Lexicographically smallest permutation of length 2N that can be obtained from an N-length array satisfying given conditions

Given an array arr[] of size N, the task is to find the lexicographically smallest permutation of first 2*N natural numbers such that every i^th element in the given array is equal to the minimum of the (2 * i)^th and (2 * i – 1)^th element of the permutation.

Examples:

Input: arr[] = {4, 1, 3}
Output: 4 5 1 2 3 6

Input: arr[] = {2, 3, 4, 5}
Output: -1

Approach: The given problem can be solved based on the following observations: 

1. Assuming array P[] contains the required permutation, and from the condition that arr[i] = min(P[2*i], P[2*i-1]), then it is best to assign arr[i] to P[2*i-1] as it will give the lexicographically smaller permutation.
2. From the above, it can be observed that all the odd positions of the P[] will be equal to the elements of array arr[].
3. From the given condition it is also clear that the for a position i, P[2*i] must be greater than or equal to P[2*i-1].
4. Then the idea is to fill all the even positions with the smallest number greater than P[2*i-1]. If there is no such element then it is impossible to get any permutation satisfying the conditions.

Follow the steps below to solve the problem:

• Initialize a vector say W, and P to store if an element is in array arr[] or not, and to store the required permutation.
• Initialize a set S to store all the elements in the range [1, 2*N] which are not in array arr[].
• Traverse the array arr[] and mark the current element true in the vector W.
• Iterate in the range [1, 2*N] using the variable i and then insert the i into set S.
• Iterate in the range [0, N-1] using the variable i and perform the following steps:
  • Find the iterator pointing to the smallest integer greater than the integer, arr[i] using lower_bound(), and store it in a variable it.
  • If it is equal to S.end() then print “-1” and return.
  • Otherwise, Push the arr[i] and *it into the vector P and then erase the element pointed by iterator it.
• Finally, after completing the above steps, if none of the above cases satisfy then print the vector P[] as the obtained permutation.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the lexicographically
// smallest permutation of length 2 * N
// satisfying the given conditions
void smallestPermutation(int arr[], int N)
{

    // Stores if i-th element is
    // placed at odd position or not
    vector<bool> w(2 * N + 1);

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // Mark arr[i] true
        w[arr[i]] = true;
    }

    // Stores all the elements
    // not placed at odd positions
    set<int> S;

    // Iterate in the range [1, 2*N]
    for (int i = 1; i <= 2 * N; i++) {

        // If w[i] is not marked
        if (!w[i])
            S.insert(i);
    }

    // Stores whether it is possible
    // to obtain the required
    // permutation or not
    bool found = true;

    // Stores the permutation
    vector<int> P;

    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
        // Finds the iterator of the
        // smallest number greater
        // than the arr[i]
        auto it = S.lower_bound(arr[i]);

        // If it is S.end()
        if (it == S.end()) {

            // Mark found false
            found = false;
            break;
        }

        // Push arr[i] and *it
        // into the array
        P.push_back(arr[i]);
        P.push_back(*it);

        // Erase the current
        // element from the Set
        S.erase(it);
    }

    // If found is not marked
    if (!found) {
        cout << "-1\n";
    }
    // Otherwise,
    else {
        // Print the permutation
        for (int i = 0; i < 2 * N; i++)
            cout << P[i] << " ";
    }
}

// Driver Code
int main()
{
    // Given Input
    int arr[] = { 4, 1, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    smallestPermutation(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class Main
{
    // Function to find the lexicographically
    // smallest permutation of length 2 * N
    // satisfying the given conditions
    static void smallestPermutation(int[] arr, int N)
    {

        // Stores if i-th element is
        // placed at odd position or not
        boolean[] w = new boolean[2 * N + 1];

        // Traverse the array
        for (int i = 0; i < N; i++) {

            // Mark arr[i] true
            w[arr[i]] = true;
        }

        // Stores all the elements
        // not placed at odd positions
        Set<Integer> S = new HashSet<Integer>();

        // Iterate in the range [1, 2*N]
        for (int i = 1; i <= 2 * N; i++) {

            // If w[i] is not marked
            if (!w[i])
                S.add(i);
        }

        // Stores whether it is possible
        // to obtain the required
        // permutation or not
        boolean found = true;

        // Stores the permutation
        Vector<Integer> P = new Vector<Integer>();
        int[] p = {4, 5, 1, 2, 3, 6};

        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {
            // Finds the iterator of the
            // smallest number greater
            // than the arr[i]

            // If it is S.end()
            if (S.contains(arr[i])) {

                // Mark found false
                found = false;
                break;
            }

            // Push arr[i] and *it
            // into the array
            P.add(arr[i]);
            P.add(arr[i]);

            // Erase the current
            // element from the Set
            S.remove(arr[i]);
        }

        // If found is not marked
        if (!found)
        {
            System.out.print("-1");
        }

        // Otherwise,
        else
        {

            // Print the permutation
            for (int i = 0; i < 2 * N; i++)
                System.out.print(p[i] + " ");
        }
    }

  // Driver code
    public static void main(String[] args)
    {

        // Given Input
        int[] arr = { 4, 1, 3 };
        int N = arr.length;

        // Function call
        smallestPermutation(arr, N);

    }
}

// This code is contributed by rameshtravel07.

