Sort permutation of N natural numbers using triple cyclic right swaps
Given an array arr[] of size N which contains the permutations of the N natural numbers, the task is to sort the permutations of N natural numbers with the help of triple cyclic right swaps.
Triple Cyclic Right Swaps: refers to the triple cyclic right shift in which –
arr[i] -> arr[j] -> arr[k] -> arr[i]
Examples:
Input: arr[] = {3, 2, 4, 1}
Output: 1
1 3 4
Explanation:
In the operation 1 the index 1, 3 and 4 are chosen and they are cyclic shifted –
arr[1] = arr[4] = 1
arr[3] = arr[1] = 3
arr[4] = arr[3] = 4
Therefore, final array will be {1, 2, 3, 4}.Input: arr[] = {2, 3, 1}
Output: 1
1 2 3
Approach: The idea is to traverse the array and find the elements of the array which is not in its actual sorted position which can be checked by if . Because there are only N natural elements in the array. Finally, find the odd length cyclic rotations required in the array to get the sorted form of the array. If there is any even length cyclic rotations required then it is not possible to sort the elements of the array.
Below is the implementation of the above approach:
C++
// C++ implementation to find the // number of operations required to // sort the elements of the array #include <bits/stdc++.h> using namespace std; #define ll long long // Function to sort the permutation // with the given operations void sortPermutation(ll arr[], ll n) { vector<pair<ll, pair<ll, ll> > > ans; vector<ll> p; // Visited array to check the // array element is at correct // position or not bool visited[200005] = { 0 }; // Loop to iterate over the elements // of the given array for (ll i = 1; i <= n; i++) { // Condition to check if the // elements is at its correct // position if (arr[i] == i) { visited[i] = 1; continue ; } else { // Condition to check if the // element is included in any // previous cyclic rotations if (!visited[i]) { ll x = i; vector<ll> v; // Loop to find the cyclic // rotations in required while (!visited[x]) { visited[x] = 1; v.push_back(x); x = arr[x]; } // Condition to check if the // cyclic rotation is a // valid rotation if ((v.size() - 3) % 2 == 0) { for (ll i = 1; i < v.size(); i += 2) { ans .push_back( make_pair( v[0], make_pair( v[i], v[i + 1]))); } continue ; } p.push_back(v[0]); p.push_back(v[v.size() - 1]); // Loop to find the index of the // cyclic rotation // for the current index for (ll i = 1; i < v.size() - 1; i += 2) { ans .push_back( make_pair( v[0], make_pair( v[i], v[i + 1]))); } } } } // Condition to if the cyclic // rotation is a valid rotation if (p.size() % 4) { cout << -1 << "\n" ; return ; } // Loop to find all the valid operations // required to sort the permutation for (ll i = 0; i < p.size(); i += 4) { ans.push_back( make_pair(p[i], make_pair(p[i + 1], p[i + 2]))); ans.push_back( make_pair(p[i + 2], make_pair(p[i], p[i + 3]))); } // Total operation required cout << ans.size() << "\n" ; for (ll i = 0; i < ans.size(); i++) { cout << ans[i].first << " " << ans[i].second.first << " " << ans[i].second.second << "\n" ; } } // Driver Code int main() { ll arr[] = { 0, 3, 2, 4, 1 }; ll n = 4; // Function Call sortPermutation(arr, n); return 0; } |
Java
/*package whatever //do not write package name here */ import java.util.