Lexicographically largest permutation by sequentially inserting Array elements at ends

Given an array arr[] of N integers, the task is to find the lexicographically largest permutation by sequentially inserting the array elements to the front or the back of another array.

Examples:

Input: arr[] = {3, 1, 2, 4}
Output: 4 3 1 2
Explanation:
The permutations that can be created by sequentially inserting the array elements to the front or the back of the container are {3, 1, 2, 4}, {1, 3, 2, 4}, {2, 3, 1, 4}, {2, 1, 3, 4}, {4, 1, 3, 2}, {4, 2, 3, 1}, {4, 2, 1, 3}, and {4, 3, 1, 2}. Out of which {4, 3, 1, 2} is the lexicographically largest permutation.

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

Approach: The given problem can be solved by using the Greedy Approach using the deque which is based on the observation that if the current array element is at least the first element of the new array, the most optimal choice always is to insert that element in front of the container in order to lexicographically maximize the permutation. Otherwise, insert the element to the end of the array. Follow the steps below to solve the given problem:

• Initialize a deque, say DQ, which stores the current state of the container.
• Initialize a variable, say mx, which stores the maximum till each index representing the 1^st element of the deque DQ.
• Traverse the given array arr[] and if the current element arr[i] >= mx, then insert arr[i] to the front of the deque DQ and update the value of mx. Otherwise, insert arr[i] to the back of the deque DQ.
• After completing the above steps, print the elements stored in the deque DQ as the resultant largest permutation.

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the lexicographically
// largest permutation by sequentially
// inserting the array elements
void largestPermutation(int arr[], int N)
{

    // Stores the current state of
    // the new array
    deque<int> p;

    // Stores the current maximum
    // element of array arr[]
    int mx = arr[0];
    p.push_back(arr[0]);

    // Iterate the array elements
    for (int i = 1; i < N; i++) {

        // If the current element is
        // smaller than the current
        // maximum, then insert
        if (arr[i] < mx)
            p.push_back(arr[i]);

        // If the current element is
        // at least the current maximum
        else {
            p.push_front(arr[i]);

            // Update the value of
            // the current maximum
            mx = arr[i];
        }
    }

    // Print resultant permutation
    for (auto i : p)
        cout << i << " ";
}

// Driver Code
int main()
{
    int arr[] = { 3, 1, 2, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);

    largestPermutation(arr, N);

    return 0;
}

// Java code for the above approach
import java.util.*;

class GFG {

    // Function to find the lexicographically
    // largest permutation by sequentially
    // inserting the array elements
    static void largestPermutation(int arr[], int N)
    {

        // Stores the current state of
        // the new array
        Deque<Integer> p = new LinkedList<Integer>();

        // Stores the current maximum
        // element of array arr[]
        int mx = arr[0];
        p.addLast(arr[0]);

        // Iterate the array elements
        for (int i = 1; i < N; i++) {

            // If the current element is
            // smaller than the current
            // maximum, then insert
            if (arr[i] < mx)
                p.addLast(arr[i]);

            // If the current element is
            // at least the current maximum
            else {
                p.addFirst(arr[i]);

                // Update the value of
                // the current maximum
                mx = arr[i];
            }
        }

        // Print resultant permutation
        for (Iterator itr = p.iterator(); itr.hasNext();) {
            System.out.print(itr.next() + " ");
        }
    }

    // Driver Code

    public static void main(String[] args)
    {
        int arr[] = { 3, 1, 2, 4 };
        int N = arr.length;

        largestPermutation(arr, N);
    }
}

// This code is contributed by Potta Lokesh

# python program for the above approach

# Function to find the lexicographically
# largest permutation by sequentially
# inserting the array elements
from typing import Deque

def largestPermutation(arr, N):

    # Stores the current state of
    # the new array
    p = Deque()

    # Stores the current maximum
    # element of array arr[]
    mx = arr[0]
    p.append(arr[0])

    # Iterate the array elements
    for i in range(1, N):

        # If the current element is
        # smaller than the current
        # maximum, then insert
        if (arr[i] < mx):
            p.append(arr[i])

        # If the current element is
        # at least the current maximum
        else:
            p.appendleft(arr[i])

            # Update the value of
            # the current maximum
            mx = arr[i]

    # Print resultant permutation
    for i in p:
        print(i, end=" ")

# Driver Code
if __name__ == "__main__":

    arr = [3, 1, 2, 4]
    N = len(arr)

    largestPermutation(arr, N)

# This code is contributed by rakeshsahni

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

  // Function to find the lexicographically
  // largest permutation by sequentially
  // inserting the array elements
  static void largestPermutation(int []arr, int N) {

    // Stores the current state of
    // the new array
    List<int> p = new List<int>();

    // Stores the current maximum
    // element of array []arr
    int mx = arr[0];
    p.Add(arr[0]);

    // Iterate the array elements
    for (int i = 1; i < N; i++) {

      // If the current element is
      // smaller than the current
      // maximum, then insert
      if (arr[i] < mx)
        p.Add(arr[i]);

      // If the current element is
      // at least the current maximum
      else {
        p.Insert(0,arr[i]);

        // Update the value of
        // the current maximum
        mx = arr[i];
      }
    }

    // Print resultant permutation
    foreach (int itr in p) {
      Console.Write(itr + " ");
    }
  }

  // Driver Code
  public static void Main(String[] args) {
    int []arr = { 3, 1, 2, 4 };
    int N = arr.Length;

    largestPermutation(arr, N);
  }
}

// This code is contributed by Rajput-Ji 

<script>
// Javascript program for the above approach

// Function to find the lexicographically
// largest permutation by sequentially
// inserting the array elements
function largestPermutation(arr, N)
{

  // Stores the current state of
  // the new array
  let p = [];

  // Stores the current maximum
  // element of array arr[]
  let mx = arr[0];
  p.push(arr[0]);

  // Iterate the array elements
  for (let i = 1; i < N; i++)
  {
  
    // If the current element is
    // smaller than the current
    // maximum, then insert
    if (arr[i] < mx) p.push(arr[i]);
    
    // If the current element is
    // at least the current maximum
    else {
      p.unshift(arr[i]);

      // Update the value of
      // the current maximum
      mx = arr[i];
    }
  }

  // Print resultant permutation
  for (i of p) document.write(i + " ");
}

// Driver Code

let arr = [3, 1, 2, 4];
let N = arr.length;

largestPermutation(arr, N);

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

4 3 1 2

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