# 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 ` `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 > qu; ` `    ``pair 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); ` ` `  `    ``// Calling function ` `    ``cout << minimumPrefixReversals(a, n); ` ` `  `    ``return` `0; ` `} `

Output:

```3
```

## 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 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 q = ``new` `Queue(); ` `         `  `        ``// 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:

```3
```

Time Complexity: O(N! * N2)

My Personal Notes arrow_drop_up The function of education is to teach one to think intensively and to think critically

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