Possible permutations at a railway track

Last Updated : 10 Nov, 2022

Given a left, right, and a spur track as shown in the below figure. There are N trucks from value 1 to N arranged in the left track. We can move directly N trucks to the right track but there can be more possibilities of moving the trucks to the right track using the spur track. We can move any truck to spur track and then move it to the right track. The task is to print all the possible permutation order in which all the N trucks can be moved from left track to right track. Note: Once a truck is moved from left truck to right/spur track then it can’t be moved to left track again. Examples:

Input: N = 2 Output: 1 2 2 1 Explanation: For the first permutation: left[] = {1, 2} right[] = {}, and spur[] = {} The truck with value 2 moved to the right track, then left[] = {1} right[] = {2}, and spur[] = {} Now moving with value 1 to the right track, then left[] = {} right[] = {1, 2}, and spur[] = {} For the second permutation: left[] = {1, 2} right[] = {}, and spur[] = {} The truck with value 2 move to the spur track, then left[] = {1} right[] = {}, and spur[] = {2} The truck with value 1 move to the right track, then left[] = {} right[] = {1}, and spur[] = {2} The truck with value 2 in the spur track move to the right track, then left[] = {} right[] = {2, 1}, and spur[] = {} Input: N = 3 Output: 1 2 3 2 1 3 3 2 1 3 1 2 2 3 1

Approach: This problem is a variation of Tower Of Hanoi and can be solved using Recursion. Below are the following cases:

Case 1: We can move the truck from left track to the spur track and recursively check for the remaining trucks on the left and spur tracks.
Case 2: We can move the truck from the spur track to right track and check for the remaining trucks of the left and spur tracks.

Below are the steps:

At each step we can move either truck from left track to spur track or from spur track to right track.
Move one truck from left track to spur track and call recursively for remaining trucks on left and spur track.
At any recursive call, if the input track is empty then, move every truck on the spur track to right track and print the current permutation on the right track

C++

// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
 
// Helper function to print all the
// possible permutation
void printPermute(vector<int>&input,
                vector<int>&spur,
                vector<int>&output)
{
 
    // If at any recursive call input
    // array is empty, then we got our
    // one of the permutation
    if(input.empty())
    {
         
        // Print the right track trucks
        for(auto &it : output) {
            cout << it << ' ';
        }
         
        // Print the spur track trucks
        for(auto &it : spur) {
            cout << it << ' ';
        }
         
        cout << endl;
    }
    else
    {
        int temp;
         
        // Pop the element from input
        // track and move it to spur
        temp=input.back();
        input.pop_back();
         
        // Case 1
        // Push the popped truck from
        // input to spur track
        spur.push_back(temp);
         
        // Recursive call for remaining
        // trucks on input, spur and
        // output track
        printPermute(input,spur,output);
         
        // remove the top truck from spur
        // track and push it in input for
        // Case 2 iteration
        spur.pop_back();
        input.push_back(temp);
         
        // Case 2
        if(!spur.empty()) {
             
            // Remove the truck from the spur
            // track and move it to the
            // output track
            temp=spur.back();
            spur.pop_back();
            output.push_back(temp);
             
            // Recursive call for remaining
            // truck on input, spur and
            // output track
            printPermute(input,spur,output);
             
            // Remove the top truck from the
            // output track and move it to
            // the spur track for the next
            // iteration
            output.pop_back();
            spur.push_back(temp);
        }
    }
}
 
// Function to print all the possible
// permutation of trucks
void possiblePermute(int n)
{
    // Array for left, spur and right track
    vector<int>spur;
    vector<int>output;
    vector<int>input;
     
    // Insert all truck value 1 to N
    for(int i = 1; i <= n; i++) {
        input.push_back(i);
    }
     
    // Helper function to find
    // possible arrangement
    printPermute(input, spur, output);
}
 
// Driver Code
int main()
{
    // Input number of truck
    int N = 4;
     
    // Function Call
    possiblePermute(N);
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Helper function to print all the
  // possible permutation
  static void printPermute(ArrayList<Integer> input,
                           ArrayList<Integer> spur,
                           ArrayList<Integer> output)
  {
 