*; class GFG { static class Pair{ long first; PairL second; Pair( long f,PairL s){ first = f; second = s; } } static class PairL{ long first; long second; PairL( long f, long s){ first = f; second = s; } } // Function to sort the permutation // with the given operations static void sortPermutation( long arr[], long n) { List<Pair> ans = new ArrayList<>(); List<Long> p = new ArrayList<>(); // Visited array to check the // array element is at correct // position or not boolean visited[] = new boolean [ 200005 ]; // Loop to iterate over the elements // of the given array for ( int i = 1 ; i <= n; i++) { // Condition to check if the // elements is at its correct // position if (arr[i] == i) { visited[i] = true ; continue ; } else { // Condition to check if the // element is included in any // previous cyclic rotations if (!visited[i]) { long x = i; List<Long> v = new ArrayList<>(); // Loop to find the cyclic // rotations in required while (!visited[( int )x]) { visited[( int )x] = true ; v.add(x); x = arr[( int )x]; } // Condition to check if the // cyclic rotation is a // valid rotation if ((v.size() - 3 ) % 2 == 0 ) { for ( int j = 1 ; j < v.size();j += 2 ) { ans.add( new Pair(v.get( 0 ), new PairL(v.get(j), v.get(j+ 1 )))); } continue ; } p.add(v.get( 0 )); p.add(v.get(v.size() - 1 )); // Loop to find the index of the // cyclic rotation // for the current index for ( int j = 1 ; j < v.size() - 1 ;j += 2 ) { ans.add( new Pair(v.get( 0 ), new PairL(v.get(j), v.get(j+ 1 )))); } } } } // Condition to if the cyclic // rotation is a valid rotation if (p.size() % 4 == 1 ) { System.out.println( "-1" ); return ; } // Loop to find all the valid operations // required to sort the permutation for ( int i = 0 ; i < p.size(); i += 4 ) { ans.add( new Pair(p.get(i), new PairL(p.get(i + 1 ), p.get(i + 2 )))); ans.add( new Pair(p.get(i+ 2 ), new PairL(p.get(i), p.get(i+ 3 )))); } // Total operation required System.out.println(ans.size()); for ( int i = 0 ; i < ans.size(); i++) { System.out.println(ans.get(i).first + " " + ans.get(i).second.first + " " + ans.get(i).second.second); } } public static void main (String[] args) { long arr[] = { 0 , 3 , 2 , 4 , 1 }; long n = 4 ; // Function Call sortPermutation(arr, n); } } // This code is contributed by aadityaburujwale. |
Python3
# Python3 implementation to find the # number of operations required to # sort the elements of the array # Function to sort the permutation # with the given operations def sortPermutation(arr, n): ans = [] p = [] # Visited array to check the # array element is at correct # position or not visited = [ 0 ] * 200005 # Loop to iterate over the elements # of the given array for i in range ( 1 , n + 1 ): # Condition to check if the # elements is at its correct # position if (arr[i] = = i): visited[i] = 1 continue else : # Condition to check if the # element is included in any # previous cyclic rotations if (visited[i] = = False ): x = i v = [] # Loop to find the cyclic # rotations in required while (visited[x] = = False ): visited[x] = 1 v.append(x) x = arr[x] # Condition to check if the # cyclic rotation is a # valid rotation if (( len (v) - 3 ) % 2 = = 0 ): for i in range ( 1 , len (v), 2 ): ans.append([v[ 0 ], v[i], v[i + 1 ]]) continue p.append(v[ 0 ]) p.append(v[ len (v) - 1 ]) # Loop to find the index of the # cyclic rotation # for the current index for i in range ( 1 , len (v) - 1 , 2 ): ans.append([v[ 0 ], v[i], v[i + 1 ]]) # Condition to if the cyclic # rotation is a valid rotation if ( len (p) % 4 ): print ( - 1 ) return # Loop to find the valid operations # required to sort the permutation for i in range ( 0 , len (p), 4 ): ans.append([p[i], p[i + 1 ], p[i + 2 ]]) ans.append(p[i [ + 2 ], p[i], p[i + 3 ]]) # Total operation required print ( len (ans)) for i in ans: print (i[ 0 ], i[ 1 ], i[ 2 ]) # Driver Code if __name__ = = '__main__' : arr = [ 0 , 3 , 2 , 4 , 1 ] n = 4 # Function Call sortPermutation(arr, n) # This code is contributed by Mohit Kumar |
C#
