Minimum number of prefix reversals to sort permutation of first N numbers
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}
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.
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 reversalsint 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 Codeint main(){ int a[] = { 1, 2, 4, 3 }; int n = sizeof(a) / sizeof(a[0]); // Calling function cout << minimumPrefixReversals(a, n); return 0;} |
Java
// Java program to find minimum// number of prefix reversals to// sort permutation of first N numbersimport 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#
// C# program to find minimum// number of prefix reversals to// sort permutation of first N numbersusing 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 |
Javascript
<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> |
Output:
3
Time Complexity: O(N! * N2)
Auxiliary Space: O(N!)
