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Sort an array where a subarray of a sorted array is in reverse order
  • Difficulty Level : Basic
  • Last Updated : 07 Apr, 2021

Given an array of N numbers where a subarray is sorted in descending order and rest of the numbers in the array are in ascending order. The task is to sort an array where a subarray of a sorted array is in reversed order. 
Examples: 
 

Input: 2 5 65 55 50 70 90 
Output: 2 5 50 55 65 70 90 
The subarray from 2nd index to 4th index is in reverse order. 
So the subarray is reversed, and the sorted array is printed. 
Input: 1 7 6 5 4 3 2 8 
Output: 1 2 3 4 5 6 7 8

 

A naive approach will be to sort the array and print the array. Time Complexity of this approach will be O(N log n).
An efficient approach will be to find and store the starting index and ending index of the reversed subarray. Since the subarray is in descending order and the rest of the elements are in ascending order, only reversing the subarray will sort the complete array. Reverse the subarray using two pointer approach
Below is the implementation of the above approach: 
 

C++




// C++ program to sort an array where
// a subarray of a sorted array
// is in reversed order
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the sorted array
// by reversing the subarray
void printSorted(int a[], int n)
{
    int front = -1, back = -1;
 
    // find the starting index of the
    // reversed subarry
    for (int i = 1; i < n; i++) {
        if (a[i] < a[i - 1]) {
            front = i - 1;
            break;
        }
    }
 
    // find the ending index of the
    // reversed subarray
    for (int i = n - 2; i >= 0; i--) {
        if (a[i] > a[i + 1]) {
            back = i + 1;
            break;
        }
    }
 
    // if no reversed subarray is present
    if (front == -1 and back == -1) {
        for (int i = 0; i < n; i++)
            cout << a[i] << " ";
        return;
    }
 
    // swap the reversed subarray
    while (front <= back) {
 
        // swaps the front and back element
        // using c++ STL
        swap(a[front], a[back]);
 
        // move the pointers one step
        // ahead and one step back
        front++;
        back--;
    }
    for (int i = 0; i < n; i++)
        cout << a[i] << " ";
}
 
// Driver Code
int main()
{
    int a[] = { 1, 7, 6, 5, 4, 3, 2, 8 };
    int n = sizeof(a) / sizeof(a[0]);
    printSorted(a, n);
    return 0;
}

Java




// Java program to sort an array where
// a subarray of a sorted array
// is in reversed order
import java.io.*;
 
class GFG
{
 
// Function to print the sorted array
// by reversing the subarray
static void printSorted(int a[], int n)
{
    int front = -1, back = -1;
 
    // find the starting index of the
    // reversed subarry
    for (int i = 1; i < n; i++)
    {
        if (a[i] < a[i - 1])
        {
            front = i - 1;
            break;
        }
    }
 
    // find the ending index of the
    // reversed subarray
    for (int i = n - 2; i >= 0; i--)
    {
        if (a[i] > a[i + 1])
        {
            back = i + 1;
            break;
        }
    }
 
    // if no reversed subarray is present
    if (front == -1 && back == -1)
    {
        for (int i = 0; i < n; i++)
            System.out.println(a[i] + " ");
        return;
    }
 
    // swap the reversed subarray
    while (front <= back)
    {
 
        // swaps the front and back element
        // using c++ STL
        int temp = a[front];
        a[front] = a[back];
        a[back] = temp;
 
        // move the pointers one step
        // ahead and one step back
        front++;
        back--;
    }
    for (int i = 0; i < n; i++)
        System.out.print(a[i] + " ");
}
 
// Driver Code
public static void main (String[] args)
{
    int a[] = { 1, 7, 6, 5, 4, 3, 2, 8 };
    int n = a.length;
    printSorted(a, n);;
}
}
 
// This code is contributed by anuj_67..

Python3




# Python 3 program to sort an array where
# a subarray of a sorted array is in
# reversed order
 
# Function to print the sorted array
# by reversing the subarray
def printSorted(a, n):
    front = -1
    back = -1
 
    # find the starting index of the
    # reversed subarry
    for i in range(1, n, 1):
        if (a[i] < a[i - 1]):
            front = i - 1
            break
 
    # find the ending index of the
    # reversed subarray
    i = n - 2
    while(i >= 0):
        if (a[i] > a[i + 1]):
            back = i + 1
            break
        i -= 1
     
    # if no reversed subarray is present
    if (front == -1 and back == -1):
        for i in range(0, n, 1):
            print(a[i], end = " ")
        return
 
    # swap the reversed subarray
    while (front <= back):
         
        # swaps the front and back element
        # using c++ STL
        temp = a[front]
        a[front] = a[back]
        a[back] = temp
 
        # move the pointers one step
        # ahead and one step back
        front += 1
        back -= 1
     
    for i in range(0, n, 1):
        print(a[i], end = " ")
  
