Related Articles

Related Articles

Sort permutation of N natural numbers using triple cyclic right swaps
  • Last Updated : 10 Jun, 2020

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 choosen 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 arr[i] != i. 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
  
    // Vistied 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;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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 = []
  
    # Vistied 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 athe 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

chevron_right


Output:

1
1 3 4

competitive-programming-img




My Personal Notes arrow_drop_up
Recommended Articles
Page :