// package whatever //do not write package name here using System; using System.Collections.Generic; public class GFG { public class PairL { public long first; public long second; public PairL( long f, long s) { first = f; second = s; } } public class Pair { public long first; public PairL second; public Pair( long f, PairL s) { first = f; second = s; } } // Function to sort the permutation // with the given operations static void sortPermutation( long [] arr, long n) { List<Pair> ans = new List<Pair>(); List< long > p = new List< long >(); // Visited array to check the // array element is at correct // position or not bool [] visited = new bool [200005]; // Loop to iterate over the elements // of the given array for ( int i = 1; i <= n; i++) { // Condition to check if the // elements is at its correct // position if (arr[i] == i) { visited[i] = true ; continue ; } else { // Condition to check if the // element is included in any // previous cyclic rotations if (!visited[i]) { long x = i; List< long > v = new List< long >(); // Loop to find the cyclic // rotations in required while (!visited[( int )x]) { visited[( int )x] = true ; v.Add(x); x = arr[( int )x]; } // Condition to check if the // cyclic rotation is a // valid rotation if ((v.Count - 3) % 2 == 0) { for ( int j = 1; j < v.Count; j += 2) { ans.Add( new Pair( v[0], new PairL(v[j], v[j + 1]))); } continue ; } p.Add(v[0]); p.Add(v[v.Count - 1]); // Loop to find the index of the // cyclic rotation // for the current index for ( int j = 1; j < v.Count - 1; j += 2) { ans.Add( new Pair( v[0], new PairL(v[j], v[j + 1]))); } } } } // Condition to if the cyclic // rotation is a valid rotation if (p.Count % 4 == 1) { Console.WriteLine( "-1" ); return ; } // Loop to find all the valid operations // required to sort the permutation for ( int i = 0; i < p.Count; i += 4) { ans.Add( new Pair( p[i], new PairL(p[i + 1], p[i + 2]))); ans.Add( new Pair(p[i + 2], new PairL(p[i], p[i + 3]))); } // Total operation required Console.WriteLine(ans.Count); for ( int i = 0; i < ans.Count; i++) { Console.WriteLine(ans[i].first + " " + ans[i].second.first + " " + ans[i].second.second); } } static public void Main() { long [] arr = { 0, 3, 2, 4, 1 }; long n = 4; // Function Call sortPermutation(arr, n); } } // contributed by akashish__ |
Javascript
// JavaScript implementation to find the // number of operations required to // sort the elements of the array const sortPermutation = (arr, n) => { let ans = []; let p = []; // Visited array to check the // array element is at correct // position or not let visited = new Array(200005).fill(0); // Loop to iterate over the elements // of the given array for (let i = 1; i <= n; i++) { // Condition to check if the // elements is at its correct // position if (arr[i] === i) { visited[i] = 1; continue ; } else { // Condition to check if the // element is included in any // previous cyclic rotations if (!visited[i]) { let x = i; let v = []; // Loop to find the cyclic // rotations in required while (!visited[x]) { visited[x] = 1; v.push(x); x = arr[x]; } // Condition to check if the // cyclic rotation is a // valid rotation if ((v.length - 3) % 2 === 0) { for (let i = 1; i < v.length; i += 2) { ans .push( [v[0], [v[i], v[i + 1]]]); } continue ; } p.push(v[0]); p.push(v[v.length - 1]); // Loop to find the index of the // cyclic rotation // for the current index for (let i = 1; i < v.length - 1; i += 2) { ans .push( [v[0], [v[i], v[i + 1]]]); } } } } // Condition to if the cyclic // rotation is a valid rotation if (p.length % 4) { console.log(-1); return ; } // Loop to find all the valid operations // required to sort the permutation for (let i = 0; i < p.length; i += 4) { ans.push( [p[i], [p[i + 1], p[i + 2]]]); ans.push( [p[i + 2], [p[i], p[i + 3]]]); } // Total operation required console.log(ans.length); for (let i = 0; i < ans.length; i++) { console.log(ans[i][0], ans[i][1][0], ans[i][1][1]); } } // Driver Code let arr = [0, 3, 2, 4, 1]; let n = 4; // Function Call sortPermutation(arr, n); // This code is contributed by akashish__ |
Output:
1 1 3 4