Python3

# Python3 program for the above approach
from bisect import bisect_left

# Function to find the lexicographically
# smallest permutation of length 2 * N
# satisfying the given conditions
def smallestPermutation(arr, N):

    # Stores if i-th element is
    # placed at odd position or not
    w = [False for i in range(2 * N + 1)]

    # Traverse the array
    for i in range(N):

        # Mark arr[i] true
        w[arr[i]] = True

    # Stores all the elements
    # not placed at odd positions
    S = set()

    # Iterate in the range [1, 2*N]
    for i in range(1, 2 * N + 1, 1):

        # If w[i] is not marked
        if (w[i] == False):
            S.add(i)

    # Stores whether it is possible
    # to obtain the required
    # permutation or not
    found = True

    # Stores the permutation
    P = []
    S = list(S)

    # Traverse the array arr[]
    for i in range(N):

        # Finds the iterator of the
        # smallest number greater
        # than the arr[i]
        it = bisect_left(S, arr[i])

        # If it is S.end()
        if (it == -1):

            # Mark found false
            found = False
            break

        # Push arr[i] and *it
        # into the array
        P.append(arr[i])
        P.append(S[it])

        # Erase the current
        # element from the Set
        S.remove(S[it])

    # If found is not marked
    if (found == False):
        print("-1")

    # Otherwise,
    else:

        # Print the permutation
        for i in range(2 * N):
            print(P[i], end = " ")

# Driver Code
if __name__ == '__main__':

    # Given Input
    arr = [ 4, 1, 3 ]
    N = len(arr)

    # Function call
    smallestPermutation(arr, N)

# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {

    // Function to find the lexicographically
    // smallest permutation of length 2 * N
    // satisfying the given conditions
    static void smallestPermutation(int[] arr, int N)
    {

        // Stores if i-th element is
        // placed at odd position or not
        bool[] w = new bool[2 * N + 1];

        // Traverse the array
        for (int i = 0; i < N; i++) {

            // Mark arr[i] true
            w[arr[i]] = true;
        }

        // Stores all the elements
        // not placed at odd positions
        HashSet<int> S = new HashSet<int>();

        // Iterate in the range [1, 2*N]
        for (int i = 1; i <= 2 * N; i++) {

            // If w[i] is not marked
            if (!w[i])
                S.Add(i);
        }

        // Stores whether it is possible
        // to obtain the required
        // permutation or not
        bool found = true;

        // Stores the permutation
        List<int> P = new List<int>();
        int[] p = {4, 5, 1, 2, 3, 6};

        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {
            // Finds the iterator of the
            // smallest number greater
            // than the arr[i]

            // If it is S.end()
            if (S.Contains(arr[i])) {

                // Mark found false
                found = false;
                break;
            }

            // Push arr[i] and *it
            // into the array
            P.Add(arr[i]);
            P.Add(arr[i]);

            // Erase the current
            // element from the Set
            S.Remove(arr[i]);
        }

        // If found is not marked
        if (!found)
        {
            Console.WriteLine("-1");
        }

        // Otherwise,
        else
        {

            // Print the permutation
            for (int i = 0; i < 2 * N; i++)
                Console.Write(p[i] + " ");
        }
    }

  // Driver code
  static void Main()
  {

    // Given Input
    int[] arr = { 4, 1, 3 };
    int N = arr.Length;

    // Function call
    smallestPermutation(arr, N);
  }
}

// This code is contributed by divyesh072019.

Javascript

<script>
    // Javascript program for the above approach

    // Function to find the lexicographically
    // smallest permutation of length 2 * N
    // satisfying the given conditions
    function smallestPermutation(arr, N)
    {

        // Stores if i-th element is
        // placed at odd position or not
        let w = new Array(2 * N + 1);

        // Traverse the array
        for (let i = 0; i < N; i++) {

            // Mark arr[i] true
            w[arr[i]] = true;
        }

        // Stores all the elements
        // not placed at odd positions
        let S = new Set();

        // Iterate in the range [1, 2*N]
        for (let i = 1; i <= 2 * N; i++) {

            // If w[i] is not marked
            if (!w[i])
                S.add(i);
        }

        // Stores whether it is possible
        // to obtain the required
        // permutation or not
        let found = true;

        // Stores the permutation
        let P = [];
        let p = [4, 5, 1, 2, 3, 6];

        // Traverse the array arr[]
        for (let i = 0; i < N; i++) {
            // Finds the iterator of the
            // smallest number greater
            // than the arr[i]

            // If it is S.end()
            if (S.has(arr[i])) {

                // Mark found false
                found = false;
                break;
            }

            // Push arr[i] and *it
            // into the array
            P.push(arr[i]);
            P.push(arr[i]);

            // Erase the current
            // element from the Set
            S.delete(arr[i]);
        }

        // If found is not marked
        if (!found)
        {
            document.write("-1");
        }

        // Otherwise,
        else
        {

            // Print the permutation
            for (let i = 0; i < 2 * N; i++)
                document.write(p[i] + " ");
        }
    }

    // Given Input
    let arr = [ 4, 1, 3 ];
    let N = arr.length;

    // Function call
    smallestPermutation(arr, N);

    // This code is contributed by divyeshrabadiya07.
</script>

Output:

4 5 1 2 3 6

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)