Minimum number of prefix reversals to sort permutation of first N numbers
Last Updated :
08 Dec, 2021
Given N numbers that 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}
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();
// perform 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;
}
// 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 generate 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# 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 generate 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
<script>
class Node
{
constructor(string,steps)
{
this.string = string;
this.steps= steps;
}
}
function minimumPrefixReversals(a)
{
// size of array
let n = a.length;
// string for initial and goal nodes
let start = "", destination = "";
// string for manipulation in while loop
let original = "",modified = "";
// node to store temporary values from front of queue
let temp = null;
// create the starting string
for (let i = 0; i < n; i++)
start += a[i];
// sort the array and prepare final destination string
a.sort(function(a,b){return a-b;});
for (let i = 0; i < n; i++)
destination += a[i];
// this queue will store all the BFS siblings
let q = [];
// place the starting node in queue
q.push(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.length != 0)
{
// put front node of queue in temporary variable
temp = q.shift();
// store the original string at this step
original = temp.string;
for (let j = 2; j <= n; j++)
{
// modified will be used to generate 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 == (destination))
{
// if string match then return
// the height of the current node
return temp.steps + 1;
}
// else put this node into queue
q.push(new Node(modified,temp.steps + 1));
}
}
// if no case match then default value
return Number.MIN_VALUE;
}
function reverse (s,index)
{
let temp = s.split("");
let i = 0;
while (i < index)
{
let c = temp[i];
temp[i] = temp[index-1];
temp[index-1] = c;
i++;index--;
}
return (temp).join("");
}
let a = [1, 2, 4, 3];
document.write(minimumPrefixReversals(a));
// This code is contributed by rag2127
</script>
# Python3 program to find
# minimum number of prefix reversals
# to sort permutation of [0] N numbers
from queue import Queue
# Function to return the minimum prefix reversals
def minimumPrefixReversals( a, n):
start = ""
destination = ""
for i in range(n):
# converts the number to a character
# and add to
start += str(a[i])
a.sort()
for i in range(n):
destination += str(a[i])
# Queue to store the pairs
# of and number of reversals
qu=Queue()
# Initially push the original
qu.put((start, 0))
# if original is the destination
if (start == destination) :
return 0
# iterate till queue is empty
while qu.not_empty :
# pair at the top
p = qu.get()
#
t = p[0]
# perform prefix reversals for all index and push
# in the queue and check for the minimal
for j in range(2,n+1) :
r = t
# reverse the till prefix j
tmpR=list(r)
tmpR[:j]=tmpR[j-1::-1]
r=''.join(tmpR)
# if after reversing the from [0] i index
# it is the destination
if (r == destination) :
return p[1] + 1
# push the number of reversals for r
qu.put((r, p[1] + 1))
# Driver Code
if __name__ == '__main__':
a = [ 1, 2, 4, 3]
n = len(a)
# Calling function
print(minimumPrefixReversals(a, n))
Time Complexity: O(N! * N2)
Auxiliary Space: O(N!)
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