    // If at any recursive call input
    // array is empty, then we got our
    // one of the permutation
    if (input.size() == 0) {
 
      // Print the right track trucks
      for (var it : output) {
        System.out.print(it + " ");
      }
 
      // Print the spur track trucks
      for (var it : spur) {
        System.out.print(it + " ");
      }
 
      System.out.print("\n");
    }
    else {
      int temp;
 
      // Pop the element from input
      // track and move it to spur
      temp = input.get(input.size() - 1);
      input.remove(input.size() - 1);
 
      // Case 1
      // Push the popped truck from
      // input to spur track
      spur.add(temp);
 
      // Recursive call for remaining
      // trucks on input, spur and
      // output track
      printPermute(input, spur, output);
 
      // remove the top truck from spur
      // track and push it in input for
      // Case 2 iteration
      spur.remove(spur.size() - 1);
      input.add(temp);
 
      // Case 2
      if (spur.size() != 0) {
 
        // Remove the truck from the spur
        // track and move it to the
        // output track
        temp = spur.get(spur.size() - 1);
        spur.remove(spur.size() - 1);
        output.add(temp);
 
        // Recursive call for remaining
        // truck on input, spur and
        // output track
        printPermute(input, spur, output);
 
        // Remove the top truck from the
        // output track and move it to
        // the spur track for the next
        // iteration
        output.remove(output.size() - 1);
        spur.add(temp);
      }
    }
  }
 
  // Function to print all the possible
  // permutation of trucks
  static void possiblePermute(int n)
  {
    // Array for left, spur and right track
    ArrayList<Integer> spur = new ArrayList<Integer>();
    ArrayList<Integer> output
      = new ArrayList<Integer>();
    ArrayList<Integer> input = new ArrayList<Integer>();
 
    // Insert all truck value 1 to N
    for (int i = 1; i <= n; i++) {
      input.add(i);
    }
 
    // Helper function to find
    // possible arrangement
    printPermute(input, spur, output);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    // Input number of truck
    int N = 4;
 
    // Function Call
    possiblePermute(N);
  }
}
 
// This code is contributed by phasing17.

Python3

# Python program for the above approach
 
# Helper function to print all the
# possible permutation
def printPermute( input1, spur, output):
 
    # If at any recursive call input1
    # array is empty, then we got our
    # one of the permutation
    if(len(input1) == 0):
      
        # Print the right track trucks
        for it in output: 
            print(it, end = ' ');
         
        # Print the spur track trucks
        for it in spur: 
            print(it, end = ' ');
         
        print()
     
    else:
     
        # Pop the element from input1
        # track and move it to spur
        temp=input1.pop();
 
        # Case 1
        # Push the popped truck from
        # input1 to spur track
        spur.append(temp);
         
        # Recursive call for remaining
        # trucks on input1, spur and
        # output track
        printPermute(input1,spur,output);
         
        # remove the top truck from spur
        # track and push it in input1 for
        # Case 2 iteration
        spur.pop();
        input1.append(temp);
         
        # Case 2
        if(len(spur) > 0): 
             
            # Remove the truck from the spur
            # track and move it to the
            # output track
            temp=spur.pop();
            output.append(temp);
             
            # Recursive call for remaining
            # truck on input1, spur and
            # output track
            printPermute(input1,spur,output);
             
            # Remove the top truck from the
            # output track and move it to
            # the spur track for the next
            # iteration
            output.pop();
            spur.append(temp);
 
# Function to print all the possible
# permutation of trucks
def possiblePermute(n):
 
    # Array for left, spur and right track
    spur = [];
    output = [];
    input1 = [];
  
    # Insert all truck value 1 to N
    for i in range(1, n + 1):
        input1.append(i);
     
    # Helper function to find
    # possible arrangement
    printPermute(input1, spur, output);
 
# Driver Code
N = 4;
     
# Function Call
possiblePermute(N);
 
# This code is contributed by phasing17

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
  // Helper function to print all the
  // possible permutation
  static void printPermute(List<int> input,
                           List<int>spur,
                           List<int>output)
  {
 
