code is contributed by Sparsh Singhal 

C#

// C# program to find minimum  
// number of prefix reversals to 
// sort permutation of first N numbers 
using System; 
using System.Collections.Generic;              
  
class GFG 
{ 
    // Node class to store a combined set of values 
    public class Node 
    { 
        public String str; 
        public int steps; 
          
        public Node(String str,int steps) 
        { 
            this.str = str; 
            this.steps= steps; 
        } 
    } 
      
    // function to find minimum prefix reversal through BFS 
    public static int minimumPrefixReversals(int[] a) 
    { 
        // size of array  
        int n = a.Length; 
          
        // string for initial and goal nodes 
        String start = "", destination = ""; 
          
        // string for manipulation in while loop 
        String original = "", modified = ""; 
          
        // node to store temporary values  
        // from front of queue 
        Node temp = null; 
          
        // create the starting string  
        for (int i = 0; i < n; i++) 
        start += a[i]; 
          
        // sort the array and prepare  
        // final destination string 
        Array.Sort(a); 
        for (int i = 0; i < n; i++) 
            destination += a[i]; 
          
        // this queue will store all the BFS siblings 
        Queue<Node> q = new Queue<Node>(); 
          
        // place the starting node in queue 
        q.Enqueue(new Node(start, 0)); 
          
        //base case:- if array is already sorted 
        if (start == destination) 
            return 0; 
          
          
        // loop until the size of queue is empty  
        while (q.Count != 0) 
        { 
            // put front node of queue in temporary variable 
            temp = q.Dequeue(); 
              
            // store the original string at this step 
            original = temp.str; 
              
            for (int j = 2; j <= n; j++) 
            { 
                // modified will be used to genrate all  
                // manipulation of original string 
                // like if original = 1342 
                // modified = 3142 , 4312 , 2431 
                  
                modified = original; 
                  
                // generate the permutation by reversing  
                modified = reverse (modified , j); 
                  
                if (modified.Equals(destination)) 
                { 
                    // if string match then return  
                    // the height of the current node 
                    return temp.steps + 1; 
                } 
                  
                // else put this node into queue 
                q.Enqueue(new Node(modified, temp.steps + 1)); 
            } 
        } 
          
        // if no case match then default value 
        return int.MinValue; 
    } 
      
    // function to reverse the string upto an index  
    public static String reverse (String s, int index) 
    { 
        char []temp = s.ToCharArray(); 
        int i = 0; 
        while (i < index) 
        { 
            char c = temp[i]; 
            temp[i] = temp[index - 1]; 
            temp[index - 1] = c; 
            i++;index--; 
        } 
        return String.Join("", temp); 
    } 
  
    // Driver code 
    public static void Main(String []args) 
    { 
        int []a = new int []{1, 2, 4, 3}; 
        Console.WriteLine(minimumPrefixReversals(a)); 
    } 
} 
  
// This code is contributed by 29AjayKumar 

Output:

Time Complexity: O(N! * N²)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

C#

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