Lexicographically smallest Permutation of Array by reversing at most one Subarray

Given an array arr[] of size N which is a permutation from 1 to N, the task is to find the lexicographically smallest permutation that can be formed by reversing at most one subarray.

Examples:

Input : arr[] = {1, 3, 4, 2, 5}
Output : 1 2 4 3 5
Explanation: The subarray from index 1 to index 3 can be reversed to get the lexicographically smallest permutation.

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

Approach: The idea to solve the problem is based on the traversal of the array. 

• In the given problem the lexicographically smallest permutation can be obtained by placing the least number at its correct place by one reversal by traversing from left and checking (i+1) is equal to arr[i]
• If it is not equal find the index of that i+1 in the array and reverse the subarray from ith index to the position (i+1).

Below is the implementation of the above approach.

C++

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

// Function to find the
// Lexicographically smallest
// Permutation by one subarray reversal
void lexsmallest(vector<int>& arr, int n)
{

    // Initialize the variables
    // To store the first and last
    // Position of the subarray
    int first = -1, flag = 0, find = -1, last = -1;

    // Traverse the array
    // And check if arr[i]!=i+1
    for (int i = 0; i < n; i++) {
        if (arr[i] != i + 1) {
            flag = 1;

            // Mark the first position
            // Of the Subarray to be reversed
            first = i;
            find = i + 1;
            break;
        }
    }

    // If flag == 0, it is the
    // Smallest permutation,
    // So print the array
    if (flag == 0) {
        for (int i = 0; i < n; i++) {
            cout << arr[i] << " ";
        }
    }

    // Check where the minimum element is present
    else {
        for (int i = 0; i < n; i++) {

            // It is the last position
            // Of the subarray to be
            // Reversed
            if (arr[i] == find) {
                last = i;
                break;
            }
        }

        // Reverse the subarray
        // And print the array
        reverse(arr.begin() + first,
                arr.begin() + last + 1);
        for (int i = 0; i < n; i++) {
            cout << arr[i] << " ";
        }
    }
}

// Driver Code
int main()
{
    // Initialize the array arr[]
    vector<int> arr = { 1, 3, 4, 2, 5 };
    int N = arr.size();

    // Function call
    lexsmallest(arr, N);
    return 0;
}

Java

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

class GFG{

// Function to find the
// Lexicographically smallest
// Permutation by one subarray reversal
static void lexsmallest(int []arr, int n)
{

    // Initialize the variables
    // To store the first and last
    // Position of the subarray
    int first = -1, flag = 0, find = -1, last = -1;

    // Traverse the array
    // And check if arr[i]!=i+1
    for (int i = 0; i < n; i++) {
        if (arr[i] != i + 1) {
            flag = 1;

            // Mark the first position
            // Of the Subarray to be reversed
            first = i;
            find = i + 1;
            break;
        }
    }

    // If flag == 0, it is the
    // Smallest permutation,
    // So print the array
    if (flag == 0) {
        for (int i = 0; i < n; i++) {
            System.out.print(arr[i]+ " ");
        }
    }

    // Check where the minimum element is present
    else {
        for (int i = 0; i < n; i++) {

            // It is the last position
            // Of the subarray to be
            // Reversed
            if (arr[i] == find) {
                last = i;
                break;
            }
        }

        // Reverse the subarray
        // And print the array
        arr = reverse(arr,first,last);
        for (int i = 0; i < n; i++) {
            System.out.print(arr[i]+ " ");
        }
    }
}
static int[] reverse(int str[], int start, int end) {

    // Temporary variable to store character
    int temp;
    while (start <= end) {
        // Swapping the first and last character
        temp = str[start];
        str[start] = str[end];
        str[end] = temp;
        start++;
        end--;
    }
    return str;
}

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

    // Initialize the array arr[]
    int []arr = { 1, 3, 4, 2, 5 };
    int N = arr.length;

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

// This code contributed by shikhasingrajput

Python3

# Python code for the above approach

# Function to find the
# Lexicographically smallest
# Permutation by one subarray reversal
def lexsmallest(arr, n):

    # Initialize the variables
    # To store the first and last
    # Position of the subarray
    first = -1
    flag = 0
    find = -1
    last = -1

    # Traverse the array
    # And check if arr[i]!=i+1
    for i in range(0, n):
        if (arr[i] != i + 1):
            flag = 1