    // If at any recursive call input
    // array is empty, then we got our
    // one of the permutation
    if(input.Count == 0)
    {
 
      // Print the right track trucks
      foreach(var it in output) {
        Console.Write(it + " ");
      }
 
      // Print the spur track trucks
      foreach(var it in spur) {
        Console.Write(it + " ");
      }
 
 
      Console.Write("\n");
    }
    else
    {
      int temp;
 
      // Pop the element from input
      // track and move it to spur
      temp=input[input.Count - 1];
      input.RemoveAt(input.Count - 1);
 
      // Case 1
      // Push the popped truck from
      // input to spur track
      spur.Add(temp);
 
      // Recursive call for remaining
      // trucks on input, spur and
      // output track
      printPermute(input,spur,output);
 
      // remove the top truck from spur
      // track and push it in input for
      // Case 2 iteration
      spur.RemoveAt(spur.Count - 1);
      input.Add(temp);
 
      // Case 2
      if(spur.Count != 0) {
 
        // Remove the truck from the spur
        // track and move it to the
        // output track
        temp=spur[spur.Count - 1];
        spur.RemoveAt(spur.Count - 1);
        output.Add(temp);
 
        // Recursive call for remaining
        // truck on input, spur and
        // output track
        printPermute(input,spur,output);
 
        // Remove the top truck from the
        // output track and move it to
        // the spur track for the next
        // iteration
        output.RemoveAt(output.Count - 1);
        spur.Add(temp);
      }
    }
  }
 
  // Function to print all the possible
  // permutation of trucks
  static void possiblePermute(int n)
  {
    // Array for left, spur and right track
    List<int>spur = new List<int>();
    List<int>output = new List<int>();
    List<int>input =  new List<int>();
 
    // Insert all truck value 1 to N
    for(int i = 1; i <= n; i++) {
      input.Add(i);
    }
 
    // Helper function to find
    // possible arrangement
    printPermute(input, spur, output);
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    // Input number of truck
    int N = 4;
 
    // Function Call
    possiblePermute(N);
  }
}
 
// This code is contributed by phasing17.

Javascript

// JS program for the above approach
 
// Helper function to print all the
// possible permutation
function printPermute( input,
                 spur, output)
{
 
    // If at any recursive call input
    // array is empty, then we got our
    // one of the permutation
    if(input.length == 0)
    {
         
        // Print the right track trucks
        for(let it of output) {
            process.stdout.write(it + ' ');
        }
         
        // Print the spur track trucks
        for(let it of spur) {
            process.stdout.write(it + ' ');
        }
         
         
        console.log(' ');
    }
    else
    {
        let temp;
         
        // Pop the element from input
        // track and move it to spur
        temp=input.pop();
 
        // Case 1
        // Push the popped truck from
        // input to spur track
        spur.push(temp);
         
        // Recursive call for remaining
        // trucks on input, spur and
        // output track
        printPermute(input,spur,output);
         
        // remove the top truck from spur
        // track and push it in input for
        // Case 2 iteration
        spur.pop();
        input.push(temp);
         
        // Case 2
        if(spur.length > 0) {
             
            // Remove the truck from the spur
            // track and move it to the
            // output track
            temp=spur.pop();
            output.push(temp);
             
            // Recursive call for remaining
            // truck on input, spur and
            // output track
            printPermute(input,spur,output);
             
            // Remove the top truck from the
            // output track and move it to
            // the spur track for the next
            // iteration
            output.pop();
            spur.push(temp);
        }
    }
}
 
// Function to print all the possible
// permutation of trucks
function possiblePermute(n)
{
    // Array for left, spur and right track
    let spur = [];
    let output = [];
    let input = [];
     
     
    // Insert all truck value 1 to N
    for(var i = 1; i <= n; i++) {
        input.push(i);
    }
     
    // Helper function to find
    // possible arrangement
    printPermute(input, spur, output);
}
 
// Driver Code
let N = 4;
     
// Function Call
possiblePermute(N);
 
// This code is contributed by phasing17

Output

Suggest improvement

Dudeney Numbers

Twin Pythagorean triplets in an array

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Possible permutations at a railway track

C++

Java

Python3

C#

Javascript

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