# Driver Code
if __name__ == '__main__':
    a = [1, 7, 6, 5, 4, 3, 2, 8]
    n = len(a)
    printSorted(a, n)
 
# This code is contributed by
# Sahil_Shelangia

C#




// C# program to sort an array where
// a subarray of a sorted array
// is in reversed order
using System;
 
class GFG
{
 
// Function to print the sorted array
// by reversing the subarray
static void printSorted(int []a, int n)
{
    int front = -1, back = -1;
 
    // find the starting index of the
    // reversed subarry
    for (int i = 1; i < n; i++)
    {
        if (a[i] < a[i - 1])
        {
            front = i - 1;
            break;
        }
    }
 
    // find the ending index of the
    // reversed subarray
    for (int i = n - 2; i >= 0; i--)
    {
        if (a[i] > a[i + 1])
        {
            back = i + 1;
            break;
        }
    }
 
    // if no reversed subarray is present
    if (front == -1 && back == -1)
    {
        for (int i = 0; i < n; i++)
        {
            Console.Write(a[i] + " ");
        }
        return;
    }
 
    // swap the reversed subarray
    while (front <= back)
    {
 
        // swaps the front and back element
        // using c++ STL
        swap(a, front, back);
 
        // move the pointers one step
        // ahead and one step back
        front++;
        back--;
    }
    for (int i = 0; i < n; i++)
    {
        Console.Write(a[i] + " ");
    }
}
 
static void swap(int[] a, int front,
                          int back)
{
    int c = a[front];
    a[front] = a[back];
    a[back] = c;
}
 
// Driver Code
public static void Main()
{
    int []a = {1, 7, 6, 5, 4, 3, 2, 8};
    int n = a.Length;
    printSorted(a, n);
}
}
 
// This code contributed by 29AjayKumar

PHP




<?php
// PHP program to sort an array where
// a subarray of a sorted array
// is in reversed order
 
// Function to print the sorted array
// by reversing the subarray
function printSorted($a, $n)
{
    $front = -1; $back = -1;
 
    // find the starting index of the
    // reversed subarry
    for ($i = 1; $i < $n; $i++)
    {
        if ($a[$i] < $a[$i - 1])
        {
            $front = $i - 1;
            break;
        }
    }
 
    // find the ending index of the
    // reversed subarray
    for ($i = $n - 2; $i >= 0; $i--)
    {
        if ($a[$i] > $a[$i + 1])
        {
            $back = $i + 1;
            break;
        }
    }
 
    // if no reversed subarray is present
    if ($front == -1 && $back == -1)
    {
        for ($i = 0; $i < $n; $i++)
            echo $a[$i] . " ";
        return;
    }
 
    // swap the reversed subarray
    while ($front <= $back)
    {
 
        // swaps the front and back element
        // using c++ STL
        $temp = $a[$front];
        $a[$front] = $a[$back];
        $a[$back] = $temp;
 
        // move the pointers one step
        // ahead and one step back
        $front++;
        $back--;
    }
    for ($i = 0; $i < $n; $i++)
        echo $a[$i] . " ";
}
 
// Driver Code
$a = array(1, 7, 6, 5, 4, 3, 2, 8);
$n = sizeof($a);
printSorted($a, $n);
 
// This code is contributed
// by Akanksha Rai

Javascript




<script>
 
// JavaScript program to sort an array where
// a subarray of a sorted array
// is in reversed order
 
    // Function to prvar the sorted array
    // by reversing the subarray
    function printSorted(a , n) {
        var front = -1, back = -1;
 
        // find the starting index of the
        // reversed subarry
        for (i = 1; i < n; i++) {
            if (a[i] < a[i - 1]) {
                front = i - 1;
                break;
            }
        }
 
        // find the ending index of the
        // reversed subarray
        for (i = n - 2; i >= 0; i--) {
            if (a[i] > a[i + 1]) {
                back = i + 1;
                break;
            }
        }
 
        // if no reversed subarray is present
        if (front == -1 && back == -1) {
            for (i = 0; i < n; i++)
                document.write(a[i] + " ");
            return;
        }
 
        // swap the reversed subarray
        while (front <= back) {
 
            // swaps the front and back element
            // using c++ STL
            var temp = a[front];
            a[front] = a[back];
            a[back] = temp;
 
            // move the pointers one step
            // ahead and one step back
            front++;
            back--;
        }
        for (i = 0; i < n; i++)
            document.write(a[i] + " ");
    }
 
    // Driver Code
     
        var a = [ 1, 7, 6, 5, 4, 3, 2, 8 ];
        var n = a.length;
        printSorted(a, n);
 
// This code is contributed by todaysgaurav
 
</script>
Output: 



1 2 3 4 5 6 7 8

 

Time Complexity: O(n)
 

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