            # Mark the first position
            # Of the Subarray to be reversed
            first = i
            find = i + 1
            break

    # If flag == 0, it is the
    # Smallest permutation,
    # So print the array
    if (flag == 0):
        for i in range(0, n):
            print(arr[i], end=" ")

    # Check where the minimum element is present
    else:
        for i in range(0, n):

            # It is the last position
            # Of the subarray to be
            # Reversed
            if (arr[i] == find):
                last = i
                break

        # Reverse the subarray
        # And print the array
        arr[first: last + 1] = arr[first: last + 1][::-1]

        print(*arr)

# Driver Code

# Initialize the array arr[]
arr = [1, 3, 4, 2, 5]
N = len(arr)

# Function call
lexsmallest(arr, N)

# This code is contributed by Samim Hossain Mondal.

C#

// C# code for the above approach
using System;

class GFG{

  // Function to find the
  // Lexicographically smallest
  // Permutation by one subarray reversal
  static void lexsmallest(int []arr, int n)
  {

    // Initialize the variables
    // To store the first and last
    // Position of the subarray
    int first = -1, flag = 0, find = -1, last = -1;

    // Traverse the array
    // And check if arr[i]!=i+1
    for (int i = 0; i < n; i++) {
      if (arr[i] != i + 1) {
        flag = 1;

        // Mark the first position
        // Of the Subarray to be reversed
        first = i;
        find = i + 1;
        break;
      }
    }

    // If flag == 0, it is the
    // Smallest permutation,
    // So print the array
    if (flag == 0) {
      for (int i = 0; i < n; i++) {
        Console.Write(arr[i]+ " ");
      }
    }

    // Check where the minimum element is present
    else {
      for (int i = 0; i < n; i++) {

        // It is the last position
        // Of the subarray to be
        // Reversed
        if (arr[i] == find) {
          last = i;
          break;
        }
      }

      // Reverse the subarray
      // And print the array
      arr = reverse(arr,first,last);
      for (int i = 0; i < n; i++) {
        Console.Write(arr[i]+ " ");
      }
    }
  }
  static int[] reverse(int[] str, int start, int end) {

    // Temporary variable to store character
    int temp;
    while (start <= end)
    {

      // Swapping the first and last character
      temp = str[start];
      str[start] = str[end];
      str[end] = temp;
      start++;
      end--;
    }
    return str;
  }

  // Driver Code
  static public void Main (){

    // Initialize the array arr[]
    int [] arr = { 1, 3, 4, 2, 5 };
    int N = arr.Length;

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

// This code is contributed by hrithikgarg03188.

Javascript

<script>
    // JavaScript code for the above approach

    // Function to find the
    // Lexicographically smallest
    // Permutation by one subarray reversal
    const lexsmallest = (arr, n) => {

        // Initialize the variables
        // To store the first and last
        // Position of the subarray
        let first = -1, flag = 0, find = -1, last = -1;

        // Traverse the array
        // And check if arr[i]!=i+1
        for (let i = 0; i < n; i++) {
            if (arr[i] != i + 1) {
                flag = 1;

                // Mark the first position
                // Of the Subarray to be reversed
                first = i;
                find = i + 1;
                break;
            }
        }

        // If flag == 0, it is the
        // Smallest permutation,
        // So print the array
        if (flag == 0) {
            for (let i = 0; i < n; i++) {
                document.write(`${arr[i]} `);
            }
        }

        // Check where the minimum element is present
        else {
            for (let i = 0; i < n; i++) {

                // It is the last position
                // Of the subarray to be
                // Reversed
                if (arr[i] == find) {
                    last = i;
                    break;
                }
            }

            // Reverse the subarray
            // And print the array
            arr.splice(first, last, ...arr.slice(first, last + 1).reverse());
            for (let i = 0; i < n; i++) {
                document.write(`${arr[i]} `);
            }
        }
    }

    // Driver Code

    // Initialize the array arr[]
    let arr = [1, 3, 4, 2, 5];
    let N = arr.length;

    // Function call
    lexsmallest(arr, N);

// This code is contributed by rakeshsahni

</script>

Output

1 2 4 3 5 

Time Complexity: O(N)
Auxiliary Space: O(1)

Related Topic: Subarrays, Subsequences, and Subsets in Array