Number of subarrays required to be rearranged to sort the given array

Given an array arr[] consisting of the first N natural numbers, the task is to find the minimum number of subarrays required to be rearranged such that the resultant array is sorted.

Examples:

Input: arr[] = {2, 1, 4, 3, 5}
Output: 1
Explanation:
Operation 1: Choose the subarray {arr[0], arr[3]}, i.e. { 2, 1, 4, 3 }. Rearrange the elements of this subarray to {1, 2, 3, 4}. The array modifies to {1, 2, 3, 4, 5}.

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

Approach: The given problem can be solved by observing the following scenarios:

• If the given array arr[] is already sorted, then print 0.
• If the first and the last element is 1 and N respectively, then only 1 subarray either arr[1, N – 2] or arr[2, N – 1] needs to be sorted. Therefore, print 1.
• f the first and the last element is N and 1 respectively, then 3 subarrays i.e., arr[0, N – 2], arr[1, N – 1], and arr[0, 1] need to be sorted. Therefore, print 3.
• Otherwise, sort the two subarrays i.e., arr[1, N – 1], and arr[0, N – 2].

Therefore, print the count of minimum number of subarrays required to be rearranged.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to count the number
// of subarrays required to be
// rearranged to sort the given array
void countSubarray(int arr[], int n)
{
    // Base Case
    int ans = 2;

    // Check if the given array is
    // already sorted
    if (is_sorted(arr, arr + n)) {
        ans = 0;
    }

    // Check if the first element of
    // array is 1 or last element is
    // equal to size of array
    else if (arr[0] == 1
             || arr[n - 1] == n) {
        ans = 1;
    }
    else if (arr[0] == n
             && arr[n - 1] == 1) {
        ans = 3;
    }

    // Print the required answer
    cout << ans;
}

// Driver Code
int main()
{
    int arr[] = { 5, 2, 3, 4, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    countSubarray(arr, N);

    return 0;
}

Java

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

class GFG{

// Function that returns 0 if a pair
// is found unsorted
static int arraySortedOrNot(int arr[], int n)
{

    // Array has one or no element or the
    // rest are already checked and approved.
    if (n == 1 || n == 0)
        return 1;

    // Unsorted pair found (Equal values allowed)
    if (arr[n - 1] < arr[n - 2])
        return 0;

    // Last pair was sorted
    // Keep on checking
    return arraySortedOrNot(arr, n - 1);
}

// Function to count the number
// of subarrays required to be
// rearranged to sort the given array
static void countSubarray(int arr[], int n)
{

    // Base Case
    int ans = 2;

    // Check if the given array is
    // already sorted
    if (arraySortedOrNot(arr, arr.length) != 0)
    {
        ans = 0;
    }

    // Check if the first element of
    // array is 1 or last element is
    // equal to size of array
    else if (arr[0] == 1 ||
             arr[n - 1] == n)
    {
        ans = 1;
    }
    else if (arr[0] == n &&
             arr[n - 1] == 1)
    {
        ans = 3;
    }

    // Print the required answer
    System.out.print(ans);
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 5, 2, 3, 4, 1 };
    int N = arr.length;

    countSubarray(arr, N);
}
}

// This code is contributed by susmitakundugoaldanga

Python3

# Python3 program for the above approach

# Function to count the number
# of subarrays required to be
# rearranged to sort the given array
def countSubarray(arr, n):

    # Base Case
    ans = 2

    # Check if the given array is
    # already sorted
    if (sorted(arr) == arr):
        ans = 0

    # Check if the first element of
    # array is 1 or last element is
    # equal to size of array
    elif (arr[0] == 1 or arr[n - 1] == n):
        ans = 1
    elif (arr[0] == n and arr[n - 1] == 1):
        ans = 3

    # Print the required answer
    print(ans)

# Driver Code
arr = [ 5, 2, 3, 4, 1 ]
N = len(arr)

countSubarray(arr, N)

# This code is contributed by amreshkumar3

C#

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

class GFG{

// Function that returns 0 if a pair
// is found unsorted
static int arraySortedOrNot(int []arr, int n)
{

    // Array has one or no element or the
    // rest are already checked and approved.
    if (n == 1 || n == 0)
        return 1;

    // Unsorted pair found (Equal values allowed)
    if (arr[n - 1] < arr[n - 2])
        return 0;

    // Last pair was sorted
    // Keep on checking
    return arraySortedOrNot(arr, n - 1);
}

// Function to count the number
// of subarrays required to be
// rearranged to sort the given array
static void countSubarray(int []arr, int n)
{

    // Base Case
    int ans = 2;

    // Check if the given array is
    // already sorted
    if (arraySortedOrNot(arr, arr.Length) != 0)
    {
        ans = 0;
    }

    // Check if the first element of
    // array is 1 or last element is
    // equal to size of array
    else if (arr[0] == 1 ||
             arr[n - 1] == n)
    {
        ans = 1;
    }
    else if (arr[0] == n &&
             arr[n - 1] == 1)
    {
        ans = 3;
    }

    // Print the required answer
    Console.Write(ans);
}

// Driver Code
public static void Main()
{
    int []arr = { 5, 2, 3, 4, 1 };
    int N = arr.Length;

    countSubarray(arr, N);
}
}

// This code is contributed by bgangwar59

Javascript

<script>

// Javascript program for the above approach

// Function that returns 0 if a pair
// is found unsorted
function arraySortedOrNot(arr, n)
{

    // Array has one or no element or the
    // rest are already checked and approved.
    if (n == 1 || n == 0)
        return 1;

    // Unsorted pair found (Equal values allowed)
    if (arr[n - 1] < arr[n - 2])
        return 0;

    // Last pair was sorted
    // Keep on checking
    return arraySortedOrNot(arr, n - 1);
}

// Function to count the number
// of subarrays required to be
// rearranged to sort the given array
function countSubarray(arr, n)
{

    // Base Case
    var ans = 2;

    // Check if the given array is
    // already sorted
    if (arraySortedOrNot(arr, arr.length) != 0)
    {
        ans = 0;
    }

    // Check if the first element of
    // array is 1 or last element is
    // equal to size of array
    else if (arr[0] == 1 ||
             arr[n - 1] == n)
    {
        ans = 1;
    }
    else if (arr[0] == n &&
             arr[n - 1] == 1)
    {
        ans = 3;
    }

    // Print the required answer
    document.write(ans);
}

// Driver Code
var arr = [ 5, 2, 3, 4, 1 ];
var N = arr.length;

countSubarray(arr, N);

// This code is contributed by SURENDRA_GANGWAR

</script>

Output:

3

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